3 Marks Question

Question 13 Marks

The mechanism of the reaction $2 NO + Br _2 \rightarrow 2 NOBr$ is :
(i) $NO + Br _2 \rightleftharpoons NOBr _2$ (fast term)
(ii) $NOBr _2+ NO \rightarrow 2 NOBr$ (slow term)
Write the rate equation for this reaction.

Answer

On the basis of the mechanism of reaction, the rate of reaction is determined from the slow term hence,
$
\text { Rate }=k\left[NOBr_2\right][NO]
$
Since, $NOBr _2$ is not a reactant but a secondary compound and the first term is reversible, hence,
Equilibrium constant $K _{ c }=\frac{\left[ NOBr _2\right]}{[ NO ]\left[ Br _2\right]}$
or $\quad\left[ NOBr _2\right]= K _{ c }[ NO ]\left[ Br _2\right]$
Put the value in eq. (i)
$
\begin{aligned}
\text { Rate } & =k K_{c}[NO]\left[Br_2\right][NO] \\
\text { Rate } & =k^{\prime}\left[NO^2\right]^2\left[Br_2\right] \\
k^{\prime} & =k \cdot K_{c}
\end{aligned}
$
Here,
Hence, the rate equation for the reaction will be as follows:
$
\text { Rate }=k^{\prime}[NO]^2\left[Br_2\right]
$

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Question 23 Marks

(a) The molecularity of the slow term of a complex reaction is the molecularity of entire reaction. Explain.
(b) High order reactions generally do not occur, why?

Answer

(a) In the mechanism of complex reactions, the slowest term is considered as the rate determining step since the value of molecularity of any reaction is not more than 3, hence the molecularity of a reaction is determined from the molecules present in slow step even if the number of molecules is a completely balanced equation is more.
(b) While a reaction occurs, there is a collision between those molecules which are oriented in a certain direction and in complex reactions, the amount of change in the concentration of molecules in the slow step is the order of the reaction. Generally, not more than three molecules takes part in the collision, hence higher order reactions do not takes place.

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Question 33 Marks

Determine the formula for the half life of a first order reaction and prove that the half life of a first order reaction does not depends on the initial concentration of the reactant.

Answer

Rate constant for first order reaction
$
k=\frac{2.303}{t} \log \frac{[R]_0}{[R]}
$
At $t _{1 / 2},[ R ]=\frac{[ R ]_0}{2}$
On putting value of $[R]$
$
k=\frac{2.303}{t_{\frac{1}{2}}} \log \frac{[R]_0}{[R]_0 / 2}
$
or $\quad t _{\frac{1}{2}}=\frac{2.303}{ k } \log 2$
$
=\frac{2.303}{k} \times 0.3010(\because \log 2=0.3010)
$
or $\quad t _{\frac{1}{2}}=\frac{0.693}{ k }$
Therefore, the value of half life for any first order reaction is fixed. That is, it does not depend on the initial concentration of reactant.

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Question 43 Marks

Differentiate between order and molecularity of a reaction.

Answer

There are following differences in the order and molecularity of a reaction :
(1) The order of a reaction is an experimental quantity while molecularity is a theoretical quantity.
(2) Order can also be zero or fractional whereas molecularity is always an integer.
(3) The mechanism of reaction is known from the order not from molecularity.
(4) Order of a reaction is applicable for both primary and complex reactions but molecularity of reaction is applicable only for primary reactions. The molecularity of complex reactions has no meaning. In complex reactions, the order is given by the slowest term and generally, the molecularity and order of the slowest term are the same.
(5) Order depends on temperature, pressure or reaction conditions but not molecularity.
(6) Order is related to concentration whereas molecularity is related to the number of molecules.

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Question 53 Marks

A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentration of both A and B are doubled?

Answer

The reaction is of first order towards A and second order towards B hence :
(i) Differential rate equation : Rate $= k [ A ]^1[B]^2$
Hence, order of total reaction = 1 + 2 = 3
(ii) On tripling the concentration of B :
$\begin{array}{l}\text { Rate }= k [ A ]^1[3 B]^2 \\ \text { Rate }=9 k [ A ]^1[B]^2\end{array}$
Hence, rate of reaction becomes 9 times.
(iii) On doubling the concentration of both A and B :
$\begin{array}{l}\text { Rate }= k [ A ]^1[B]^2 \\ \text { Rate }= k [2 A]^1[2 B]^2 \\ \text { Rate }=8 k [ A ]^1[B]^2\end{array}$
Hence, rate of reaction becomes 8 times.

