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Chemical Kinetics

JEE Chemistry · 84 questions · Page 5 of 9 · Click an option or "Show Solution" to reveal answer

Q41
The reaction 2X → \to B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be:
A 18.0 h
B 9.0 h
C 7.2 h
D 12.0 h
Correct Answer
Option A
Solution

For zero order reaction, t1/2 =

a02k{{{a_0}} \over {2k}}

⇒\Rightarrow k =

a02t1/2{{{a_0}} \over {2{t_{1/2}}}}

=

0.22×6{{0.2} \over {2 \times 6}}

= 1.67 ×\times 10-2 mol L–1h–1 For zero order reaction, A0 - At = kt ⇒\Rightarrow 0.5 - 0.2 = 1.67 ×\times 10-2 t ⇒\Rightarrow t =

0.31.67×10−2{{0.3} \over {1.67 \times {{10}^{ - 2}}}}

= 18 h

Q42
For the reaction, 2A + B → \to products, when the concentrations of A and B both were doubled, the rate of the reaction increased from 0.3 mol L−-1s−-1 to 2.4 mol L−-1s−-1. When the concentration of A alone is doubled, the rate increased from 0.3 mol L−-1s−-1 to 0.6 mol L−-1s−-1.
A Total order of the reaction is 4
B Order of the reaction with respect to B is 2
C Order of the reaction with respect to B is 1
D Order of the reaction with respect to A is 2
Correct Answer
Option B
Solution
r=K[A]x[B]yr = K{\left[ A \right]^x}{\left[ B \right]^y}
⇒8=23=2x+y\Rightarrow 8 = {2^3} = {2^{x + y}}
⇒x+y=3 ...(1)\Rightarrow x + y = 3\,...(1)
⇒2=2x\Rightarrow 2 = {2^x}
⇒x=1,y=2\Rightarrow x = 1,y = 2

Order w.r.t. A = 1 Order w.r.t. B = 2

Q43
A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B(in s) : (Use ln 2 = 0.693)
A 180
B 120
C 300
D 900
Correct Answer
Option D
Solution

At = A0.e-k1t Bt = B0.e-k2t k1 =

ln⁡2300{{\ln 2} \over {300}}

k2 =

ln⁡2180{{\ln 2} \over {180}}

Given, A0 = B0 and At and Bt are related as [A] = 4[B] ∴\therefore A0.e-k1t = 4B0.e-k2t ⇒\Rightarrow

e(ln⁡2180−ln⁡2300)t{e^{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}}

= 4 ⇒\Rightarrow

(ln⁡2180−ln⁡2300)t{\left( {{{\ln 2} \over {180}} - {{\ln 2} \over {300}}} \right)t}

= ln 4 = 2ln 2 ⇒\Rightarrow

(1180−l300)t{\left( {{1 \over {180}} - {{\mathop{\rm l}\nolimits} \over {300}}} \right)t}

= 2 ⇒\Rightarrow t =

2×180×300120{{2 \times 180 \times 300} \over {120}}

= 900 sec

Q44
The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is :
A + 6RT
B – 6 (2.303)RT
C – 6RT
D + 6(2.303)RT
Correct Answer
Option B
Solution

The rate constant of a reaction without catalyst is

k=Ae−EaRTk = A{e^{ - {{{E_a}} \over {RT}}}}

The rate constant in presence of catalyst is given by

k′=Ae−Ea′RTk' = A{e^{ - {{E{'_a}} \over {RT}}}}

∴\therefore

k′k=e−(Ea′−Ea)RT{{k'} \over k} = {e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}

⇒\Rightarrow 106 =

e−(Ea′−Ea)RT{e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}

⇒\Rightarrow ln 106 =

−(Ea′−Ea)RT{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}

⇒\Rightarrow

Ea′−Ea{E{'_a} - {E_a}}

= -RTln 106 ⇒\Rightarrow

Ea′−Ea{E{'_a} - {E_a}}

= -6RT×\times2.303

Q45
For the reaction 2H2(g) + 2NO(g) → \to N2(g) + 2H2O(g) the observed rate expression is, rate = Kf[NO]2[H2]. The rate expression for the reverse reaction is :
A Kb[N2][H2O]
B Kb[N2][H2O]2/[H2]
C Kb[N2][H2O]2/[NO]
D Kb[N2][H2O]2
Correct Answer
Option B
Solution

