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Chemical Kinetics

JEE Chemistry · 84 questions · Page 6 of 9 · Click an option or "Show Solution" to reveal answer

Q51
For a reaction, N2O5( g)→2NO2( g)+12O2( g)\mathrm{N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\dfrac{1}{2} \mathrm{O}_{2(\mathrm{~g})} in a constant volume container, no products were present initially. The final pressure of the system when 50%50 \% of reaction gets completed is
A 5/25 / 2 times of initial pressure
B 7/27 / 2 times of initial pressure
C 7/47 / 4 times of initial pressure
D 5 times of initial pressure
Correct Answer
Option C
Solution
x=P02Ptotal =P0−P02+P0+P04=74P0 Option (4) \begin{aligned} & \quad x=\frac{P_0}{2} \\ & P_{\text{total }}=P_0-\frac{P_0}{2}+P_0+\frac{P_0}{4}=\frac{7}{4} P_0 \\ & \text{ Option (4) } \end{aligned}
Q52
In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are t1t_1 and t2t_2 (s), respectively. The ratio t1/t2t_1/t_2 will be:
A 43\dfrac{4}{3}
B 32\dfrac{3}{2}
C 34\dfrac{3}{4}
D 23\dfrac{2}{3}
Correct Answer
Option D
Solution

For a first-order reaction, the integrated rate law is given by:

ln⁡[A][A]0=−kt\ln \frac{[A]}{[A]_0} = -kt

where:

[A]0[A]_0

is the initial concentration,

[A][A]

is the concentration at time

tt

,

kk

is the rate constant. Let's find the times

t1t_1

and

t2t_2

corresponding to when the concentration becomes

14[A]0\frac{1}{4}[A]_0

and

18[A]0\frac{1}{8}[A]_0

respectively. For concentration

14[A]0\frac{1}{4}[A]_0

: Substitute into the integrated rate law:

ln⁡(14)=−kt1\ln\left(\frac{1}{4}\right) = -kt_1

Recognizing that

ln⁡(14)=−ln⁡4\ln\left(\frac{1}{4}\right) = -\ln 4

, we have:

−ln⁡4=−kt1⟹t1=ln⁡4k-\ln 4 = -kt_1 \quad \Longrightarrow \quad t_1 = \frac{\ln 4}{k}

For concentration

18[A]0\frac{1}{8}[A]_0

: Substitute into the integrated rate law:

ln⁡(18)=−kt2\ln\left(\frac{1}{8}\right) = -kt_2

Recognizing that

ln⁡(18)=−ln⁡8\ln\left(\frac{1}{8}\right) = -\ln 8

, we have:

−ln⁡8=−kt2⟹t2=ln⁡8k-\ln 8 = -kt_2 \quad \Longrightarrow \quad t_2 = \frac{\ln 8}{k}

Now, to find the ratio

t1t2\frac{t_1}{t_2}

:

t1t2=ln⁡4kln⁡8k=ln⁡4ln⁡8\frac{t_1}{t_2} = \frac{\frac{\ln 4}{k}}{\frac{\ln 8}{k}} = \frac{\ln 4}{\ln 8}

We can simplify further by expressing the logarithms in terms of

ln⁡2\ln 2

:

ln⁡4=ln⁡(22)=2ln⁡2\ln 4 = \ln (2^2) = 2\ln 2
ln⁡8=ln⁡(23)=3ln⁡2\ln 8 = \ln (2^3) = 3\ln 2

Thus,

t1t2=2ln⁡23ln⁡2=23\frac{t_1}{t_2} = \frac{2\ln 2}{3\ln 2} = \frac{2}{3}

So, the ratio

t1t2\frac{t_1}{t_2}

is

23\frac{2}{3}

. This corresponds to Option D.