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Question 63 Marks

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Answer

Generally, by increasing the temperature, the rate of reaction increases. The rate of reaction is expressed in terms of rate constant. Therefore on increasing the temperature, the value of rate constant also increases.
When the temperature of a chemical reaction is increased by 10°C, the rate constant almost doubles.
Hence, temperature coefficient $=\frac{ k _{( t +10)}}{ k _{ t }} \approx 2$
The dependence of rate of reaction can be explained from the Arrhenius equation.
$k = Ae ^{- E _{ a } / RT }$
Here, A is Arrhenius factor or frequency factor, it is also called pre-exponential factor. It is constant for a particular reaction. Here R is gas constant and $E_{a}$ is activation energy which is expressed in J mol^-1.
An activated complex is formed between the reactant and the product, the energy required for its formation is called activation energy.

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Question 73 Marks

The rate of a reaction quadruples when the temperature changes from 293 K to 313 Κ. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Answer

According to Arrhenius equation,
$\begin{array}{l}\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right] \\ E_a=2.303 R\left[\frac{T_1 T_2}{T_2-T_1}\right] \log \frac{k_2}{k_1}\end{array}$
Given :
$\begin{array}{l}\log \frac{k_2}{k_1}=4, T_1=293 K, T_2=313 K \\ E_a=2.303 \times 8.314 \times\left[\frac{293 \times 313}{313-293}\right]\left(\log \frac{4}{1}\right) \\ E_a=19.1471 \times \frac{91709}{20} \times 0.6021 \\ E_a=19.1471 \times 4585.45 \times 0.6021 \\ E_a=52863.2 J mol ^{-1}\end{array}$
Hence, activation energy,
$E _{ a }=52.86 KJ mol ^{-1}$

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Question 83 Marks

The decomposition of NH₃ on platnium surface is zero order reaction. What are the rates of production of N₂ and H₂ if = 2.5 × 10^-4 mol L^-1 s^-1?

Answer

The reaction of decomposition of Ammonia is as follows :
$2 NH _3 \rightarrow N_2+3 H _2$
Rate of reaction $=\frac{ d \left[ N _2\right]}{ dt }=$ $ k (\text {concentration})^{\circ}$
Since order or reaction = 0
Hence, $\frac{ d \left[ N _2\right]}{ dt }=2.5 \times 10^{-4} mol L ^{-1} s^{-1} \times 1$
hence, rate of formation of N₂
$=\frac{ d \left[ N _2\right]}{ dt }=2.5 \times 10^{-4} mol L ^{-1} s^{-1}$
and rate of formation of H₂
$=\frac{ d \left[ H _2\right]}{ dt }=3 \times \frac{ d \left[ N _2\right]}{ dt }$
$=3 \times 2.5 \times 10^{-4} mol L ^{-1} s^{-1}$
$=7.5 \times 10^{-4} mol L ^{-1} s^{-1}$

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Question 93 Marks

The decomposition of hydrocarbon follows the equation
$k =\left(4.5 \times 10^{11} s^{-1}\right) e ^{-28000 K / T }$ Calculate E_a.

Answer

According to Arrhenius equation,
$k = A \cdot e ^{- E _{ a } / RT }$$\quad \quad \ldots \ldots(1)$
Given: $k =\left(4.5 \times 10^{-11} s^{-1}\right) e ^{-28000 K / T }$$\quad \quad \ldots \ldots(2)$
Comparing equation (1) and (2),
$\begin{aligned}-\frac{E_a}{R T} & =\frac{-28000 K}{T} \\ \frac{E_a}{R} & =28000 K\end{aligned}$
$\begin{array}{l}E_a=R \times 28000 K \\ E_a=8.314 JK ^{-1} mol^{-1} \times 28,000 K \\ E_a=232792 J mol ^{-1}\end{array}$
Hence, activation energy, E_a = 232.79 k J mol^-1

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Question 103 Marks

Consider a certain reaction A → Products with k = 2 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concnetration of A is 1.0 mol L^-1.

Answer

According to the question, the reaction is of first order.
Hence, $k =\frac{2.303}{ t } \log \frac{[ R ]_0}{[ R ]}$
$ k = 2.0 \times 10^{-2} s^{-1},$ $t =100 s$, ${[ R ]_0=1.0 mol L ^{-1}, [ R ]=? }$
On putting values,
$ 2.0 \times 10^{-2} =\frac{2.303}{100} \log \frac{1}{[ R ]} $
$\log \frac{1}{[ R ]} =\frac{2 \times 10^{-2} \times 100}{2.303}$
$\log \frac{1}{[R]}=\frac{2}{2.303}=0.8684$
$\frac{1}{[R]}=$ Antilog 0.8684
$\frac{1}{[ R ]}=7.386 $
${[ R ]=\frac{1}{7.386}=0.135 M }$
Therefore, after 100s, the concentration of A will remain 0.135 Μ.

