Question 13 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. $\left[R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.$, log $\left.4=0.602\right]$.
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 23 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 $\Delta 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 $\Delta \mathrm{G}$ is +ve , then it is coupled with some other reaction which has $\Delta \mathrm{G}$-ve and large.
$\text { e.g } 2 \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s}) \rightarrow 4 \mathrm{Fe}(\mathrm{~s})+3 \mathrm{O}_2(\mathrm{~g}) \Delta \mathrm{G}^{\circ}=\mathrm{X} \mathrm{kJmol}^{-1}$
Thus for the overall reaction to become spontaneous, this reaction is coupled with carbon monoxide oxidation reaction.
$6 \mathrm{CO}(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2 \Delta \mathrm{G}^{\circ}=-\mathrm{Y} \mathrm{kJmol}^{-1}$
Where $Y>X$. The large negative value of $\Delta G^{\circ}$ for this reaction shows that $\mathrm{Fe}_2 \mathrm{O}_3$ can be reduce to Fe by CO .
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/atm |
| 0 |
0.30 |
| 300 |
0.50 |
Calculate the rate constant
(Given: log 2 = 0.301, log 3=0.4771, log 4 =0.6021) AnswerGiven: Initial pressure, $P_o = 0.30\ atm$
$P_t = 0.50\ 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:
$SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g)$
|
Experiment
|
Time/s$^{–1}$
|
Total pressure/atm
|
|
$1$
|
$0$
|
$0.4$
|
|
$2$
|
$100$
|
$0.7$
|
Calculate the rate constant.
$($Given: $\log 4 = 0.6021, \log 2 = 0.3010).$ Answer$SO_2Cl_2 \rightarrow SO_2+Cl_2$
|
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.
$[R=8.314 J K^{-1} MOL^{-1}, \log 4=0.6021]$
AnswerGiven if rate at 293K is R thus at 313K rate becomes 4R
$\log\frac{\text{k}_{2}}{\text{k}_{1}}=\frac{\text{E}_{a}}{\text{2.303 R}}\Bigg[\frac{\text{T}_{2}-\text{T}_{1}}{\text{T}_{1}\times\text{T}_{2}}\Bigg]$
$\log\frac{\text{4R}}{\text{R}}=\frac{\text{E}_{a}}{\text{2.303 }\times{8.314}}\Bigg[\frac{\text{313}-\text{293}}{\text{293}\times\text{313}}\Bigg]$
$\log\text{4}=\frac{\text{E}_{a}}{\text{19.1471 }}\Bigg[\frac{\text{20}}{\text{91709}}\Bigg]$
$\text{0.6021}=\frac{\text{E}_{a}}{\text{19.1471 }}\Bigg[\frac{\text{20}}{\text{91709}}\Bigg]$
$\frac{\text{0.6021}\times{19.1471}\times{91709}}{\text{20 }}=\text{E}_{a}$
Ea = 52863∙2177J or 52∙863KJ.
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 $1/10th$ 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 $3.8 \times 10^{-2}s$.
View full question & answer→Question 93 Marks
For the first order thermal decomposition reaction, following data were obtained:
| $C_2H_5Cl(g)$ |
$\xrightarrow{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }}$ |
$C_2H_4(g) + HCl(g)$ |
| Time/sec |
|
Total pressure/atm |
| 0 |
|
0.30 |
| 300 |
|
0.50 |
Calculate the rate constant.
(Given : log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021) Answer$\text{k}=\frac{2.303}{\text{t}}\text { }\log\text{ }\frac{\text{p}_{\text{i}}}{2\text{p}_{\text{i}}-\text{p}_{\text{t}}}$
$=\frac{2.303}{300}\text{ }\log\text{ }\frac{0.3}{2\times0.3-0.5}$
$=\frac{2.303}{300}\text{ }\log\text{ }3$
$=\frac{2.303\times0.4771}{300}$
$= 0.0036\ atm^{-1}$ or $0.004\ atm^{-1}$ (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 $\log 2 = 0.3010 \log 4 = 0.6021)$ 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
|
|
$1$
$2$
|
$0$
$100$
|
$0.4$
$0.7$
|
Calculate the rate constant.
$($Given$: \log 4 = 0.6021, \log 2 = 0.3010)$ Answer
$SO_2Cl_2\rightarrow SO_2+ Cl_2$
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 $H_2O_2$ 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: $R = 8.314 J K^{–1} mol^{–1}$)
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 $= 177.7M^{–2} min^{–1}\times (0.25)^2 \times (0.25)M^3$ 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° 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 $0.0051\ min^{–1}$. 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 $NH_3$ on platinum surface,$\text{2NH}_{3}\text{(g)}\text{ }^{Pt}_\rightarrow\text{N}_{2}\text{(g)}+\text{3H}_{2}\text{(g)}$ 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}_\rightarrow\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°C$ to $100°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(\text{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 $27.66kJ mol^{-1}$.