Q53
A(g) → B(g) + C(g) is a first order reaction. Time t ∞ Psystem Pt P∞ The reaction was started with reactant A only. Which of the following expressions is correct for rate constant k?
A k=1tln⁡p∞pt\mathrm{k}=\dfrac{1}{\mathrm{t}} \ln \dfrac{\mathrm{p}_{\infty}}{\mathrm{p}_{\mathrm{t}}}
B k=1tln⁡p∞2(p∞−pt)\mathrm{k}=\dfrac{1}{\mathrm{t}} \ln \dfrac{\mathrm{p}_{\infty}}{2\left(\mathrm{p}_{\infty}-\mathrm{p}_{\mathrm{t}}\right)}
C k=1tln⁡2(p∞−pt)pt\mathrm{k}=\dfrac{1}{\mathrm{t}} \ln \dfrac{2\left(\mathrm{p}_{\infty}-\mathrm{p}_{\mathrm{t}}\right)}{\mathrm{p}_{\mathrm{t}}}
D k=1tln⁡p∞(p∞−pt)\mathrm{k}=\dfrac{1}{\mathrm{t}} \ln \dfrac{\mathrm{p}_{\infty}}{\left(\mathrm{p}_{\infty}-\mathrm{p}_{\mathrm{t}}\right)}
Correct Answer
Option B
Solution
Pt=Po+x⇒x=Pt−Po=Pt−P∞2P∞=2Po⇒Po=P∞2k=1tln⁡PoPo−xk=1tln⁡P∞2(P∞−Pt)\begin{aligned} & \mathrm{P}_{\mathrm{t}}=\mathrm{P}^{\mathrm{o}}+\mathrm{x} \Rightarrow \mathrm{x}=\mathrm{P}_{\mathrm{t}}-\mathrm{P}^{\mathrm{o}}=\mathrm{P}_{\mathrm{t}}-\frac{\mathrm{P}_{\infty}}{2} \\ & \mathrm{P}_{\infty}=2 \mathrm{P}^{\mathrm{o}} \Rightarrow \mathrm{P}^{\mathrm{o}}=\frac{\mathrm{P}_{\infty}}{2} \\ & \mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}^{\mathrm{o}}}{\mathrm{P}^o-\mathrm{x}} \\ & \mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{P}_{\infty}}{2\left(\mathrm{P}_{\infty}-\mathrm{P}_{\mathrm{t}}\right)} \end{aligned}
Q54
For A2+B2⇌2AB\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB} Ea\mathrm{E}_{\mathrm{a}} for forward and backward reaction are 180 and 200 kJ mol−1200 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively If catalyst lowers Ea\mathrm{E}_{\mathrm{a}} for both reaction by 100 kJ mol−1100 \mathrm{~kJ} \mathrm{~mol}^{-1}. Which of the following statement is correct?
A Catalyst does not alter the Gibbs energy change of a reaction.
B The enthalpy change for the reaction is +20 kJ mol−1+20 \mathrm{~kJ} \mathrm{~mol}^{-1}.
C Catalyst can cause non-spontaneous reactions to occur.
D The enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.
Correct Answer
Option A
Solution

A2+B2⇌2AB\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}

Ef=180 kJ mol−1Eb=200 kJ mol−1ΔH=Ef−Eb=−20 kJ mol−1 In presence of catalyst : Ef=180−100=80 kJ mol−1Eb=200−100=100 kJ mol−1\begin{aligned} &\begin{aligned} & \mathrm{E}_{\mathrm{f}}=180 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{E}_{\mathrm{b}}=200 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}=-20 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}\\ &\text{ In presence of catalyst : }\\ &\begin{aligned} & \mathrm{E}_{\mathrm{f}}=180-100=80 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{E}_{\mathrm{b}}=200-100=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned}
Q55
Consider the following statements related to temperature dependence of rate constants. Identify the correct statements. A. The Arrhenius equation holds true only for an elementary homogenous reaction. B. The unit of AA is same as that of kk in Arrhenius equation. C. At a given temperature, a low activation energy means a fast reaction. D. A and Ea as used in Arrhenius equation depend on temperature. E. When Ea≫RT,A\mathrm{Ea} \gg \mathrm{RT}, \mathrm{A} and Ea become interdependent. Choose the correct answer from the options given below:
A A, C and D Only
B B, D and E Only
C B and C Only
D A and B Only
Correct Answer
Option C
Solution

Let’s examine each statement in the light of

k=A e−EaRT .k = A\,e^{-\frac{E_a}{RT}}\,.

