Question 13 Marks
How is the concept of coupling reactions useful in explaining the occurrence of non spontaneous thermochemical reactions? Explain giving an example.
AnswerThere are reactions for which the value of $\triangle G$ is not negative i.e they are non spontaneous.
However such reactions can be made spontaneous if they are coupled with reactions having very large negative Gibb’s energy change of reaction.
When $\triangle G$ is $+ve,$ then it is coupled with some other reaction which has $\triangle G –ve$ and large.
$e.g \ \ce{2Fe_2 O_3 (s) \rightarrow 4Fe(s) + 3O_2 (g) \triangle G^\circ = X KJ mol^{-1}}$
Thus for the overall reaction to become spontaneous, this reaction is coupled with carbon monoxide oxidation reaction.
$ \ce{6CO(g) + 3O_2(g) \rightarrow 6CO_2 \triangle G^\circ = -Y KJ mol^{-1}}$
Where $Y > X.$ The large negative value of $\triangle G^\circ$ for this reaction shows that $\ce{Fe_2O_3}$ can be reduce to $Fe$ by $CO$ .
View full question & answer→Question 23 Marks
A certain reaction is $50\%$ complete in $20$ minutes at $300 K$ and the same reaction is again $50\%$ complete in $5$ minutes at $350 K.$ Calculate the activation energy if it is a first order reaction. $\ce{[R = 8.314 JK^{-1}mol^{-1}, \log 4 = 0.602]}.$
Answer$\log\frac{\text{k}_{2}}{\text{k}_{1}}=\frac{\text{E}_\text{a}}{\text{2.303R}}\bigg(\frac{\text{T}_{2} \ - \ \text{T}_{1}}{\text{T}_{1}\text{T}_{2}}\bigg)$
$\log\frac{\text{k}_{2}}{\text{k}_{1}}=\frac{\text{E}_\text{a}}{\text{2.303}\times\text{8.314 JK}^{-1}\text{mol}^{-1}}\bigg(\frac{\text{350-300}}{\text{350}\times\text{300 K}}\bigg)$
$\log\frac{\frac{0.693}{5}}{\frac{0.693}{20}}=\frac{\text{Ea}}{\text{2.303}\times\text{8.314}}\times\frac{\text{50}}{\text{350}\times\text{300}}$
$\log{4}=\frac{\text{E}_\text{a}}{\text{2.303}\times\text{8.314}}\times\frac{\text{50}}{\text{350}\times{300}}\text{ J mol}^{-1}$
$\text{E}_\text{a}=\text{2.303}\times\text{8.314}\times\text{300}\times\text{7}\times\text{0.6021 J} \text{mol}^{-1}$
$\text{= 24210 J mol}^{-1}$
$=\text{24.210 kJ mol}^{-1}.$
View full question & answer→Question 33 Marks
For the first order thermal decomposition reaction, the following data were obtained: $\text{C}_{2}\text{H}_{5}\text{Cl}\text{(g)}\rightarrow \text{C}_{2}\text{H}_{4}\text{(g)} + \text{HCl}\text{(g)} $
| Time/$\sec$ |
Total pressure$/\ce{atm}$ |
| $0$ |
$0.30$ |
| $300$ |
$0.50$ |
Calculate the rate constant $($Given: $\ce{\log 2 = 0.301, \log 3=0.4771, \log 4 =0.6021)}$ AnswerGiven: Initial pressure, $P_o = 0.30 \ \ce{atm}$
$P_t = 0.50 \ \ce{atm}$
$t = 300 s$
Rate constant, $\text{k} = \frac{2.303}{t}\text{log}\frac{\text{p}_{o}}{2\text{p}_{o} - \text{p}_{t}}$
$ = \frac{2.303}{300\text{ s }}\text{log}\frac{0.30}{2\times0.30-0.50}$
$ = \frac{2.303}{300\text{ s }}\text{log}\frac{0.30}{0.60 - 0.50}$
$ = \frac{2.303}{300\text{ s}}\text{log}\frac{0.30}{0.10}$
$ = \frac{2.303}{300\text{ s}}\text{log }3$
$=\frac{2.303}{300\text{ s }}\times0.4771$
$ = \frac{1.099}{300\text{ s }}$
$ = 0.0036\text{ s}^{-1} / 3.66\times10^{-3}\text{ s}^{-1}.$
View full question & answer→Question 43 Marks
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given : log 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)
Answer$t=\frac{2.303}{k}\log\frac{[A]o}{[A]}$$20\text{ }\text{min}=\frac{2.303}{k}\text{ }\log\frac{100}{75}\text{ }\text{ }\text{ }\text{ }\dots(1)$
$t=\frac{2.303}{k}\text{ }\log\frac{100}{25}\text{ }\text{ }\text{ }\text{ }\dots(2)$
Divide (1) equation by (2)
$\frac{20}{t}=\frac{\frac{2.303}{k}\text{ }\log\frac{100}{75}}{\frac{2.303}{k}\text{ }\log\frac{100}{25}}$
$=\frac{\log4/3}{\log4}$
$20/\text{t}=0.1250/0.6021$
$\text{t}=96.3\text{ }\text{min}$
View full question & answer→Question 53 Marks
The following data were obtained during the first order thermal decomposition of $SO_2Cl_2$ at a constant volume:
$\ce{SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)}$
|
Experiment
|
Time/s$^{–1}$
|
Total pressure/atm
|
| $1$ |
$0$ |
$0.4$ |
| $2$ |
$100$ |
$0.7$ |
Calculate the rate constant.
$($Given: $\ce{log4 = 0.6021, log2 = 0.3010).}$ Answer$\ce{SO_2Cl_2 \rightarrow SO_2+Cl_2}$
|
At $t$
|
$=$ |
$0s$ |
$0.4 \ atm$ |
$0 \ atm$ |
$0 \ atm$ |
|
At $t$
|
$=$ |
$100s$ |
$(0.4 – x) \ atm$ |
$x \ atm$ |
$x \ atm$ |
$Pt = 0.4 – x + x + x$
$Pt = 0.4 + x$
$0.7 = 0.4 + x$
$x = 0.3$
$\text{k}=\frac{\text{2.303}}{\text{t}}\log\frac{\text{pi}}{\text{2pi-pt}}$
$\text{k}=\frac{\text{2.303}}{\text{t}}\log\frac{\text{0.4}}{\text{0.8-0.7}}$
$\text{k}=\frac{\text{2.303}}{\text{100s}}\log\frac{\text{0.4}}{\text{0.1}}$
$\text{k}=\frac{\text{2.303}}{\text{100s}}\times0.6021=1.39\times10^{-2}\text{s}^{-1}$ View full question & answer→Question 63 Marks
The rate of a reaction becomes four times when the temperature changes from $293 K$ to $313K.$ Calculate the energy of activation $(E_a)$ of the reaction assuming that it does not change with temperature. $\ce{[R=8.314 JK^{-1} MOL^{-1}, \log 4=0.6021]}$
AnswerGiven if rate at $293K$ is $R$ thus at $313K$ rate becomes $4R$
$\log\frac{\text{k}_{2}}{\text{k}_{1}}=\frac{\text{E}_{a}}{\text{2.303 R}}\Bigg[\frac{\text{T}_{2}-\text{T}_{1}}{\text{T}_{1}\times\text{T}_{2}}\Bigg]$
$\log\frac{\text{4R}}{\text{R}}=\frac{\text{E}_{a}}{\text{2.303 }\times{8.314}}\Bigg[\frac{\text{313}-\text{293}}{\text{293}\times\text{313}}\Bigg]$
$\log\text{4}=\frac{\text{E}_{a}}{\text{19.1471 }}\Bigg[\frac{\text{20}}{\text{91709}}\Bigg]$
$\text{0.6021}=\frac{\text{E}_{a}}{\text{19.1471 }}\Bigg[\frac{\text{20}}{\text{91709}}\Bigg]$
$\frac{\text{0.6021}\times{19.1471}\times{91709}}{\text{20 }}=\text{E}_{a}$
$\text{Ea = 52863.2177J}$ or $\text{52.863 KJ}.$
View full question & answer→Question 73 Marks
A first order reaction has a rate constant of $0.0051\ min^{–1}$. If we begin with $0.10 M$ concentration of the reactant, what concentration of the reactant will be left after $3$ hours?
Answer$\text{t}=\frac{2.303}{\text{k}}\times\log\frac{[\text{A}]_{0}}{[\text{A}]}$
$3\times60\text{min}=\frac{2.303}{0.0051\text{min}^{-1}}\log\frac{0.10}{[\text{A}]}$
$\log\frac{0.10}{[\text{A}]}=\frac{180\times0.0051}{2.303}$
$\log\frac{0.10}{[\text{A}]}=0.399$
$[\text{A}] = 0.04 \text{M}.$
View full question & answer→Question 83 Marks
The rate constant for a first order reaction is $60 s^{-1}$ How much time will it take to reduce the concentration of the reactant to $1/10^{th}$ of its initial value?
Answer$\text{t}=\frac{2.303}{\text{k}}\times\log\frac{[\text{A}]_{0}}{[\text{A}]_{t}}$
$=\frac{2.303}{60\text{s}^{-1}}\times\log\frac{1}{1/10}$
$= 0.038s$ or $3.8 x 10^{-2}s.$
View full question & answer→Question 93 Marks
For the first order thermal decomposition reaction, following data were obtained:
| $\ce{C_2H_5Cl(g)}$ |
$\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }}$ |
$\ce{C_2H_4(g) + HCl(g)}$ |
| Time/$\sec$ |
|
Total pressure/$\ce{atm}$ |
| $0$ |
|
$0.30$ |
| $300$ |
|
$0.50$ |
Calculate the rate constant. $($Given $\ce{: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)}$ Answer$\text{k}=\frac{2.303}{\text{t}}\text { }\log\text{ }\frac{\text{p}_{\text{i}}}{2\text{p}_{\text{i}}-\text{p}_{\text{t}}}$
$=\frac{2.303}{300}\text{ }\log\text{ }\frac{0.3}{2\times0.3-0.5}$
$=\frac{2.303}{300}\text{ }\log\text{ }3$
$=\frac{2.303\times0.4771}{300}$
$= 0.0036 \ \ce{atm^{-1}}$ or $0.004 \ \ce{atm^{-1}} ($approx$.)$
View full question & answer→Question 103 Marks
Following data are obtained for the reaction:
$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2} O_2$
| $t/s$ |
$0$ |
$300$ |
$600$ |
| $[N2O5]/mol L^{–1}$ |
$1.6 \times 10^{–2}$ |
$0.8 \times 10^{–2}$ |
$0.4 \times 10^{–2}$ |
- Show that it follows first order reaction.
- Calculate the half$-$life.