View full question & answer→Question 193 Marks
The decomposition of $\mathrm{NH}_3$ on platinum surface is zero order reaction. If rate constant $(\mathrm{k})$ is $4 \times 10^{-3} \mathrm{Ms}^{-1}$, how long will it take to reduce the initial concentration of $\mathrm{NH}_3$ from $0.1 M$ to $0.064 M$.
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\text{ seconds}$
View full question & answer→Question 203 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 213 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 223 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? $[R = 8.31J K–^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 233 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 243 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 $=\frac{\text{k}}{2.303}\text{ or k}=-2.303\times\text{slope}$
$\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 253 Marks
The rate for the reaction $R→P$ is rate = k[R]. It has been shown graphically below. What is rate constant for the reaction?
AnswerFrom the graph:
$\text { Case I: Rate }=\mathrm{k}|\mathrm{~A}|$
$1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k}\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)$
$\therefore \mathrm{k}=\frac{1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}}{0.1 \mathrm{~mol} \mathrm{~L}^{-1}}=0.1 \mathrm{~s}^{-1}$
Case II:
$3 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k}\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)$
$\mathrm{k}=\frac{3 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}}{0.3 \mathrm{~mol} \mathrm{~L}^{-1}}=0.1 \mathrm{~s}^{-1}$
Hence, $K=0.1 \mathrm{~s}^{-1}$.
View full question & answer→Question 263 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?
Answeri. Threshold energy for the reaction $=300 \mathrm{KJ} \mathrm{~mol}^{-1}$.
ii. Activation energy for the forward reaction $=300-150=150 \mathrm{KJ} \mathrm{~mol}^{-1}$.
iii. Activation energy for the backward reaction $=300-100=200 \mathrm{KJ} \mathrm{~mol}^{-1}$.
iv. Enthalpy change for the forward reaction $\Delta_t \mathrm{H}=100-150=-50 \mathrm{KJ} \mathrm{~mol}^{-1}$.
View full question & answer→Question 273 Marks
The rate constant for the decomposition of ethyl iodide
$C_2H_5l(g) → C_2H_4(g)+Hl(g)$
at $600K$ is $1.60 \times 10^{-5} s^{-1}$. Its energy of activation is $209kJ/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)$$\text{k}_2=6.36\times10^{-3}\text{ s}^{-1}$
View full question & answer→Question 283 Marks
Rate constant k for first order reaction has been found to be $2.54 \times 10^{-3} s^{-1}$. Calculate its three-fourth life.
Answer$\text{t}=\frac{2.303}{\text{k}}\log \frac{\text{|R|}_0}{\text{|R|}}\dots(\text{i)}$
$\text{k}=2.54\times10^{-3}\text{s}^{-1};[\text{R}]=\frac{|R|}{4}$
Substituting these value in equation (i), we get
$\text{t}_{\frac{3}{4}}=\frac{2.303}{2.54\times10^{-3}}\log\frac{ \frac{\text{|R|}_0}{\text{|R|}_0}}{4}=0.9066\times10^3\log4$
$\text{t}_{\frac{3}{4}}=0.9066\times10^3\times0.6021\text{ s}=5.46\times10^2\text{s}.$
View full question & answer→Question 293 Marks
The activation energy for the reaction,
$2 HI(g) \rightarrow H_2 + I_2(g)$
is $209.5 kJ mol^{–1}$^ at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Answer$2HI (g) \rightarrow H + I_2(g)$
Activation energy,$ E_a = 209.5 kJ mol^{−1}$
Multiply by 1000 to convert in j
$E_a= 209500 J mol^{−1}$
Temperature, $T = 581 K$
Gas constant$, R = 8.314 JK^{−1} mol^{−1}$^
According to Arhenious equatio
$K = A e^{–Ea/RT}$
In this formula term $e^{–Ea/RT}r$epresent the number of molecules which have energy equal or more than activation energy,
Number of molecules $= e^{–Ea/RT}$
Plug the values we get
Number of molecules
$=\text{e}-\frac{209500}{8.314\times581}$
$= e^{-43.4}$^
$=\frac{1}{\text{e}^{-43.4}}$
taking antilog of we get
$= 1.47 x 10^{-19}$
View full question & answer→Question 303 Marks
$^{238}_{92}\text{U}$ Change to $^{206}_{92}\text{Pb}$ by successive radioactive decay.