A.

“Arrhenius holds true only for an elementary homogeneous reaction.”

• False – it’s an empirical law and is routinely used (and found valid) for complex, composite or even heterogeneous processes.

B.

“The unit of AA is the same as that of kk.”

• True – since e−Ea/RTe^{-E_a/RT} is dimensionless, AA must carry whatever units kk has.

C.

“At a given temperature, a low activation energy means a fast reaction.”

• True – smaller EaE_a makes exp⁡(−Ea/RT)\exp(-E_a/RT) larger, hence a larger kk.

D. “AA and EaE_a as used in Arrhenius depend on temperature.”

• False – in the simple Arrhenius model both are treated as constants (over modest TT-ranges).

E.

“When Ea≫RTE_a \gg RT, AA and EaE_a become interdependent.”

• False – although in some data sets you observe a correlation (compensation effect), the basic Arrhenius form still treats them as independent parameters.

Therefore the only correct statements are B and C.

Answer: B and C only.

Q56
For the non – stoichiometre reaction 2A + B →\to C + D, the following kinetic data were obtained in three separate experiments, all at 298 K. .tg .tg Initial Concentration (A) Initial Concentration (B) Initial rate of formation of C (mol L-1 s-1) 0.1 M 0.1 M 1.2 x 10-3 0.1 M 0.2 M 1.2 x 10-3 0.2 M 0.1 M 2.4 x 10-3 The rate law for the formation of C is:
A dcdt=k[A][B]2{{dc} \over {dt}} = k[A]{[B]^2}
B dcdt=k[A]{{dc} \over {dt}} = k[A]
C dcdt=k[A][B]{{dc} \over {dt}} = k[A]{[B]}
D dcdt=k[A2][B]{{dc} \over {dt}} = k[A^2]{[B]}
Correct Answer
Option B
Solution

Let rate of reaction

=d[C]t=k[A]x[B]y= {{d\left[ C \right]} \over t} = k{\left[ A \right]^x}{\left[ B \right]^y}

Now from the given data

1.2×10−3=k[0.1]x[0.1]y     ...(i)1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,\,...\left( i \right)
1.2×10−3=k[0.1]x[0.2]y     ...(ii)1.2 \times {10^{ - 3}} = k{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}\,\,\,\,\,...\left( {ii} \right)
2.4×10−3=k[0.2]x[0.1]y    ...(iii)2.4 \times {10^{ - 3}} = k{\left[ {0.2} \right]^x}{\left[ {0.1} \right]^y}\,\,\,\,...\left( {iii} \right)

Dividing equation

(i)(i)

by

(ii)(ii)
⇒    1.2×10−31.2×10−3=k[0.1]x[0.1]yk[0.1]x[0.2]y\Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {1.2 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}}}

We find,

y=0y=0

Now dividing equation

(i)(i)

by

(iii)(iii)
⇒    1.2×10−32.4×10−3=k[0.1]x[0.1]yk[0.2]x[0.1]y\Rightarrow \,\,\,\,{{1.2 \times {{10}^{ - 3}}} \over {2.4 \times {{10}^{ - 3}}}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.2} \right]}^x}{{\left[ {0.1} \right]}^y}}}