$($Given $\ce{log 2 = 0.3010 log 4 = 0.6021)}$ Answer
- $\text{k}=\frac{2.303}{\text{t}}\text{ }\log\frac{\text{[A]}o}{[\text{A}]}$
$\text{k}=\frac{2.303}{300}\text{ }\log\frac{1.6\times10^{-2}}{0.8\times10^{-2}}$
$\text{k}=\frac{2.303}{300}\text{ }\log2=2.31\times10^{-3}\text{S}^{-1}$
$\text{At}\text{ }600\text{s},\text{ }\text{k}=\frac{2.303}{\text{t}}\text{ }\log\frac{\text{[A]}o}{\text{[A]}}$
$=\frac{2.303}{600}\text{ }\log\ \frac{1.6\times10^{-2}}{0.4\times10^{-2}}$
$\text{k}=2.31\times10^{-3}\text{S}^{-1}$
k is constant when using first order equation, therefore, it follows first order kinetics.
OR
In equal time interval, half of the reactant gets converted into product and the rate of reaction is independent of concentration of reactant, so it is a first order reaction.
- $t_{1/2} = 0.693/k$
$= 0.693/2.31 \times 10^{-3}$
$= 300 s$ View full question & answer→Question 113 Marks
The following data were obtained during the first order thermal decomposition of $SO_2Cl_2$ at a constant volume:
$SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)$
| Experiment |
Time/s$^{-1}$ |
Total pressure/atm |
| $12$ |
$0100$ |
$0.40.7$ |
Calculate the rate constant.
(Given: $\ce{log4 = 0.6021, log2 = 0.3010)}$ Answer$\ce{SO_2Cl_2\rightarrow SO_{2 }+ Cl_2}$
$At t = 0s0.4 \ atm \ 0 \ atm\ 0 \ atm$
$At t = 100s (0.4 – x) \ atm \ x \ atm \ x \ atm$
$Pt = 0.4 – x + x + x$
$Pt = 0.4 + x$
$0.7 = 0.4 + x$
$x = 0.3$
$k=\frac{\text{2.303}}{\text{t}}\log\frac{\text{pi}}{\text{}2pi-pt}$
$k=\frac{\text{2.303}}{\text{t}}\log\frac{\text{0.4}}{\text{0.8-0.7}}$
$k=\frac{\text{2.303}}{\text{100}}\log\frac{\text{0.4}}{\text{0.1}}$
$k=\frac{\text{2.303}}{\text{100}}\times0.6021=1.39\times10^{-2}\text{s}^{-1}$
View full question & answer→Question 123 Marks
The rate constant for the first order decomposition of $\ce{H2O2}$ is given by the following equation:$\log{k}=14.2-\frac{\text{1.0}\times\text{10}^{4}}{\text{T}}K$
Calculate $E_a$ for this reaction and rate constant $k$ if its half$-$life period be $200$ minutes.
$($Given: $R = 8.314\ JK^{–1}\ mol^{–1})$
Answer$\log{k}=\log{A}-\text{E}_{a}\text{/2.303 RT}$
$\text{E}_{a}=\text{/2.303RT=1.0}\times\text{10}^{4}\text{K/T}$
$\text{E}_{a}=1.0\times\text{10}^{4}\times\text{2.303}\times\text{8.314}$
$=191471.4\ J/mol$
$t_{1/2}=0.693/k$
$k=0.693/200\ min$
$=0.0034\ min^{-1}.$
View full question & answer→Question 133 Marks
For the reaction
$2NO_{(g)} + Cl_{2(g)} \rightarrow 2 NOCl_{(g)}$
The following data were collected. All the measurements were taken at 263K:
| Experiment. No. |
Initial $[NO] (M)$ |
Initial $[Cl_2] (M)$ |
Initial rate of disappearance of $Cl_2(M/min)$ |
| $1$ |
$0.15$ |
$0.15$ |
$0.60$ |
| $2$ |
$0.15$ |
$0.30$ |
$1.20$ |
| $3$ |
$0.30$ |
$0.15$ |
$2.40$ |
| $4$ |
$0.25$ |
$0.25$ |
$?$ |
- Write the expression for rate law.
- Calculate the value of rate constant and specify its units.
- What is the initial rate of disappearance of $Cl_2$ in exp. $4?$
Answer
- $Rate = k[NO]^2[Cl_2].$
- $0.60M \ min^{–1 }= k[0.15]^2[0.15]M^2.$
$k = 177.7M^{–2} \ min^{–1}$
- Initial rate of disappearance of $Cl_2$ in exp.$ 4$
Rate $= k[NO]^2[Cl_2]$
Rate $= 177.7M^{–2} min^{–1}\times (0.25)^2 \times (0.25)M^3$
Rate $= 2.8M min^{–1}.$ View full question & answer→Question 143 Marks
Nitrogen pentoxide decomposes according to equation: $\text{2N}_{2}\text{O}_{5}\text{(g)}\rightarrow\text{4NO}_{2}\text{(g)}+\text{O}_{2}\text{(g)}.$
This first order reactionwas allowed to proceed at $40^\circ C$ and the data below were collected:
| $[N_2O_5](M)$ |
Time $(min)$ |
| $0.400$ |
$0.00$ |
| $0.289$ |
$20.0$ |
| $0.209$ |
$40.0$ |
| $0.151$ |
$60.0$ |
| $0.109$ |
$80.0$ |
- Calculate the rate constant. Include units with your answer.
- What will be the concentration of $N_2O_5$ after $100$ minutes?
- Calculate the initial rate of reaction.
Answer
- $\text{k}=\frac{2.303}{t}\log\frac{[A_{o}]}{[A]}$
$\text{k}=\frac{2.303}{\text{20 min}}\log\frac{0.400}{0.289}$
$\text{k} = 0.0163\text{ min}^{–1}$
- $\text{k}=\frac{2.303}{\text{t}}\log\frac{[A_o]}{[A]}$
$\text{0.0163}=\frac{2.303}{\text{100}}\log\frac{0.400}{[A]}$
$[\text{A}] = 0.078\text{M}$
- $\text{Initial rate R = k}[\text{N}_2\text{O}_5]$
$= 0.0163\text{ min}^{–1} \times (0.400\text{ M})$
$= 0.00652\text{M min}^{–1}.$ View full question & answer→Question 153 Marks
A first order reaction has a rate constant of $0.005\ min^{–1}$. If we begin with $0.10 M$ concentration of the reactant, what concentration of reactant will remain in solution after $3$ hours?
Answer$\text{t}=\frac{\text{2.303}}{\text{k}}\times\log\frac{\text{[A]}_{o}}{\text{[A]}}$
$\text{3}\times\text{60min}=\frac{\text{2.303}}{\text{0.0051 min}^{-1}}$ $\frac{\text{log0.10}}{\text{[A]}}$
$\log\frac{\text{0.10}}{\text{[A]}}=\frac{\text{180}\times\text{0.0051}}{\text{2.303}}$
$\log\frac{\text{0.10}}{\text{[A]}}=\text{0.399}$
$\text{[A]}=\text{0.04M}$
View full question & answer→Question 163 Marks
The decomposition of $\ce{NH_3}$ on platinum surface $,\text{2NH}_{3}\text{(g)}\text{ }^{Pt}_\longrightarrow\text{N}_{2}\text{(g)}+\text{3H}_{2}\text{(g)}$ is a zero order reaction with $k=\text{2.5}\times\text{10}^{-4}\text{Ms}^{-1}$ . what are the rates of production of $N_2$ and $H_2?$
Answer$\text{2NH}_{3}\text{(g)}\text{ }^{Pt}_\longrightarrow\text{ N}_{2}\text{(g)}+\text{3H}_{2}\text{(g)}$
$\frac{\text{-d[NH}_{3}]}{\text{dt}}=\text{k[NH}_{3}]^{o}=\text{2.5}\times\text{10}^{-4}\text{Ms}^{-1}$
$-\frac{\text{1}}{\text{2}}\frac{\text{d[NH}_{3}]}{\text{dt}}=+\frac{\text{d[N}_{2}]}{\text{dt}}=+\frac{\text{1}}{\text{3}}\frac{\text{d[H}_{2}]}{\text{dt}}$
Rate of Production of $N_2=+\frac{\text{d[N}_{2}]}{\text{dt}}=-\frac{\text{1}}{\text{2}}\frac{\text{d[NH}_{3}]}{\text{dt}}$
=$\frac{\text{1}}{\text{2}}\times\text{(2.5}\times\text{10}^{-4}\text{Ms}^{-1})=\text{1.25}\times\text{10}^{-4}\text{Ms}^{-1}$
Rate of production of hydrogen= $\frac{\text{d[H}_{2}]}{\text{dt}}=-\frac{\text{3}}{\text{2}}\frac{\text{d[NH}_{3}]}{\text{dt}}$
=$\frac{\text{3}}{\text{2}}\times\text{(2.5}\times\text{10}^{-4}\text{ Ms}^{-1})$
$=3.75 \times 10^{-4} Ms^{-1} $
View full question & answer→Question 173 Marks
The rate of a particular reaction triples when temperature changes from $50^\circ C$ to $100^\circ C$. Calculate the activation energy of the reaction.$[\log 3 = 0.4771; R = 8.314 \ JK^{-1} \ mol^{-1}].$
Answer$T_1=50^oC=323K T_{2 }= 100^oC = 373K$ let rate constant $= k_1$ at $323 K.$
let rate constant $= k_2$ at $373 K$
$\log\frac{\text{k}_{2}}{\text{k}_{1}}=\frac{\text{Ea}}{\text{2.303R}}\bigg(\frac{\text{T}_{2}-\text{T}_{1}}{\text{T}_{1}\text{T}_{2}}\bigg)$
when $T$ is $373K, k_2=3k_1$_
$\log\frac{\text{3k}_{1}}{\text{k}_{1}}=\frac{\text{Ea}}{\text{2.303}\times\text{8.314}}\bigg(\frac{\text{373}\text{-}\text{323}}{\text{323}\times\text{373}}\bigg)$
$\text{0.4771}=\frac{\text{Ea}}{\text{2.303}\times\text{8.314}}\times\frac{\text{50}}{\text{323}\times\text{373 J mol}^{-1}}$
$\text{Ea}=\text{22011.76 J mol}^{-1}=\text{22.012kJ mol}^{-1}.$
View full question & answer→Question 183 Marks
A first order reaction is $50\%$ completed in $40$ minutes at $300K$ and in $20$ minutes at $320K$. Calculate the activation energy of the reaction.