A sample of urenium ore was anlaysed and found to contain $1.0\ g$ of $^{238}\text{U}$ and $0.1\ g$ of $^{206}\text{Pb}$ had accumulated due to decay of $^{238}\text{U}$, find out the age of ore. (Half- life of $^{238}\text{U}$$= 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 313 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 $1.5 \times 10^4 s^{-1}$ ?
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, $k_1 = 4.5 \times 10^3 s^{-1}$
$T_1 = 273 + 10 = 283 K$
$k_2 = 1.5 \times 104 s^{-1}$
$E_a = 60 kJ mol^{-1} = 6.0 \times 10^4 J mol^{-1}$
$\text{Then, log}\frac{1.5\times10^4}{4.5\times10^3}=\frac{6.0\times10^4\text{Jmol}^{-1}}{2.303\times8.314\ \text{JK}^{-1}\text{mol}^{-1}}\bigg(\frac{\text{T}_2-283}{283\text{T}_2}\bigg)$
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 323 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 |
$[A]/mol\ L^{-1}$ |
$[B]/mol\ L^{-1}$ |
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 333 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 343 Marks
Calculate the half-life of a first order reaction from their rate constants given below:
- $200$ s$^{–1}$
- $2$ min$^{–1}$
- $4$ 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)
- $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$
$=\frac{0.693}{2\ \text{min}^{-1}}$
$= 0.35$ min (approximately)
- $\text{Half life, t}_\frac{1}{2}=\frac{0.693}{\text{k}}$
$=\frac{0.693}{4\ \text{years}^{-1}}$
$= 0.173$ years (approximately) View full question & answer→Question 353 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 363 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 373 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)
$=\text{anti log}(\bar{1}.8929)$
$= 0.7814$µg
Therefore, $= 0.7814$µg 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 383 Marks
For the decomposition of azoisopropane to hexane and nitrogen at $543$ K, the following data are obtained.
| t (sec) |
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 393 Marks
A first order gas reaction $A_2B_2(g) → 2A(g) + 2B(g)$ 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 403 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 $= 300KJ\ mol^{-1}.$
- Activation energy for the forward reaction $= 300 - 150 = 150KJ\ mol^{-1}.$
- Activation energy for the backward reaction $= 300 - 100 = 200KJ\ mol^{-1}.$
- Enthalpy change for the forward reaction $\Delta_\text{t}\text{H} = 100 - 150 = - 50KJ\ mol^{-1}.$
View full question & answer→Question 413 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 423 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 × 3 × 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 433 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 443 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 453 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.
$\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \log 4=0.602\right]$
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 463 Marks
A solution of $H_2O_2$ when titrated against $KMnO_4$ solution at different intervals of time gave the following results:
| Time (minutes) |
$0$ |
$10$ |
$20$ |
| Volume of $KMnO_4$ (mL) |
$23.8$ |
$14.7$ |
$9.1$ |
Show that decomposition of $H_2O_2$ 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 473 Marks
The rate of reaction, $2NO + Cl_2→ 2NOCl$ 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 $= x + y = 2+1 = 3$
View full question & answer→Question 483 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 493 Marks
For the reaction:
$2A + B \rightarrow A_2B$
The rate$ = k[A][B]^2 $with $k = 2.0 \times 10^{–6} mol^{–2} L^2 s^{–1}$. Calculate the initial rate of the reaction when $[A] = 0.1 mol L^{–1}, [B] = 0.2 mol L^{–1}$. Calculate the rate of reaction after [A] is reduced to 0.06 mol $L^{–1}.$
AnswerThe initial rate of the reaction is
Rate$= k[A][B]2$
$= (2.0 \times 10^{-6} mol^{-2} L^2 s^{-1})(0.1mol L^{-1})(0.2 mol L^{-1})^2$
$= 8.0 \times 10^{-9} mool^{-2} L^2 s^{-1}$
When $[A]$ is reduced from $0.1 mol L ^{-1}$ to $0.06 mol^{-1}$, the concentration of $A$ reacted $=(0.1-0.06) mol L ^{-1}=0.004 mol L ^{-1}$
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, $[B] = (0.2 - 0.02) mol L^{-1}$
$= 0.18 mol L^{-1}$
After [A] is reduced to $0.06 mol^{-1}$, the rate of the r euction is given by,
Rate $= k[A][B]_2$
$= (2.0 \times 10-6 mol^{-2} L^2 s^{-1})(0.06 mol L^{-1})(0.18 mol L^{-1})^2$
$= 3.89 mol L^{-1} s^{-1}$
View full question & answer→Question 503 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$k =2.418 \times 10^{-5} s^{-1}$
$T = 546 K$
$E_a = 179.9kJ mol^{-1}$
$= 179.9 \times 10^3 J mol^{-1}$
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 = antilog $(12.5917)$
$= 3.9 \times 10^{12} s^{−1} $(approximately)
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