We find,

x=1x=1

Hence

d[C]dt=k[A]1[B]0{{d\left[ C \right]} \over {dt}} = k{\left[ A \right]^1}{\left[ B \right]^0}
Q57
For the reaction 2A + B → \to C, the values of initial rate at diffrent reactant concentrations are given in the table below. The rate law for the reaction is : .tg .tg [A] (mol L-1) [B] (mol L-1) Initial Rate (mol L-1s-1) 0.05 0.05 0.045 0.10 0.05 0.090 0.20 0.10 0.72
A Rate = k[A][B]2
B Rate = k[A]2[B]2
C Rate = k[A]2[B]
D Rate = k[A][B]
Correct Answer
Option A
Solution

Rate law for the reaction, 2A + B →\to C Rate law (R) = k[A]x[B]y From experiment 1 : R1 = 0.045 = k[0.05]x [0.05]y ............(i) From experiment 2 : R2 = 0.090 = k[0.1]x [0.05]y ...............(ii) From experiment 3 : R3 = 0.72 = k[0.2]x [0.1]y ...................(iii) Divide equation (ii) by equation (i),

R2R1=0.090.045=k[0.1]x[0.05]yk[0.05]x[0.05]y{{{R_2}} \over {{R_1}}} = {{0.09} \over {0.045}} = {{k{{\left[ {0.1} \right]}^x}{{\left[ {0.05} \right]}^y}} \over {k{{\left[ {0.05} \right]}^x}{{\left[ {0.05} \right]}^y}}}

⇒\Rightarrow 2 = (2)x ⇒\Rightarrow x = 1 Divide equation (iii) by equation (ii),

R3R2=0.720.09=k[0.2]x[0.1]yk[0.1]x[0.05]y{{{R_3}} \over {{R_2}}} = {{0.72} \over {0.09}} = {{k{{\left[ {0.2} \right]}^x}{{\left[ {0.1} \right]}^y}} \over {k{{\left[ {0.1} \right]}^x}{{\left[ {0.05} \right]}^y}}}

⇒\Rightarrow 8 = 2x 2y ⇒\Rightarrow 8 = 21 2y [as x = 1] ⇒\Rightarrow 2y = 4 ⇒\Rightarrow y = 2 ∴\therefore Rate law (R) = k[A]1[B]2

Q58
Reaction A(g)→2 B( g)+C(g)\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) is a first order reaction. It was started with pure A .tg .tg t/min Pressure of system at time t/mm Hg 10 160 ∞\infty 240 Which of the following option is incorrect?
A Initial pressure of A is 80 mm Hg
B The reaction never goes to completion
C Partial pressure of A after 10 minute is 40 mm Hg
D Rate constant of the reaction is 1.693 min−11.693 \mathrm{~min}^{-1}
Correct Answer
Option D
Solution
P∞=3P0=240P0=80 mm of HgKt=ln⁡(P∞−P0P∞−Pt)K×10=ln⁡(240−80240−160)K=ln⁡210=0.0693 min−1 Option (3) is incorrect \begin{aligned} &\begin{aligned} & \mathrm{P}_{\infty}=3 \mathrm{P}_0=240 \\ & \quad \mathrm{P}_0=80 \mathrm{~mm} \text{ of } \mathrm{Hg} \\ & \mathrm{Kt}=\ln \left(\frac{\mathrm{P}_{\infty}-\mathrm{P}_0}{\mathrm{P}_{\infty}-\mathrm{Pt}}\right) \\ & \mathrm{K} \times 10=\ln \left(\frac{240-80}{240-160}\right) \\ & \mathrm{K}=\frac{\ln 2}{10}=0.0693 \mathrm{~min}^{-1} \end{aligned}\\ &\text{ Option (3) is incorrect } \end{aligned}
Q59
The following results were obtained during kinetic studies of the reaction ; 2A + B → \to Products .tg .tg Experiment [A] (in mol L−-1) [b] (in mol L−-1) Initial Rate of reaction (In mol L−-1 min−-1) I 0.10 0.20 6.93 G 10−-3 II 0.10 0.25 6.93 G 10−-3 III 0.20 0.30 1.386 G 10−-2 The time (in minutes) required to consume half of A is :
A 5
B 10
C 1
D 100
Correct Answer
Option B
Solution

To determine the time required to consume half of A, we must first determine the rate law for the reaction based on the provided kinetic data.