$\big($Given$:\log2=0.3010,\log4=0.6021,\text{R}=8.314\ \text{JK}^{-1}\text{mol}^{-1}\big)$
AnswerGiven,
$\text{t}_{\frac{1}{2}}=40\text{ min}.$
$\text{T}_1=300\text{k}$
$\text{t}_{\frac{1}{2}}=20\text{ min}.$
$\text{T}_2=320\text{k}$
Half$-$life $\text{t}_{\frac{1}{2}}$ for the first-order reaction is,
$\text{t}_{\frac{1}{2}}=\frac{0.693}{\text{k}_1}\ \text{and}\ \text{t}_{\frac{1}{2}}=\frac{0.693}{\text{k}_2}$
$\text{k}_1=\frac{0.693}{40}\ \text{and}\ \text{k}_2=\frac{0.693}{20}$
$\frac{\text{k}_2}{\text{k}_1}=\frac{\frac{0.693}{20}}{\frac{0.693}{10}}=\frac{40}{20}=2$
$\log2=\frac{\text{E}_\text{a}}{2.303\times8.314}\Big[\frac{320-300}{300\times320}\Big]$
$\frac{2.303\times8.314\times300\times320\log2}{20}=\text{E}_\text{a}$
$\text{E}_\text{a}=\frac{2.303\times8.314\times300\times320\times0.3010}{20}$
$=27663.79\text{J mol}^{-1}$
$\text{E}_\text{a}=27.66379\ \text{kJ mol}^{-1}$
Activation energy of the reaction is $27.66\ kJ\ mol^{-1}.$
View full question & answer→Question 193 Marks
The decomposition of $NH_3$ on platinum surface is zero order reaction. If rate constant $(k)$ is $4 \times 10^{-3} Ms^{-1},$ how long will it take to reduce the initial concentration of $NH_3$ from $0.1M$ to $0.064M$.
AnswerGiven,
$\text{k}=4\times10^{-3}\text{Ms}^{-1}$
$[\text{A}_0]=0.1\text{M}$
$[\text{A}]=0.064\text{M}$
For a zero$-$order reaction,
$\text{k}=\frac{1}{\text{t}}\big\{[\text{A}_0]-[\text{A}]\big\}$
$4\times10^{-3}\text{ Ms}^{-1}=\frac{1}{\text{t}}\big\{[0.1]-[0.064]\big\}$
$\text{t}=\frac{0.1-0.64}{4\times10^{-3}}$
$\text{t}=0.009\times10^{3}$
$\text{t}=9$ seconds
View full question & answer→Question 203 Marks
How can you determine the rate law of the following reaction?
$2\text{NO}(\text{g})+\text{O}_2(\text{g})\xrightarrow{\ \ \ \ \ }2\text{NO}_2(\text{g})$
AnswerThe rate law can be determined by measuring the rate of this reaction as a function of initial concentration by keeping the concentration of one of the reactants constant and changing the concentration of other reactant or by changing the concentration of both the reactants. From the concentration dependence of rate, rate law can be determined.
View full question & answer→Question 213 Marks
Answer the following questions on the basis of the given plot of potential energy vs reaction coordinate$:$
- What is the threshold energy for the reaction?
- What is the activation energy for forward reaction?
- What is the activation energy for backward reaction?
- What is enthalpy change for the forward reaction?
Answer
- Threshold energy for the reaction $= 300 \ KJ \ mol^{-1}.$
- Activation energy for the forward reaction $= 300 - 150 = 150 \ KJ \ mol^{-1}.$
- Activation energy for the backward reaction $= 300 - 100 = 200 \ KJ \ mol^{-1}.$
- Enthalpy change for the forward reaction $\Delta_\text{t}\text{H} = 100 - 150 = - 50 \ KJ \ mol^{-1}.$
View full question & answer→Question 223 Marks
Answer the following questions on the basis of the given plot of potential energy vs reaction coordinate$:$
- What is the threshold energy for the reaction?
- What is the activation energy for forward reaction?
- What is the activation energy for backward reaction?
- What is enthalpy change for the forward reaction?
Answer
- Threshold energy for the reaction $= 300 \ KJ \ mol^{-1}.$
- Activation energy for the forward reaction $= 300 - 150 = 150 \ KJ \ mol^{-1}.$
- Activation energy for the backward reaction $= 300 - 100 = 200 \ KJ \ mol^{-1}.$
- Enthalpy change for the forward reaction $\Delta_\text{t}\text{H} = 100 - 150 = - 50 \ KJ \ mol^{-1}.$
View full question & answer→Question 233 Marks
Match the items of Column I and Column II.
|
|
Column I
|
|
Column II
|
|
(i)
|
Diamond.
|
(a)
|
Short interval of time.
|
|
(ii)
|
Instantaneous rate.
|
(b)
|
Ordinarily rate of conversion is imperceptible.
|
|
(iii)
|
Average rate.
|
(c)
|
Long duration of time.
|
Answer
|
|
Column I
|
|
Column II
|
|
(i)
|
Diamond.
|
(b)
|
Ordinarily rate of conversion is imperceptible.
|
|
(ii)
|
Instantaneous rate.
|
(a)
|
Short interval of time.
|
|
(iii)
|
Average rate.
|
(c)
|
Long duration of time.
|
Explanation:
- Rate of conversion of diamond is imperceptible because it requires high activation energy.
- Instantaneous rate of a reaction is rate of a reaction at a particular moment of time.
- Average rate is obtained by dividing the change in concentration of any one of the reactant of product by the time taken for the change i.e. $\frac{\Delta\text{x}}{\Delta\text{t}}.$
View full question & answer→Question 243 Marks
A reaction is first order in A and second order in B.
- Write the differential rate equation.
- How is the rate affected on increasing the concentration of B three times?
- How is the rate affected when the concentrations of both A and B are doubled?
Answer
- Differential rate equation of reaction is
$\frac{\text{dx}}{\text{dt}}=\text{k}[\text{A}]^1[\text{B}]^2=\text{k}[\text{A}]\ [\text{B}]^2$
- When cone. of B is tripled, it means cone. of B becomes (3 × B)
$\therefore$ New rate of reaction,
$\frac{\text{dx'}}{\text{dt}}=\text{k}[\text{A}]\ [\text{3B}]^2=9\text{k}[\text{A}]\ [\text{B}]^2=\bigg(\frac{\text{dx}}{\text{dt}}\bigg)$
i.e., rate of reaction will become 9 times.
- When cone. of A is doubled and that of B is also doubled, then cone. of A becomes [2A] and that of B becomes [2B] rate of reaction,
$\frac{\text{dx''}}{\text{dt}}=\text{k}[\text{2A}]\ [\text{3B}]^2=8\text{k}[\text{A}]\ [\text{B}]^2$
i.e., the rate of reaction will become 8 times the rate as in (1). View full question & answer→Question 253 Marks
Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions.
$\text{2H}_2\text{O}_2\xrightarrow{\text{OH}^-}2\text{H}_2\text{O}+\text{O}_2$
This reaction takes place in two step as given below:
Step-I $\text{H}_2\text{O}_2+\text{I}^-\rightarrow\text{H}_2\text{O}_2+\text{IO}^-\text{ (slow)}$
Step-II $\text{H}_2\text{O}_2+\text{IO}^-\rightarrow\text{H}_2\text{O}_2+\text{I}+\text{O}_2\text{ (fast})$
- Write the rate law expression and determine the order of reaction $w.r.t. H_2O_2.$
- What is the molecularity of each individual step?
Answer
- Rate $= K[H_2O_2]^1 [I^-]^1$
- Order of reaction $w.r.t.H_2O_2 = 1.$
- Molecularity of step $I = 2$ amd step $II = 2.$
View full question & answer→Question 263 Marks
After 24 hrs. Only 0.125gm out of the initial quantity of 1gm of a radioactive isotope remains behind. What is its half life period?
Answer$\text{Here, [R]}_0 =\text{ g, [R]} = 0.125\text{g, t = 24h.}$$\text{k}=\frac{2.303}{\text{t}}\log\frac{|\text{R|}_0}{\text{|R|}} $
$\text{k}=\frac{2.303}{24}\log\frac{1}{0.125}$
$\text{k}=\frac{2.303}{24}\log8$
$\text{k}=\frac{2.303}{24}\times0.9031$
$\text{k}=0.866 \text{ h}^{-1}$
$\text{t}_{\frac{1}{2}}=\frac{0.693}{\text{k}}$
$\text{t}_{\frac{1}{2}}=\frac{0.693}{0.0866\text{ h}^{-1}}\text{ or t}_{\frac{1}{2}}=8\text{ h}.$
View full question & answer→Question 273 Marks
In a pseudo first order hydrolysis of ester in water, the following results were obtained:
| $t/s$ |
$0$ |
$30$ |
$60$ |
$90$ |
| $[\text{Ester}]/mol\ L^{-1}$ |
$0.55$ |
$0.31$ |
$0.17$ |
$0.085$ |
- Calculate the average rate of reaction between the time interval $30$ to $60$ seconds.
- Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer
- Average rate of reaction between interval of time $30$ to $60$ second is given by
Average rate $=\frac{\triangle\text{x}}{\triangle\text{t}}=\frac{\text{C}_2-\text{C}_1}{\triangle\text{t}}$
$=\frac{0.17-0.3}{60-30}$
$=\frac{0.14}{30}=-0.00467$
$= -4.67 \times 10^{-3} Ms^{-1}$
Minus sign shows that rate of reaction is decreasing with time as cone. of ester is decreasing with time.
- Pseudo first order rate constant $k$ is given by
$\text{k}=\frac{2.303}{\text{t}}\ \text{log}\ \frac{\text{a}}{\text{a-x}}$
Where $a$ is initial cone. and $(a - x)$ cone. after time $t$.
Here $a= 0.55 M$.
| $\text{time t}$ in $(\sec.)$ |
$\text{conc. of ester/M}\\\ \ \ \ \ \ \ \ (\text{a - x})$ |
$\text{k}=\frac{2.303}{\text{t}}\ \text{log}\ \frac{\text{a}}{\text{a - x}}$ |
| $30$ |
$0.31$ |
$\text{k}_1=\frac{2.303}{30}\ \text{log}\ \frac{0.55}{\text{0.31}}\\=1.91\times10^{-2}\text{s}^{-1}$ |
| $60$ |
$0.17$ |
$\text{k}_2=\frac{2.303}{60}\ \text{log}\ \frac{0.55}{\text{0.17}}\\=1.95\times10^{-2}\text{s}^{-1}$ |
| $90$ |
$0.085$ |
$\text{k}_3=\frac{2.303}{90}\ \text{log}\ \frac{0.55}{\text{0.085}}\\=2.06\times10^{-2}\text{s}^{-1}$ |
The nearly equal values of $k$ confirms that reaction is of first order.
The actual value of rate constant is the average of three values of $k$.