Keq =

kfkb{{{k_f}} \over {{k_b}}}

=

[N2][H2O]2[H2]2[NO]2{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {{{\left[ {{H_2}} \right]}^2}{{\left[ {NO} \right]}^2}}}
kf[H2][NO]2=kb[N2][H2O]2[H2]{k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2} = {k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}

At equilibrium rf = rb Given rf =

kf[H2][NO]2{k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2}

∴\therefore rb =

kb[N2][H2O]2[H2]{k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}
Q46
For kinetic study of the reaction of iodide ion with H2O2\mathrm{H}_{2} \mathrm{O}_{2} at room temperature : (A) Always use freshly prepared starch solution. (B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution. (C) Record the time immediately after the appearance of blue colour. (D) Record the time immediately before the appearance of blue colour. (E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution. Choose the correct answer from the options given below :
A (A),(B),(C)(\mathrm{A}),(\mathrm{B}),(\mathrm{C}) only
B (A),(D),(E)(\mathrm{A}),(\mathrm{D}),(\mathrm{E}) only
C (D),(E)(\mathrm{D}),(\mathrm{E}) only
D (A),(B),(E)(\mathrm{A}),(\mathrm{B}),(\mathrm{E}) only
Correct Answer
Option A
Solution

To minimize contamination, use freshly prepared starch solution to determine end point.

As KI\mathrm{KI} is used in excess to consume all the H2O2\mathrm{H}_{2} \mathrm{O}_{2} the concentration of sodium thiosulphate solution is less than KI\mathrm{KI} solution.

After appearance of blue colour record the time immediately.

Q47
Half life of zero order reaction A→\mathrm{A} \rightarrow product is 1 hour, when initial concentration of reactant is 2.0 mol L−12.0 \mathrm{~mol} \mathrm{~L}{ }^{-1}. The time required to decrease concentration of A from 0.50 to 0.25 mol L−10.25 \mathrm{~mol} \mathrm{~L}^{-1} is :
A 0.5 hour
B 15 min
C 60 min
D 4 hour
Correct Answer
Option B
Solution

For zero order reaction

 Half life =Ao2k60min⁡=22kk=160M/min⁡\begin{aligned} & \text{ Half life }=\frac{\mathrm{A}_{\mathrm{o}}}{2 \mathrm{k}} \\ & 60 \min =\frac{2}{2 \mathrm{k}} \\ & \mathrm{k}=\frac{1}{60} \mathrm{M} / \min \end{aligned}

Now

At=Ao−ktt=Ao−Atk=0.5−0.251/600.25×60t=15 min\begin{aligned} & A_t=A_o-k t \\ & t=\frac{A_o-A_t}{k} \\ &=\frac{0.5-0.25}{1 / 60} \\ & 0.25 \times 60 \\ & t=15 \mathrm{~min} \end{aligned}
Q48
Integrated rate law equation for a first order gas phase reaction is given by (where Pi\mathrm{P}_{\mathrm{i}} is initial pressure and Pt\mathrm{P}_{\mathrm{t}} is total pressure at time tt)
A k=2.303t×log⁡Pi(2Pi−Pt)k=\dfrac{2.303}{t} \times \log \dfrac{P_i}{\left(2 P_i-P_t\right)}
B k=2.303t×log⁡(2Pi−Pt)Pi\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \times \log \dfrac{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}{\mathrm{P}_{\mathrm{i}}}
C k=2.303t×Pi(2Pi−Pt)k=\dfrac{2.303}{t} \times \dfrac{P_i}{\left(2 P_i-P_t\right)}
D k=2.303t×log⁡2Pi(2Pi−Pt)\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \times \log \dfrac{2 \mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}
Correct Answer
Option A
Solution
A→B+CPi00Pi−xxx\begin{array}{llll} \mathrm{A} \rightarrow & \mathrm{B} & + & \mathrm{C} \\ \mathrm{P}_{\mathrm{i}} & 0 & & 0 \\ \mathrm{P}_{\mathrm{i}}-\mathrm{x} & \mathrm{x} & & \mathrm{x} \end{array}
Pt=Pi+xPi−x=Pi−Pt+Pi=2Pi−PtK=2.303tlog⁡Pi2Pi−Pt\begin{aligned} & \mathrm{P_t=P_i+x} \\ & \mathrm{P_i-x=P_i-P_t+P_i} \\ & \mathrm{=2 P_i-P_t} \\ & \mathrm{K=\frac{2.303}{t} \log \frac{P_i}{2 P_i-P_t}} \end{aligned}
Q49
For a reaction A→K1 B→K2CA \xrightarrow{\mathrm{K}_1} \mathrm{~B} \xrightarrow{\mathrm{K}_2} \mathrm{C} If the rate of formation of B is set to be zero then the concentration of B is given by :
A (K1−K2)[A](\mathrm{K}_1-\mathrm{K}_2)[\mathrm{A}]
B (K1+K2)[A](\mathrm{K}_1+\mathrm{K}_2)[\mathrm{A}]
C K1K2[ A]\mathrm{K}_1 \mathrm{K}_2[\mathrm{~A}]
D (K1/K2)[A](\mathrm{K}_1 / \mathrm{K}_2)[\mathrm{A}]
Correct Answer
Option D
Solution