From the data, we can write the general rate law for the reaction as:

Rate=k[A]m[B]n\text{Rate} = k [A]^m [B]^n

We can now use the experimental data to find the values of the exponents m m and n n , as well as the rate constant k k .

From Experiment I and II, we can see that the initial concentration of A remains constant at 0.10 M while the concentration of B changes.

Experiment I:Rate=k[0.10]m[0.20]n=6.93×10−3 mol L−1min−1Experiment II:Rate=k[0.10]m[0.25]n=6.93×10−3 mol L−1min−1\begin{align*} \text{Experiment I:} \quad \text{Rate} = k [0.10]^m [0.20]^n &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \\ \text{Experiment II:} \quad \text{Rate} = k [0.10]^m [0.25]^n &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \end{align*}

By comparing the rates from Experiment I and II:

[0.10]m[0.25]n[0.10]m[0.20]n=6.93×10−36.93×10−3\frac{[0.10]^m [0.25]^n}{[0.10]^m [0.20]^n} = \frac{6.93 \times 10^{-3}}{6.93 \times 10^{-3}}
[0.25]n[0.20]n=1  ⟹  (0.250.20)n=1  ⟹  n=0\frac{[0.25]^n}{[0.20]^n} = 1 \implies \left(\frac{0.25}{0.20}\right)^n = 1 \implies n = 0

Thus, the reaction is zero order with respect to B.

Next, we compare Experiment I and III to determine the order with respect to A:

Experiment I:Rate=k[0.10]m[0.20]0=6.93×10−3 mol L−1min−1Experiment III:Rate=k[0.20]m[0.30]0=1.386×10−2 mol L−1min−1\begin{align*} \text{Experiment I:} \quad \text{Rate} = k [0.10]^m [0.20]^0 &= 6.93 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \\ \text{Experiment III:} \quad \text{Rate} = k [0.20]^m [0.30]^0 &= 1.386 \times 10^{-2} \, \text{mol L}^{-1} \text{min}^{-1} \end{align*}
k[0.20]mk[0.10]m=1.386×10−26.93×10−3\frac{k [0.20]^m}{k [0.10]^m} = \frac{1.386 \times 10^{-2}}{6.93 \times 10^{-3}}
(0.200.10)m=2  ⟹  (2)m=2  ⟹  m=1\left( \frac{0.20}{0.10} \right)^m = 2 \implies (2)^m = 2 \implies m = 1

Therefore, the reaction is first-order with respect to A. Now, we can write the rate law as:

Rate=k[A]1\text{Rate} = k [A]^1

Let's calculate the rate constant k k using data from any experiment (say Experiment I):

6.93×10−3=k×0.106.93 \times 10^{-3} = k \times 0.10
k=6.93×10−30.10=6.93×10−2 min−1k = \frac{6.93 \times 10^{-3}}{0.10} = 6.93 \times 10^{-2} \, \text{min}^{-1}

To determine the half-life (time required to consume half of A) for a first-order reaction, we use the formula:

t1/2=0.693kt_{1/2} = \frac{0.693}{k}

Substituting the value of k k :

t1/2=0.6936.93×10−2=10 minutest_{1/2} = \frac{0.693}{6.93 \times 10^{-2}} = 10 \, \text{minutes}

Therefore, the time required to consume half of A is: Option B: 10

Q60
It is true that :
A A first order reaction is always a single step reaction
B A zero order reaction is a multistep reaction
C A zero order reaction is a single step reaction
D A second order reaction is always a multistep reaction
Correct Answer
Option B
Solution

Zero order reaction has complex mechanism. Zero order reaction is a multistep reaction.

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