Therefore, rate constant of reaction
$\text{k}=\frac{\text{k}_1+\text{k}_2+\text{k}_3}{3}$
$=\frac{(1.91+1.96+2.06)\times10^{-2}}{3}$
$= 1.97 \times 10^{-2} s^{-1}$ View full question & answer→Question 283 Marks
Calculate the half$-$life of a first order reaction from their rate constants given below:
- $\ce{200 s^{–1}}$
- $\ce{2 min^{–1}}$
- $\ce{4 years^{–1}}$
Answer
- $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$
$=\frac{0.693}{200\ \text{min}^{-1}}$
$= 3.4 \times 10^{-3} s($approximately$)$
- $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$
$=\frac{0.693}{2\ \text{min}^{-1}}$
$\ce{= 0.35 min} ($approximately$)$
- $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$
$=\frac{0.693}{4\ \text{years}^{-1}}$
$\ce{= 0.173 years} ($approximately$)$ View full question & answer→Question 293 Marks
$A + 2B \rightarrow 3C + 2D$. The rate of disappearance of $B$ is $1 \times 10^{-2} mol L^{-1}s ^{-1}$ What will be:
- Rate of the reaction.
- Rate of change in concentration of $A $and $C?$
Answer
- $\text{As}\frac{\text{-d}|\text{B|}}{\text{dt}}=1\times10^{-2}\text{mol L}^{-1}\text{ s}^{-1}$
$ \therefore\text{ Rate}= -\frac{1}{2} \frac{\text{d |B|}}{\text{dt}}=\frac{1}{2}\times1\times10^{-2}=0.5\times10^{-2}\text{mol L}^{-1}\text{s}^{-1}$
- $\text{Rate}= -\frac{\text{d|A|}}{\text{dt}}= -\frac{1}{2} \frac{\text{d|B|}}{\text{dt}}=+\frac{1}{3}\frac{\text{d|C|}}{\text{dt}}$
Rate of change in concentration of $A$
$=-\frac{\text{d|A|}}{\text{dt}}=-\frac{1}{2}\frac{\text{d|B|}}{\text{dt}}$
$=0.5\times10^{-2}\text{mol L}^{-1}\text{s}^{-1}$
Rate of change in concentration of $C$
$=+ \frac{\text{d|C|}}{\text{dt}}=- \frac{3}{2}\frac{\text{d|B|}}{\text{dt}}\times{1}\times10^{-2}\\ =1.5\times10^{-2}\text{mol L}^{-1}\text{s}^{-2}$ View full question & answer→Question 303 Marks
Observe the graph in diagram and answer the following questions.
- If slope is equal to $-2.0 \times 10^{-6} \sec^{-1} ,$ what will be the value of rate constant?
- How does the half$-$life of zero order reaction relate to its rate constant?
Answer
- Slope $=\frac{\text{k}}{2.303}$ or $k=-2.303\times$ slope
$\therefore \text{ k}=-2.303\times(-2.0\times10^{-6}\text{s}^{-1})$$\text{k}=4.606\times10^{-6}\text{s}^{-1}$
- For a zero order reaction:
$\text{t}=\frac{|\text{R}_0|-|\text{R}|}{\text{k}}$
$\text{At t}=\text{t}_{1/2},[\text{R}]=\frac{|\text{R|}_0}{2}$
$\therefore \text{t}_{1/2}=\frac{|\text{R|}_0-\frac{\text{|R|}}{2}}{\text{k}}\text{ or} \text{ t}_{1/2}=\frac{\text{|R|}_0}{2\text{k}}$ View full question & answer→Question 313 Marks
$^{238}_{92}\text{U}$ Change to $^{206}_{92}\text{Pb}$ by successive radioactive decay. A sample of urenium ore was anlaysed and found to contain $1.0 g$ of $^{238}\text{U}$ and $0.1 g$ of $^{206}\text{Pb}$ had accumulated due to decay of $^{238}\text{U}$, find out the age of ore. $($Half$-$ life of $^{238}\text{U}$$= 4.5\times10^9$ years$).$
Answer$[A]_{0 }=$ Initial amount of $^{238}\text{U}$ = amount of $^{238}\text{U}$ left at time of $^{238}\text{U}$
decayed$[A]_0 = 1.0 +$ amount of $^{238}\text{U}$ decayed
Now, amount of $^{238}\text{U}$ decayed $=\frac{1.0\times238}{206}\text{ g}=0.1155 {\text{ g}}$
$\therefore\text{ [A]}_0 = 1.0 \text{ g + 0.1155 g = 1.1155 g}$
Determination of k: $\text{k}=\frac{0.693}{\text{t}_{1/2}}=\frac{0.693}{4.5\times10^9}=0.154\times 10^{-9}\text{ year}^{-1}$ Determination of times: $\text{t}=\frac{2.303}{\text{k}}\log\frac{|\text{A}|_0}{|\text{A}|}$
Substituting the values of $[A]_{0 =}1.1155 g$ and $k = 0.154\times10^{-9}\text{year}^{-1}$ $\text{t}= \frac{2.303}{0.154\times10^9} \log \frac{1.1155}{1}$
$=0.7099\times10^9 \text{ year}$
$=0.7099\times10^8 \text{ year}$
View full question & answer→Question 323 Marks
A solution of $\ce{H_2O_2}$ when titrated against $\ce{KMnO_4}$ solution at different intervals of time gave the following results:
| Time $($minutes$)$ |
$0$ |
$10$ |
$20$ |
| Volume of $\ce{KMnO_4 (mL)}$ |
$23.8$ |
$14.7$ |
$9.1$ |
Show that decomposition of $\ce{H_2O_2}$ is first order reaction. Answer
- $\text{k}=\frac{2.303}{14.7}\log\frac{2.303}{10}=\frac{2.303}{10}\times0.2093=0.048\min^{-1}$
- $\text{k}=\frac{2.303}{20}\log\frac{2.303}{9.1}=\frac{2.303}{20}\times0.4176=0.048\min^{-1}$
Since the value of $k$ comes out to be constant in both the cases.
therefore the reaction is of first order. View full question & answer→Question 333 Marks
A graph between ln $k$ and $\frac{1}{\text{t}}$ for a reaction is given. Here $k$ is rate constant and $T$ is temperature in Kelvin.If $OA = a$ and $OB = b,$ answer the following:
- What is the activation energy $(Ea)$ of the reaction?
- What is the frequency factor $(A)$ for the reaction?
AnswerAccording to Arrhenius equation, In $\text{k}=-\frac{\text{E}_\text{a}}{\text{RT}}+\text{ In A}$
- Slope $=-\frac{\text{OB}}{\text{OA}}=-\frac{\text{b}}{\text{a}}=-\frac{\text{E}_\text{a}}{\text{R}}\text{ E}_\text{a}=\frac{\text{b}}{\text{a}}\text{ R}$
- Intercept on $y-$axis $= OB = b =$ In $A$ or $A = e^b.$
View full question & answer→Question 343 Marks
Differentiate between rate of reaction and reaction rate constant.
Answer
| S. No |
Rate of Reaction |
Reaction Rate Constant |
| $(i)$ |
Rate of reaction is the change in concentration of a reactant or product in a unit interval of time. |
It is the rate of reaction when the molar concentration of each of the reactants is unity. |
| $(ii)$ |
The rate of reaction at any instant of time depends upon the molar concentrations of the reactants at that time. |
The rate constant does not depend upon the concentrations of the reactants. |
| $(iii)$ |
Its units are always mol litre$^{–1}$ time$^{–1}.$ |
Its units depend upon the order of reaction. |
View full question & answer→Question 353 Marks
Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.
AnswerThermodynamics feasibility of a reaction depends on Gibbs free energy i.e., AG must be negative for spontaneous process. Kinetic feasibility depends on the activation energy of reaction, the lesser is the activation energy, the greater is the feasibility of reaction, i.e.,
$\text{Diamond}\xrightarrow{\ \ \ \ \ \ }\text{Graphite }\Delta\text{G}=-\text{ve}$
This process is thermodynamically feasible but it is very slow due to its high activation energy.
View full question & answer→Question 363 Marks
For a reaction, the energy of activation is zero. What is the value of rate constant at $300K,$ if $k = 1.6 \times 10^6 s –^1$ at $280K$? $\ce{[R = 8.31JK–^1 mol–^1]}.$
AnswerGiven $,\text{T}_1=280\text{ K},\text{k}_1=1.6\times10^6\text{s}^{-1},\text{k}_2=?,\text{E}_\text{a}=0\text{ T}_2=300 \text{ K}.$
By Arrhenius equation,
$\log \frac{\text{k}_2}{\text{k}_1}=\frac{\text{E}_a}{2.303\text{ R}}\bigg[\frac{\text{T}_2-\text{T}_1}{\text{T}_2\text{T}_1}\bigg]$
$\text{As, Ea} = 0$
$\therefore$$\log\frac{\text{k}_2}{\text{k}_1}=0$
$\text{or }\frac{\text{k}_2}{\text{k}_1}=1 \text{ or}\text{ k}_2=\text{k}_1$0
Thus, the rate constant at $300K$ is $1.6\times10^6\text{s}^{-1}.$
View full question & answer→Question 373 Marks
In a reaction between $A$ and $B$, the initial rate of reaction $(r_0)$ was measured for different initial concentrations of $A$ and $B$ as given below:
| $A/mol L^{-1}$ |
$0.20$ |
$0.20$ |
$0.40$ |
| $B/mol L^{-1}$ |
$0.30$ |
$0.10$ |
$0.05$ |
| $r_0/mol L^{-1} s^{-1}$ |
$5.07 \times 10^{-5}$ |
$5.07 \times 10^{-5}$ |
$1.43 \times 10^{-4}$ |
What is the order of the reaction with respect to $A$ and $B?$ AnswerIn a reaction $A$ and $B$, Let order of reaction $w.r.t. A$ is $x$ and $w.r.t. B$ is y. Then the rate of reaction can be written as
rate $= k[A]^x [B]^y$
From given table data, $1$ and $2$ we can write
$5.07 \times 10^{-5} = k[10.20]^x [0.30]^y ....(i)$
$5.07 \times 10^{-5} = k[0.20]^x [0.10]^y ....(ii)$
Dividing $(ii)$ by $(i),$ we get
$\frac{5.07\times10^{-5}}{5.07\times10^{-5}}=\frac{\text{k}[0.20]^\text{x}\ [0.10]^\text{y}}{\text{k}[0.20]^\text{x}\ [0.30]^\text{y}}$
$\text{or}\ \ \ 1=\Big(\frac{0.10}{0.30}\Big)^\text{y}\ \text{or}\ \ \ \ \text{y}=0$
From given table data, $2$ and $3$ we can write
$5.07 \times 10^{-5} = k[0.20]^x [0.20]^y$
$=\text{k}[0.20]^\text{y}\times1\ \ \ [\therefore\ \ \text{y}=1]\ ...\text{(iii)}$
$7.06 \times 10^{-5} = k[0.20]^x [0.05]^y$
$=\text{k}[0.40]^\text{x}\times1\ \ ...(\text{iv})$
Dividing $(iv)$ by $(iii)$, we get
$\frac{5.07\times10^{-5}}{5.07\times10^{-5}}=\frac{\text{k}[0.20]^\text{x}}{\text{k}[0.20]^\text{x}}=\bigg[\frac{0.4}{0.2}\bigg]^\text{x}=(2)^2$
or $(2)^x = 3/2 = 1.5$
or $x = 0.5$
Thus the order of reaction w.r.t. A is $\frac{1}{2}$ and $w.r.t. B$ is zero.