To determine the concentration of B \mathrm{B} when the rate of formation of B \mathrm{B} is set to zero, let's consider the reaction sequence:

A→K1B→K2C\mathrm{A} \xrightarrow{\mathrm{K}_1} \mathrm{B} \xrightarrow{\mathrm{K}_2} \mathrm{C}

The rate of formation of B \mathrm{B} from A \mathrm{A} is given by:

Rate of formation of B=K1[A]\text{Rate of formation of } \mathrm{B} = \mathrm{K}_1[\mathrm{A}]

The rate of consumption of B \mathrm{B} to form C \mathrm{C} is given by:

Rate of consumption of B=K2[B]\text{Rate of consumption of } \mathrm{B} = \mathrm{K}_2[\mathrm{B}]

At steady state, where the rate of formation of B \mathrm{B} is set to zero, the rate of formation of B \mathrm{B} is equal to the rate of its consumption.

Therefore, we have:

K1[A]=K2[B]\mathrm{K}_1[\mathrm{A}] = \mathrm{K}_2[\mathrm{B}]

Solving for [B] [\mathrm{B}] , we get:

[B]=K1K2[A][\mathrm{B}] = \frac{\mathrm{K}_1}{\mathrm{K}_2} [\mathrm{A}]

Therefore, the correct concentration of B \mathrm{B} is: Option D:

(K1K2)[A]\left(\frac{\mathrm{K}_1}{\mathrm{K}_2}\right)[\mathrm{A}]
Q50
Consider an elementary reaction A( g)+B( g)→C( g)+D( g) \mathrm{A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{~g})+\mathrm{D}(\mathrm{~g}) If the volume of reaction mixture is suddenly reduced to 13\dfrac{1}{3} of its initial volume, the reaction rate will become ' x′x^{\prime} times of the original reaction rate. The value of xx is :
A 3
B 9
C 13\dfrac{1}{3}
D 19\dfrac{1}{9}
Correct Answer
Option B
Solution
R1=K[ A]1[ B]1R1=K[nA V]1[nB V]1R2=K[3nA V]1[3nB V]1R2=9R1\begin{aligned} & \mathrm{R}_1=\mathrm{K}[\mathrm{~A}]^1[\mathrm{~B}]^1 \\ & \mathrm{R}_1=\mathrm{K}\left[\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{~V}}\right]^1\left[\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{~V}}\right]^1 \\ & \mathrm{R}_2=\mathrm{K}\left[\frac{3 \mathrm{n}_{\mathrm{A}}}{\mathrm{~V}}\right]^1\left[\frac{3 \mathrm{n}_{\mathrm{B}}}{\mathrm{~V}}\right]^1 \\ & \mathrm{R}_2=9 \mathrm{R}_1 \end{aligned}
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