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Question 113 Marks

For the reaction :
2A + B → A₂B
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.

Answer

For the reaction = 2A + B → A₂B
Initial rate = k[A][B]²
$k =2.0 \times 10^{-6} mol^{-2} L^2 s^{-1}$
Given : $[ A ]=0.1 mol L ^{-1}$ and $[ B ]=0.2 mol L ^{-1}$
Hence, initial rate , $=2.0 \times 10^{-6} \times 0.1 \times(0.2)^2$
$\begin{array}{l}=2 \times 10^{-6} \times 0.1 \times 0.04 \\ =8 \times 10^{-9} mol L ^{-1} s^{-1}\end{array}$
When the concentration of A reduces to 0.06 mol L^-1 i.e., if 0.04 mol of B reacts with 0.1 mol of A then the reaction :
According to stoichiometry,
$2 A+ B \rightarrow A _2 B$
Initial concentration A = 0.1 B = 0.2
Concentration at time t (0.1 - 0.04)[0.2 - (0.02)]
Hence, [A] = 0.06 M and [B]= 0.18M
In this case, rate of reaction = k[A][B]²
$=2 \times 10^{-6} \times 0.06 \times(0.18)^2$
Rate = 3.888 × 10^-9
$=3.89 \times 10^{-9} mol L ^{-1} s^{-1}$

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Question 123 Marks

A first order reaction takes 40 min for 30% decomposition. Calculate $t_{1 / 2}$.

Answer

The reaction is progressing at 30%
Hence, assuming [R]₀ = 1
[R] = 1 - 0.3 = 0.7 and t = 40 minute
Hence, rate constant,
$k=\frac{2.303}{t} \log \frac{[R]_0}{[R]}$
$\begin{array}{l}k=\frac{2.303}{40} \log \frac{1}{0.7} \\ k=0.05757 \log \frac{10}{7} \\ k=0.05757(\log 10-\log 7)\end{array}$
$\begin{array}{l} k =0.05757(1-0.8451) \\ k =0.05757 \times(0.1549)\end{array}$
$\begin{array}{l} k =8.917 \times 10^{-3} \text { minute }^{-1} \\ k =8.92 \times 10^{-3} \text { minute }^{-1}\end{array}$
$t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{8.92 \times 10^{-3}}$
$\begin{array}{l} t _{1 / 2}=0.07769 \times 10^3 \text { minute } \\ t _{1 / 2}=77.7 \text { minute }\end{array}$

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Question 133 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.

Answer

Time for a first order reaction,
$t =\frac{2.303}{ k } \log \frac{[ R ]_0}{[ R ]}$
After time (t) = R = 0.01 [R]₀ because 99% reaction is taking place :
Time taken to complete 99% reaction :
$t _{0.99}=\frac{2.303}{ k }=\log \frac{[ R ]_0}{0.01[ R ]_0}=\frac{2.303}{ k } \log 10^2$
On completion of 90% reaction, [R] = 0.1 [R₀]
Hence, time taken to complete 90% reaction,
$t _{0.90}=\frac{2.303}{ k }=\log \frac{[ R ]_0}{0.1[ R ]_0}=\frac{2.303}{ k } \log 10$
Hence, $\frac{t_{0.99}}{t_{0.90}}=\frac{2.303}{k} \log 10^2 \times \frac{k}{2.303} \times \frac{1}{\log 10}$
$\frac{t_{0.99}}{t_{0.90}}=\frac{\log 10^2}{\log 10}=\frac{2}{1}$
$\frac{ t _{0.99}}{ t _{0.90}}=\frac{2}{1}$
This proves that in first order, reaction the time taken to complete 99% of reaction is double the time taken to complete 90% of the reaction.

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Question 143 Marks

The half-life for radioactive decay of ¹⁴C is 5730 yeras. An archeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

Answer

Half life, $t_{1 / 2}=5730$ year
Hence, rate constant (exponential),k or $\lambda=\frac{0.693}{ t _{1 / 2}}$
$k =\frac{0.693}{5730}=1.209 \times 10^{-4} year ^{-1}$
Since the reaction of radioactive disintegration is of first order hence the rate constant or exponent
$k=\frac{2.303}{t} \log \frac{\left[R_0\right]}{[R]}$
Since, 20% decomposition is taking place, hence ¹⁴C at t = 80%
$\left[ R _0\right]=100$ and $[ R ]=80$
hence, $t =\frac{2.303}{ k } \log \frac{100}{80}$
$t =\frac{2.303}{1.209 \times 10^{-4}} \log 1.25$
$\begin{aligned} t & =\frac{2.303}{1.209 \times 10^{-4}} \times(0.0969) \\ & =0.1845 \times 10^4=1845 \text { year }\end{aligned}$

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