View full question & answer→Question 383 Marks
The rate for the reaction $R\rightarrow P$ is rate $= k[R].$ It has been shown graphically below. What is rate constant for the reaction?
AnswerFrom the graph:
Case $I:$ Rate $= k|A|$
$\ce{1 \times 10^{-2}mol L^{-1} s-^1 = k (0.1 mol L^{-1})}$
$\therefore \text{ k}=\frac{1\times10^{-2} \text{ mol L}^{-1}\text{s}^{-1}}{0.1\text{ mol L}^{-1}}=0.1\text{ s}^{-1}$
Case $II:$
$\ce{3 \times 10^{-2}mol L^{-1} s-^1 = k (0.3 mol L^{-1})}$
$ \text{ k}=\frac{3\times10^{-2} \text{ mol L}^{-1}\text{s}^{-1}}{0.3\text{ mol L}^{-1}}=0.1\text{ s}^{-1}$
Hence, $\ce{K = 0.1 s^{-1}}.$
View full question & answer→Question 393 Marks
The rate constant for the decomposition of ethyl iodide $\ce{C_2H_5l(g) \rightarrow C_2H_4(g)+Hl(g)}$
at $600K$ is $1.60 \times 10^{-5} s^{-1} .$ Its energy of activation is $\ce{209 kJ/mol}.$ Calculate the rate constant of the reaction at $700K.$
AnswerWe know that, $\log \text{k}_2-\log \text{k}_1= \frac{\text{E}_\text{a}}{2.303 \text{ R}} \bigg[\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg]$
$\log\text{ k}_2=\log\text{k}_1+\frac{\text{E}_a}{2.303 \text{ R}}\bigg[\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg]$
$\log \text{k}_2=\log(1.60\times10^{-5}\text{s})+\frac{2090000\text{ J}\text{ mol}^{-1}}{2.303\times8.314\text{ J}\text{ mol}^{-1} \text{ K}^{-1}} \bigg[\frac{1}{600\text{ K}}-\frac{1}{700\text{ K}}\bigg]$
$\log\text{k}_2=-4.796+2.599$
$= -2.197\text{ or}\text{ k}_2=\text{Antilog}(\overline3.803)$
$\text{k}_2=6.36\times10^{-3}\text{ s}^{-1}$
View full question & answer→Question 403 Marks
Rate constant $k$ for first order reaction has been found to be $2.54 \times 10^{-3} s^{-1}.$ Calculate its three$-$fourth life.
Answer$\text{t}=\frac{2.303}{\text{k}}\log \frac{\text{|R|}_0}{\text{|R|}}\dots(\text{i)}$
$\text{k}=2.54\times10^{-3}\text{s}^{-1};[\text{R}]=\frac{|R|}{4}$
Substituting these value in equation $(i),$ we get
$\text{t}_{\frac{3}{4}}=\frac{2.303}{2.54\times10^{-3}}\log\frac{ \frac{\text{|R|}_0}{\text{|R|}_0}}{4}=0.9066\times10^3\log4$
$\text{t}_{\frac{3}{4}}=0.9066\times10^3\times0.6021\text{ s}=5.46\times10^2\text{s}.$
View full question & answer→Question 413 Marks
For which type of reactions, order and molecularity have the same value?
AnswerElementary reactions have same value of order and molecularity.
Elementary reaction is chemical reaction in which one or more reactants racts directly to form product in a single step with single transition state.
Example:
$\text{A + B}\xrightarrow{\ \ \ \ \ }\text{Products}$
Rate equation: $[A]^1[B]^1$ Order of reaction is sum of the cofficients of rate equation.
$\therefore$ Order of reaction $= 2$ Molecularity $= 2$
View full question & answer→Question 423 Marks
Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?
AnswerBalanced chemical equation often leads to incorrect order or rate law. For example the following reaction seems to be a tenth order reaction.
$\text{KClO}_3+6\text{FeSO}_4+3\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ }\text{KCl}+3\text{H}_2\text{O}+3\text{Fe}_2(\text{SO}_4)_3$
This is actually a second order reaction. Actually the reaction is complex and occurs in several steps. The order of such reaction is determined by the slowest step in the reaction mechanism. Order is determined experimentally and is confined to the dependence of observed rate of reaction on the concentration of reactants.
View full question & answer→Question 433 Marks
Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions?
AnswerA complex reaction occurs through several elementary reactions. Numbers of molecules involved in each elementary reaction may be different i.e., the molecularity of each step may be different. Therefore, the molecularity of overall complex reaction is meaningless. On the other hand, order of a complex reaction is experimentally determined by the slowest step in its mechanism and is therefore, applicable even in the case of complex reactions.
View full question & answer→Question 443 Marks
During nuclear explosion, one of the products is $^{90}$ Sr with half$-$life of $28.1$ years. If $1\mu g of ^{90}$ Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after $10$ years and $60$ years if it is not lost metabolically.
Answer$\text{Here, k}=\frac{0.693}{\text{t}{1/2}}=\frac{0.693}{28.1}\text{y}^{-1}$
It is known that, $\text{t}=\frac{2.303}{\text{k}}\text{log}\frac{[\text{R}]_\circ}{\text{[R]}}$
$10=\frac{2.303}{\frac{0.693}{28.1}}\text{log}\frac{1}{\text{[R]}}$
$10=\frac{2.303}{\frac{0.693}{28.1}}(-\text{log[R])}$
$\text{log[R}]=-\frac{10\times0.693}{2.303\times28.1}$
$[R] =$ anti $log( 0.1071)$
$=\text{anti log}(\bar{1}.8929)$
$= 0.7814µg$
Therefore, $= 0.7814µg$ of $^{90}$ Sr will remain after $10$ years.
$\text{Again, t}=\frac{2.303}{\text{k}}\text{log}\frac{[\text{R}]_\circ}{\text{[R]}}$
$60=\frac{2.303}{\frac{0.693}{28.1}}\text{log}\frac{1}{\text{[R]}}$
$\text{log[R}]=-\frac{60\times0.693}{2.303\times28.1}$
$[R]$ anti $log(-0.6425)$
$=\text{anti log}(\bar{1}.3575)$
$= 0.2278µg$
Therefore, $0.2278µg$ of $^{90}$ Sr will remain after $60$ years.
View full question & answer→Question 453 Marks
For the decomposition of azoisopropane to hexane and nitrogen at $543 K$, the following data are obtained.
| $\ce{t (\sec)}$ |
$\ce{P(mm of Hg)}$ |
| $0$ |
$35.0$ |
| $360$ |
$54.0$ |
| $720$ |
$63.0$ |
Calculate the rate constant. AnswerThe decomposition of azoisopropane to hexane and nitrogen at $543 K$ is represented by the following equation.
$(\text{CH}_{3})_{2} \text{CHN = NCH ( CH}_{3})_{2\text{g}} \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\text{N}_{2\text{(g})} + \text{C}_{6}\text{H}_{14\text{(g})}\\ \ \ \text{At t= 0 }\ \ \ \ \ \ \ \ \ \ \ \ \text{p}_{0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0 \\ \ \text{At t= 0 }\ \ \ \ \ \ \ \ \ \ \ \text{p}_{0-\text{p}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{p} \ \ \ \ \ \ \ \ \ \ \ \ \text{p}$
After time, t, total pressure, $P_t = (P_0 - p) + p + p$
$P_t = P_0 + p$
$p = P_t - P_0$
Therefore, $P_0 - p = P_0 - (P_t - P_0)$
$= 2P_0 - P_t$
For a first order reaction,
$\text{k}=\frac{2.303}{\text{t}}\text{log}\frac{\text{P}_\circ}{\text{P}_\circ-\text{P}}$
$=\frac{2.303}{\text{t}}\text{log}\frac{\text{P}_\circ}{2\text{P}_\circ-\text{P}_\text{t}}$
$\text{When t}=360\ \text{s, k}=\frac{2.303}{360\ \text{s}}\text{log}\frac{35.0}{2\times35.0-54.0}$
$= 2.175 \times 10^{-3} s^{-1}$
$\text{When t}=720\ \text{s, k}=\frac{2.303}{720\ \text{s}}\text{log}\frac{35.0}{2\times35.0-63.0}$
View full question & answer→Question 463 Marks
A first order gas reaction $\ce{A_2B_2(g) \rightarrow 2A(g) + 2B(g)}$ at the temperature $400^\circ C$ has the rate constant $K = 2.0 \times 10^{-4}s^{-1}$ . What percentage of $A_2B_2$ is decomposed on heating for $900$ seconds.
Answer$\text{k}=\frac{2.303}{\text{t}}\log \frac{\text{|R|}_0}{\text{|R|}}$
$2.0\times10^{-4}\text{s}^{-1}=\text{k}=\frac{2.303}{900}\log \frac{\text{|R|}_0}{\text{|R|}}$
$\log \frac{\text{|R|}_0}{\text{|R|}}=\frac{2.0\times10^{-4}\times900}{2.303}=0.0781$
$\log \frac{\text{|R|}_0}{\text{|R|}}=-0.0781$
$ \frac{\text{|R|}_0}{\text{|R|}}=\text{Antilog }\overline{ 1}.9219$
$ \frac{\text{|R|}_0}{\text{|R|}}=0.835$
$\text{[R]} = 0.835\text{ [R]}_0$$\text{If [R]}_0 = 100\text{, then.}$
$\text{[R] }= 83.5.$
View full question & answer→Question 473 Marks
The rate for the reaction $R\rightarrow P$ is rate $= k[R].$ It has been shown graphically below. What is rate constant for the reaction?
AnswerFrom the graph:
Case $I:$ Rate $= k|A|$
$1 \times 10^{-2}\text{ mol L}^{-1} s-^1 = k (0.1 \text{ mol L}^{-1})$
$\therefore \text{ k}=\frac{1\times10^{-2} \text{ mol L}^{-1}\text{s}^{-1}}{0.1\text{ mol L}^{-1}}=0.1\text{ s}^{-1}$
Case $II:$
$3 \times 10^{-2}\text{ mol L}^{-1} s-^1 = k (0.3 \text{ mol L}^{-1})$
$ \text{ k}=\frac{3\times10^{-2} \text{ mol L}^{-1}\text{s}^{-1}}{0.3\text{ mol L}^{-1}}=0.1\text{ s}^{-1}$
Hence, $K = 0.1 s^{-1}.$
View full question & answer→Question 483 Marks
In a reaction if the concentration of reactant $A$ is tripled, the rate of reaction becomes twenty seven times. What is the order of the reaction?
AnswerThe rate of reaction is directly proportional to the concentration of reactants.
Then acc to the information, if the reactant get tripled, then the rate becomes $27$ times.
$3 \times 3 \times 3 = 27$ times
It will be $3$ power $3.$
Order can be defined as the power that will be raised in the conc. of reactant in the rate law equation.
So, here the order will be $3,$ it’s a $3^{rd}$ order reaction.
View full question & answer→Question 493 Marks
Following reaction takes place in one step: $\text{2NO(g)+O}_2\text{(g)}\rightarrow\text{2NO}_2\text{(g)}$
How will the rate of the above reaction change if the volume of the reaction vessel is reduced to one$-$third of its original volume? Will there be any change in the order of the reaction with reduced volume?
Answer$\therefore$ Rate $=\text{k[NO]}^2[\text{O}_2]$
Let initially, moles of $NO = a,$ moles of $O_{2 }= b,$ volume of vesse $l= V.$
Then $\text{[NO]}=\frac{\text{a}}{\text{V}}\text{M},[\text{O}_2]=\frac{\text{b}}{\text{V}}\text{M}$
Rate $\text{(r}_1)=\text{k}\bigg(\frac{\text{a}}{\text{V}}\bigg)^2\bigg(\frac{\text{b}}{\text{V}}\bigg)=\text{k}\frac{\text{a}^2\text{b}}{\text{V}^3}\dots(\text{i)}$
Now, new volume $=\frac{\text{V}}{3 }$
$\therefore$ New concentrations:$ [\text{NO]}=\frac{\text{a}}{\frac{\text{V}}{3}}=\frac{3}{\text{V}}$
$[\text{O}_2]= \frac{\text{b}}{\frac{\text{V}}{3}}=\frac{\text{3a}}{\text{V}}$
$\therefore$ New rate $(r_2)= \text{k}\bigg(\frac{\text{3a}}{\text{V}}\bigg)^2\bigg(\frac{3\text{b}}{\text{V}}\bigg)= \frac{27\text{ka}^2\text{b}}{\text{V}^3}\dots \text{(ii)}$
$\therefore \frac{\text{r}_2}{\text{r}_2}=27 \text{ or}\text{ r}_2=27\text { r}_1$.
i.e., rate becomes $27$ times.
Thus, there is no effact on the order of reaction.
View full question & answer→Question 503 Marks
A certain reaction is $50\%$ complete in $20$ minutes at $300K$ and the same reaction is again $50\%$ complete in $5$ minutes at $350K$. Calculate the activation energy if it is a first order reaction.
$[R = 8.314\ J K^{–1} mol^{–1} ; \log 4 = 0.602]$
AnswerFor a first order reaction, $\text{k}=\frac{0.693}{\text{t}_{1/2}}$ $ \text{T}_1 = 300\text{ K},\text{k}_1=\frac{0.693}{20}=3.456\times10^{-2}\min^{-1}$
$\text{T}_2 = 350\text{K}\text{ k}_2=\frac{0.693}{5}=1.386\times10^{-1}\min{-1}$
$\log \frac{\text{k}_2}{\text{k}_1}=\frac{\text{E}_a}{2.303\text{ R}}\bigg[\frac{\text{T}_2-\text{T}_1}{\text{T}_2\text{T}_1}\bigg]$
$\log \frac{1.386\times10^{-1}}{3.465\times10^{-2}}=\frac{\text{E}_\text{a}}{2.303\times8.314} (\frac{350-300}{350\times300})$
$\log 4=\frac{\text{E}_\text{a}}{2.303\times8.314 } (\frac{50}{350\times300})$$0.602=\frac{E_\text{a}}{19.147}(\frac{50}{350\times300})$
$\text{E}_\text{a}=\frac{0.602\times19.147\times350\times300}{50}$
$=24205.63\text{ J}\text{ mol}^{-1}=24.206\text{ KJ mol}^{-1}$
View full question & answer→Question 513 Marks
The activation energy for the reaction,
$\ce{2 HI(g) \rightarrow H2 + I2(g)}$ is $209.5\ kJ\ mol^{–1}$ at $581K$.
Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Answer$2HI (g) \rightarrow H + I_2(g)$
Activation energy$, E_a = 209.5 \ kJ\ mol^{−1}$
Multiply by $1000$ to convert in $j$
$E_{a }= 209500\ J\ mol^{−1}$
Temperature$, T = 581 K$
Gas constant$, R = 8.314 JK^{−1} mol^{−1}$
According to Arhenious equation
$K = A e^{–Ea/RT}$
In this formula term $e^{–Ea/RT}$ represent the number of molecules which have energy equal or more than activation energy,
Number of molecules $= e^{–Ea/RT}$
Plug the values we get
Number of molecules
$=\text{e}-\frac{209500}{8.314\times581}$
$= e^{-43.4}$
$=\frac{1}{\text{e}^{-43.4}}$
taking antilog of we get
$= 1.47 \times 10^{-19}$
View full question & answer→Question 523 Marks
For the reaction: $2A + B \rightarrow A_2B$ The rate $= k[A][B]^2$ with $k = 2.0 \times 10^{–6} \text{ mol}^{–2} L^2 s^{–1}.$ Calculate the initial rate of the reaction when $[A] = 0.1 \text{ mol L}^{–1}, [B] = 0.2 \text{ mol L}^{–1}. $Calculate the rate of reaction after $[A]$ is reduced to $0.06 \text{ mol L}^{–1}.$
AnswerThe initial rate of the reaction is
Rate$= k[A][B]2$
$= (2.0 \times 10^{-6} \text{ mol}^{-2} L^2 s^{-1})(0.1\text{ mol L}^{-1})(0.2 \text{ mol L}^{-1})^2$
$= 8.0 \times 10^{-9} \text{ mol}^{-2} L^2 s^{-1}$
When $[A]$ is reduced from $0.1 \text{ mol L}^{-1}$ to $0.06 \text{ mol}^{-1},$ the concentration of $A$ reacted
$= (0.1 - 0.06) \text{mol L}^{-1} = 0.004 \text{mol L}^{-1}$
Therefore, concentration of $B$ reacted $=\frac{1}{2}\times0.04\ \text{ mol L}^{-1}-0.02\ \text{ mol L}^{-1}$
Then, concentration of $B$ available, $[B] = (0.2 - 0.02) \text{ mol L}^{-1}$
$= 0.18 \text{mol L}^{-1}$
After $[A]$ is reduced to $0.06 \text{ mol}^{-1},$ the rate of the r euction is given by,
Rate $= k[A][B]_2$
$= (2.0 \times 10-6 \text{ mol}^{-2} L^2 s^{-1})(0.06 \text{ mol L}^{-1})(0.18 \text{ mol L}^{-1})^2$
$= 3.89 \text{mol L}^{-1} s^{-1}$
View full question & answer→Question 533 Marks
The rate constant for the decomposition of hydrocarbons is $2.418 \times 10^{–5}s^{–1}$ at $546 K$. If the energy of activation is $179.9\ kJ/mol,$ what will be the value of pre$-$exponential factor.
Answer$k =2.418 \times 10^{-5} s^{-1}$
$T = 546 K$
$E_a = 179.9\ kJ\ mol^{-1}$
$=179.9 \times 10^3 J\ mol^{-1}$
According to the Arrhenius equation,
$k = Ae^{-E_a/RT}$
$\text{In k}=\text{In A}-\frac{\text{E}_\text{a}}{\text{RT}}$
$\text{log k}=\text{log A}-\frac{\text{E}_\text{a}}{2.303\text{RT}}$
$\text{log A}=\text{log k}+\frac{\text{E}_\text{a}}{2.303\text{RT}}$
$=\text{log}(2.418\times10^{-5}\text{s}^{-1})+\frac{179.9\times10^3\text{J mol}^{-1}}{2.303\times8.314\text{Jk}^{-1}\text{mol}^{-1}\times546\text{K}}$
$= (0.3835 - 5) + 17.2082$
$= 12.5917$
Therefore$, A = \text{antilog}\ (12.5917)$
$= 3.9 \times 10^{12} s^{−1} ($approximately$)$
View full question & answer→Question 543 Marks
The following results have been obtained during the kinetic studies of the reaction: $2A + B \rightarrow C + D$
| Experiment |
$[A]/mol\ L^{-1}$ |
$[B]/mol\ L^{-1}$ |
Initial rate of formation of
$D/mol\ L^{-1} min^{-1}$ |
| $I$ |
$0.1$ |
$0.1$ |
$6.0 \times 10^{-3}$ |
| $II$ |
$0.3$ |
$0.2$ |
$7.2 \times 10^{-2}$ |
| $III$ |
$0.3$ |
$0.4$ |
$2.88 \times 10^{-1}$ |
| $IV$ |
$0.4$ |
$0.1$ |
$2.40 \times 10^{-2}$ |
Determine the rate law and the rate constant for the reaction. AnswerLet the order of the reaction with respect to $A$ be $x$ and with respect to $B$ be $y$.
Therefore, rate of the reaction is given by,
Rate $= k[A]^x [B]^y$
According to the question,
$6.0 \times 10^{-3} = k[0.1]^x [0.1]^y ....(i)$
$7.2 \times 10^{-2} = k[0.3]^x [0.2]^y ....(ii)$
$2.88 \times 10^{-1} = k[0.3]^x [0.4]^y ....(iii)$
$2.40 \times 10^{-2} = k[0.4]^x [0.1]^y ....(iv)$
Dividing equation $(iv)$ by $(i),$ we obtain
$\frac{2.40\times10^{-2}}{6.0\times10^{-3}}=\frac{\text{k}[0.4]^\text{x}[0.1]^\text{y}}{\text{k}[0.1]^\text{x}[0.1]^\text{y}}$
$4=\frac{[0.4]\text{x}}{[0.1]^\text{x}}$
$4=\Big(\frac{0.4}{0.1}\Big)^\text{x}$
$(4)^1 = 4^x$
$x = 1$
Dividing equation $(iii)$ by $(ii),$ we obtain
$\frac{2.88\times10^{-1}}{7.2\times10^{-2}}=\frac{\text{k}[0.3]^\text{x}[0.4]^\text{y}}{\text{k}[0.3]^\text{x}[0.2]^\text{y}}$
$4=\Big(\frac{0.4}{0.2}\Big)^\text{y}$
$4 = 2^y$
$2^2 = 2^y$
$y = 2$
Therefore, the rate law is
Rate $= k[A] [B]^2$
$\text{k}=\frac{\text{Rate}}{\text{[A][B]}^2}$
From experiment $I,$ we obtain
$\text{k}=\frac{6.0\times10^{-3}\ \text{mol L}^{-1}\text{min}^{-1}}{(0.1\ \text{mol L}^{-1})(0.1\ \text{mol L}^{-1})^2}$
$= 6.0\ L^2\ mol^{-2} min^{-1}$
From experiment III, we obtain
$\text{k}=\frac{2.88\times10^{-1}\ \text{mol L}^{-1}\text{min}^{-1}}{(0.3\ \text{mol L}^{-1})(0.4\ \text{mol L}^{-1})^2}$
$= 6.0\ L^2\ mol^{-2} min^{-1}$
From experiment IV, we obtain
$\text{k}=\frac{2.40\times10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}}{(0.4\ \text{mol L}^{-1})(0.1\ \text{mol L}^{-1})^2}$
$= 6.0 L^2 mol^{-2} min^{-1}$
Therefore, rate constant$, k = 6.0 \ L^2\ mol^{-2} min^{-1}$
View full question & answer→Question 553 Marks
Explain the difference between instantaneous rate of a reaction and average rate of a reaction.
AnswerAverage rate of a reaction is the change in concentration of reactants or products and the time taken for that change to occur.
Average rate $=-\frac{\Delta[\text{R}]}{\Delta\text{t}}=+\frac{\Delta[\text{P}]}{\Delta\text{t}}$
It occurs for a long interval of time. It can be determined for multistep as well as elementary reactions.
Instantaneous rate is obtained when we consider the average rate at the smallest time interval dt (i.e., when $\Delta\text{t}$ approaches zero). Hence, for an infinitesimally small dt intantaneous rate is given by:
Intantaneous rate $=-\frac{\text{d[R]}}{\text{dt}}=+\frac{\text{d[P]}}{\text{dt}}$
It occurs for a short span of time.
It cannot be determined for multistep and elementry reaction.
View full question & answer→Question 563 Marks
For a first order reaction, show that time required for $99\%$ completion is twice the time required for the completion of $90\%$ of reaction.
AnswerFor a first order reaction, the time required for $99\%$ completion is
$\text{t}_1=\frac{2.303}{\text{k}}\text{log}\frac{100}{100-99}$
$=\frac{2.303}{\text{k}}\text{log}{100}$
$=2\times\frac{2.303}{\text{k}}$
For a first order reaction, the time required for $90\%$ completion is
$\text{t}_2=\frac{2.303}{\text{k}}\text{log}\frac{100}{100-99}$
$=\frac{2.303}{\text{k}}\text{log}{10}$
$=\frac{2.303}{\text{k}}$
Therefore$, t_1 = 2t_2$
Hence, the time required for $99\%$ completion of a first order reaction is twice the time required for the completion of $90\%$ of the reaction.
View full question & answer→Question 573 Marks
The decomposition of $A$ into product has value of $k$ as $4.5 \times 10^3 s^{–1}$ at $10^\circ C$ and energy of activation $60\ kJ\ mol^{–1}$. At what temperature would $k$ be $1.5 \times 10^4s^{–1}$?
AnswerFrom Arrhenius equation, we obtain $\text{log}\frac{\text{k}_2}{\text{k}_1}=\frac{\text{E}_\text{a}}{2.303\text{R}}\bigg(\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg)$
Also$, k_1 = 4.5 \times 10^3 s^{-1}$
$T_1 = 273 + 10 = 283 K$
$k_2 = 1.5 \times 104 s^{-1}$
$E_a = 60\ kJ\ mol^{-1} = 6.0 \times 10^4\ J\ mol^{-1}$
Then$, \log\frac{1.5\times10^4}{4.5\times10^3}=\frac{6.0\times10^4\text{Jmol}^{-1}}{2.303\times8.314\ \text{JK}^{-1}\text{mol}^{-1}}\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)$
$0.5229 = 3133.6279 \bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)$
$\frac{0.5229\times283\text{T}_2}{3133.627}=\text{T}_2-283$
$0.0472 T_2 = T_2 - 283$
$0.9528T_2 = 283$
$T_2\ 297.019\ K\ ($approximately$)$
$= 297 K$
$= 24^\circ C$
Hence$, k$ would be $1.5 \times 10^4 s^{-1}$ at $24^\circ C$.
View full question & answer→Question 583 Marks
The following data were obtained during the first order thermal decomposition of $\ce{N2O5(g)}$ at a constant volume:
$\ce{2N2O5(g) \rightarrow 2N2O4(g) + O2(g)}$
| S. No. |
$\text{Time/s}$ |
$\text{Total pressure/atm}$ |
| $1.$ |
$0$ |
$0.5$ |
| $2.$ |
$100$ |
$0.512$ |
Calculate the rate constant. AnswerLet the pressure of $\ce{N2O5(g)}$ decrease by $2x\ \text{atm}$. As two moles of $\ce{N2O5}$ decompose to give two moles of $\ce{N2O4(g)}$ and one mole of $\ce{O2(g)},$ the pressure of $\ce{N2O4(g)}$ increases by $2x\ \text{atm}$ and that of $\ce{O2(g)}$ increases by $x\ \text{atm}$.
| $\ce{2N2O(g) \rightarrow 2N2O4(g) + O2(g)}$ |
| At $t= 0$ |
$0.5\ \text{atm}$ |
$0\ \text{atm}$ |
$0\ \text{atm}$ |
| At time $t$ |
$(0.5 - 2x)\ \text{atm}$ |
$2x\ \text{atm}$ |
$X\ \text{atm}$ |
$\text{p}_\text{t}=\text{PN}_2\text{O}_5+\text{PN}_2\text{O}_4+\text{PO}_2$
$=(0.5-2\text{x})+2\text{x}+\text{x}=0.5+\text{x}$
$\text{x}=\text{p}_\text{t}-0.5$
$\text{PN}_2\text{O}_5=0.5-2\text{x}=0.5-2(\text{p}_\text{t}-0.5)=1.5-2\text{p}_\text{t}$
$= (0.5 – 2 \text{x}) + 2 \text{x} + \text{x} = 0.5 + \text{x}$
$\text{x = pt} – 0.5$
$\text{PN}_2\text{O}_5=0.5-2\text{x}=0.5-2(\text{p}_\text{t}-0.5)=1.5-2\text{p}_\text{t}$
$\text{At t}=100\text{ s};\text{p}_\text{t}=0.512\ \text{atm},$
$\text{PN}_2\text{O}_5=1.5-2\times0.512=0.476\ \text{atm}$
Thus, $\text{k}=\frac{2.303}{\text{t}}\log\frac{\text{p}_\text{i}}{\text{p}_\text{A}}$
$=\frac{2.303}{100\text{ s}}\log\frac{0.5\text{ \text{atm}}}{0.476\ \text{atm}}$
$=\frac{2.303}{100\text{ s}}\times0.0216=4.98\times10^{-4}\text{s}^{-1}$ View full question & answer→Question 593 Marks
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
AnswerThe rate constant of reaction increases with increase of temperature. This increase is generally two fold to five fold for $10$ Krise in temperature. This is explained on the basis of collision theory. The main parts of collision theory are as follows:
- For a reaction to occur, there must be collision between the reacting species.
- Only a certain fraction of total collisions are effective in forming the products.
- For effective collisions, the molecule must possess the sufficient energy $($equal or greater than threshold energy$)$ as well as proper orientation.
On the basis above conclusions, the rate of reaction is given by
Rate $= f × 2 ($where $f$ is the effective collisions and is total number of collisions per unit volume per second$)$
Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant $k$ was proposed by Arrhenius. The equation, called Arrhenius equation is usually written in the form,
$\text{K}=\text{Ae}^{\text{-E}_\text{a}/\text{RT}}\ ....(\text{i})$
where $A$ is a constant called frequency factor (because it gives the frequency of binary collisions of the reacting molecules per second per litre, $E_0$ is the energy of activation, Risa gas constant and Tis the absolute temperature. The factor $e^{-E_a/RT}$ gives the fraction of molecules having energy equal to or greater than the activation energy, $Ea.$
The energy of activation $(E_a)$ is an important quantity and it is characteristic of the reaction. Using the above equation, its value can be calculated.
Taking logarithm or both sides of equation $(i),$ we get,
$\text{In k}=\text{In A}-\frac{\text{E}_\text{a}}{\text{RT}_1}$
If the value of the rate constant at temper$-$atures $T_1$ and $T_2$ are $k_1$ and $k_2$ respectively, then we have
$\text{In k}_1=\text{In A}-\frac{\text{E}_\text{a}}{\text{RT}_1}\ ....(\text{ii} )$
$\text{In k}_2=\text{In A}-\frac{\text{E}_\text{a}}{\text{RT}_2}\ ....(\text{iii} )$
Subtracting eqn. $(i)$ from eqn. $(ii),$ we get
$\text{In k}_2=\text{In k}_1=\frac{\text{-E}_\text{a}}{\text{RT}_2}+\frac{\text{E}_\text{a}}{\text{RT}_1}$
$=\frac{\text{E}_\text{a}}{\text{RT}_1}+\frac{\text{E}_\text{a}}{\text{RT}_2}$
$\text{or}\ \ \ \text{In}\frac{\text{k}_2}{\text{k}_1}=\frac{\text{E}_\text{a}}{\text{R}}+\bigg(\frac{1}{\text{T}_1}-\frac{1}{\text{T}_2}\bigg)$
$=\frac{\text{E}_\text{a}}{\text{R}}\bigg(\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg)$
$\text{or}\ \ \ \text{log}\frac{\text{k}_2}{\text{k}_1}=\frac{\text{E}_\text{a}}{\frac{2.303}{\text{R}}}\bigg(\frac{\text{T}_2-\text{T}_1}{\text{T}_1\text{T}_2}\bigg)$
Thus knowing the values of the constant $k_1$ and $k_2$ at two different temperature $T_1$ and $T_2,$ the value of $E_a$ can be calculated. View full question & answer→Question 603 Marks
The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$.
Fill in the blanks in the following table:
| Experiment |
$[A]/mol\ L^{-1}$ |
$[B]/mol\ L^{-1}$ |
$\text{Initial rate/mol L}^{-1} min^{-1}$ |
| $I$ |
$0.1$ |
$0.1$ |
$2.0 \times 10^{-2}$ |
| $II$ |
$-$ |
$0.2$ |
$4.0 \times 10^{-2}$ |
| $III$ |
$0.4$ |
$0.4$ |
$-$ |
| $IV$ |
$-$ |
$0.2$ |
$2.0 \times 10^{-2}$ |
AnswerThe given reaction is of the first order with respect to $A$ and of zero order with respect to $B$.
Therefore, the rate of the reaction is given by,
Rate $= k[A]^1 [B]^0$
Rate $= k[A]$
From experiment $I,$ we obtain
$2.0 \times 10^{-2}\ mol\ L^{-1} min^{-1} = k(0.1\ mol\ L^{-1})$
$k = 0.2\ min^{-1}$
From experiment $II,$ we obtain
$4.0 \times 10^{-2}\ mol\ L^{-1} min^{-1} = 0.2\ min^{-1} [A]$
$[A] = 0.2\ mol\ L^{-1}$
From experiment $III,$ we obtain
Rate $= 0.2\ min^{-1} \times 0.4\ mol\ L^{-1}$
$= 0.08\ mol\ L^{-1} min^{-1}$
From experiment $IV,$ we obtain
$2.0 \times 10^{-2}\ mol\ L^{-1}\ min^{-1} = 0.2\ min^{-1}[A]$
$[A] = 0.1\ mol\ L^{-1}$
View full question & answer→Question 613 Marks
The following data were obtained during the first order thermal decomposition of $\ce{SO2Cl2}$ at a constant volume.
$\ce{SO2Cl2 (g) \rightarrow SO2 (g) + Cl2 (g)}$
| Experiment |
$\ce{Time/s}^{-1}$ |
$\ce{Total pressure/atm}$ |
| $1$ |
$0$ |
$0.5$ |
| $2$ |
$100$ |
$0.6$ |
Calculate the rate of the reaction when total pressure is $0.65\ \ce{atm}.$ AnswerThe thermal decomposition of $\ce{SO2Cl2}$ at a constant volume is represented by the following equation.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{SO}_{2}\text{Cl}_{2}\text{(g)} \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\text{SO}_{2\text{(g})} + \text{Cl}_{2\text{(g)}}$
$\ \ \text{At t= 0 }\ \ \ \ \ \ \ \ \ \ \ \ \text{p}_{0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ 0 $
$\text{At t= 0 }\ \ \ \ \ \ \ \ \ \ \ \ \text{p}_{0-\text{p}}\ \ \ \ \ \ \text{p} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{p}\ \ $
After time$, t,$ total pressure$, P_t = (P_0 - p) + p + p$
$P_t = P_0 + p$
$p =P_t - P_0$
Therefore,
$P_0 - p = P_0 - (P_t - P_0)$
$= 2P_0 - P_t$
For a first order reaction,
$\text{k}=\frac{2.303}{\text{t}}\text{log}\frac{\text{P}_\circ}{\text{P}_\circ-\text{p}}$
$=\frac{2.303}{\text{t}}\text{log}\frac{\text{P}_\circ}{2\text{P}_\circ-\text{p}_\text{t}}$
$\text{When t}=100\ \text{s, k}=\frac{2.303}{100\ \text{s}}\text{log}\frac{0.5}{2\times0.5-0.6}$
$= 2.231 \times 10^{-3} s^{-1}$
When $P_t = 0.65 \ \text{atm},$
$P_0 + p = 0.65$
$p = 0.65 - P_0$
$= 0.65 - 0.5$
$= 0.15 \ \text{atm}$
Therefore, when the total pressure is $0.65 \ \text{atm}$, pressure of $\ce{SOCl2}$ is
$\ce{P_{SOCl2}} = P_0 - p$
$= 0.5 - 0.15$
$= 0.35 \ \text{atm}$
Therefore, the rate of equation, when total pressure is $0.65 \ \text{atm},$ is given by,
Rate $= k(\ce{P_{SOCl2}})$
$= (2.23 \times 10^{-3} s^{-1})(0.35 \ \text{atm})$
$= 7.8 \times 10^{-4} \ \text{atm} \ s^{-1}$
View full question & answer→Question 623 Marks
For a reaction $\text{A + B}\xrightarrow{\ \ \ \ \ }\text{Products},$ the rate law is - $\text{Rate}=\text{k}[\text{A}][\text{B}]^\frac{3}{2}$ Can the reaction be an elementary reaction? Explain.
AnswerDuring an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is $\frac{3}{2}.$ This indicates that the reaction is not an elementary reaction. Moreover for elementary reactions. the reaction order, the molecularity and the stoichiometric coefficient are the same, although only numerically, because they are different concept.
View full question & answer→Question 633 Marks
Nitric oxide, $NO$, reacts with oxygen to produce nitrogen dioxide.
$\text{2NO(g)+O}_2(\text{g)}\rightarrow\text{2NO}_2\text{(g)}$
The rate law for this reaction is:
$\text{Rate = k}[\text{NO]}^2[\text{O]}_2$
Propose a machanism for the reaction.
AnswerThe probable proposed mechanism may be, $\text{NO+O}_2\rightarrow \text{NO}_3 \text{ (fast)}\ \dots\text{Step I}\\ $ $\text{NO}_3+ \text{NO }\rightarrow\text{NO}_2+\text{NO}_2 \text{ (slow)}\ \dots\text{Step II}{}$Since slowest reaction is the rate determining step, therefore
$\text{Rate = k}_1[\text{NO}_3]\text{ [NO]}$ $\text{k}=\frac{\text{|NO}_3|}{|\text{NO| |O}_2|}$ $[\text{NO}_3]=\text{k[NO] [O}_2]$ $\text{Rate = k}_1\text{K[NO] [O}_2]\text{ [NO]}=\text{k[NO]}^2[\text{O}_2]$, where $K =k_1.K$
View full question & answer→Question 643 Marks
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with $t_{1/2} = 3.00$ hours. What fraction of sample of sucrose remains after $8$ hours?
AnswerFor a first order reaction, $k =\frac{2.303}{\text{t}}\text{log}\frac{\text{[R]}_0}{\text{[R]}}$
It is given that, $t_{1/2 }= 3.00$ hours
$\text{Therefore, k}=\frac{0.693}{\text{t}_{1/2}}$
$=\frac{0.693}{3}\text{h}^{-1}$
$= 0.231 h^{-1}$
$\text{Then},0.231\text{h}^{-1}=\frac{2.303}{\text{8h}}\text{log}\frac{\text{[R]}_0}{\text{[R]}}$
$=\text{log}\frac{\text{[R]}_0}{\text{[R]}}=\frac{0.231\ \text{h}^{-1}\times8\text{h}}{\text{2.303}}$
$\frac{\text{[R]}_0}{\text{[R]}}=\text{anti log}(0.8024)$
$\frac{\text{[R]}_0}{\text{[R]}}=6.3445$
$\frac{\text{[R]}_0}{\text{[R]}}=0.1576\ (\text{approx})$
$= 0.158$
Hence, the fraction of sample of sucrose that remains after $8$ hours is $0.158.$
View full question & answer→Question 653 Marks
The rate of reaction, $\ce{2NO + Cl_{2 }\rightarrow 2NOCl}$ is doubled when concentration of $Cl_2$ is doubled and it becomes eight times when concentration of both $NO$ and $Cl_2$ are doubled. Deduce the order of the reaction.
AnswerLet $r = k [NO]x [Cl_2] y …(i)$
$2r = k [NO]x [2Cl_2] y …(ii)$
$8r = k [2NO]x [2Cl_2] y …(iii)$
Dividing $(iii)$ by $(ii),$
we get$\frac{8\text{r}}{2\text{r}}= \frac{\text{k|2 NO|}^{\text{x}} |2 \text{Cl}_2|^{\text{y}}}{\text{k | NO |}^\text{x}\text{ |2 Cl}| ^\text{y}}$
$2^2=[2]^\text{x}$
$\text{X}=2$
Putting the value of $x$ in $(in) (i)$ and $(ii)$ ,we get
$\text{r = k[NO]}^2\text{[Cl}_2|^\text{y}$
$2\text{r = k[NO]}^2\text{ [2Cl}_2]^\text{y}$
$\frac{2\text{r}}{\text{r}}=\frac{|\text{2 Cl}_2|^\text{y}}{|\text{Cl}_2|^\text{y}}$
$2 = [2]^y$
$y = 1$
Rate$= k [NO]^2[Cl2]^1$^
Overall order of reaction $= x + y = 2+1 = 3$
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Consider a certain reaction $A \rightarrow$ Products with $k = 2.0 \times 10^{–2}s^{–1}$. Calculate the concentration of A remaining after $100 s$ if the initial concentration of A is $1.0 \ mol \ L^{–1}.$
Answer$k =2.0 \times 10^{-5} s^{-1}$
$T = 100 s$
$[A]_0 1.0 \ mol \ L^{-1}$
Since the unit of k is $s^{−1},$ the given reaction is a first order reaction.
Therefore, k $=\frac{2.303}{\text{t}}\text{log}\frac{\text{[A]}_0}{\text{[A]}}$
$2.0\times10^{-2}\text{s}^{-1}=\frac{2.303}{\text{log s}}\text{log}\frac{1.0}{\text{[A]}}$
$2.0\times10^{-2}\text{s}^{-1}=\frac{2.303}{\text{log s}}(\text{-log}{\text{[A]}})$
$-\text{log [A]}=\frac{2.0\times10^{-2}\times100}{2.303}$
$\text{[A]}=\text{anti log}\bigg(-\frac{2.0\times10^{-2}\times100}{2.303}\bigg)$
$= 0.135 \ mol \ L^{−1} ($approximately$)$
Hence, the remaining concentration of $A$ is $0.135 \ mol \ L^{-1}$
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What is a secondary cell or battery? Explain the charging mechanism of over all reaction in a lead storage battery.
AnswerSecondary Cell
A secondary cell is an electrochemical cell that can be recharged after use by passing an electric current through it in the opposite direction. Unlike primary cells, the chemical reactions in secondary cells are reversible, allowing them to be reused multiple times.
Charging Mechanism of Lead Storage Battery
When the battery is connected to an external DC source for charging, it acts as an electrolytic cell. The chemical reactions that occurred during discharge are reversed.
Overall Charging Reaction:
$2 PbSO_4(s)+2 H_2 O(l) \xrightarrow{\text { Electrolysis }} Pb(s)+PbO_2(s)+2 H_2 SO_4(aq)$
1. At Cathode: The Lead sulphate ( $PbSO _4$ ) deposited on the negative electrode is reduced back to metallic Lead $(P b)$.
$PbSO_4(s)+2 e^{-} \rightarrow Pb(s)+SO_4^{2-}(a q)$
2. At Anode: The Lead sulphate on the positive electrode is oxidized back to Lead dioxide ( $PbO _2$ ).
$PbSO _4(s)+2 H _2 O \rightarrow PbO _2(s)+ SO _4^{2-}(a q)+4 H ^{+}+2 e^{-}$
As a result, the concentration of sulphuric acid ( $H _2 SO _4$ ) increases, restoring the battery to its original state.
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Draw a diagram of galvanie cell and explain the following reaction
$Zn ( s )+2 Ag ^{+}( aq ) \longrightarrow Zn ^{2+}( aq )+2 Ag ( s )$ which electrode is electronegative in this reaction? what are the charge carriers in a cell? write the reactions occurring at each electrode.
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