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Chemical Kinetics

JEE Chemistry · 84 questions · Page 7 of 9 · Click an option or "Show Solution" to reveal answer

Q61
The differential rate law for the reaction H2 + I2 →\to 2HI is
A −d[H2]dt - {{d\left[ {{H_2}} \right]} \over {dt}} = −d[I2]dt - {{d\left[ {{I_2}} \right]} \over {dt}} = −d[HI]dt - {{d\left[ {{HI}} \right]} \over {dt}}
B d[H2]dt {{d\left[ {{H_2}} \right]} \over {dt}} = d[I2]dt {{d\left[ {{I_2}} \right]} \over {dt}} = d[HI]dt {{d\left[ {{HI}} \right]} \over {dt}}
C 12d[H2]dt{1 \over 2}{{d\left[ {{H_2}} \right]} \over {dt}} = 12d[I2]dt{1 \over 2}{{d\left[ {{I_2}} \right]} \over {dt}} = −d[HI]dt - {{d\left[ {{HI}} \right]} \over {dt}}
D −2d[H2]dt - 2{{d\left[ {{H_2}} \right]} \over {dt}} = −2d[I2]dt - 2{{d\left[ {{I_2}} \right]} \over {dt}} = d[HI]dt{{d\left[ {{HI}} \right]} \over {dt}}
Correct Answer
Option D
Solution

rate of appearance of

HI=12d[HI]dtHI = {1 \over 2}{{d\left[ {HI} \right]} \over {dt}}

rate of formation of

H2=−d[H2]dt{H_2} = {{ - d\left[ {{H_2}} \right]} \over {dt}}

rate of formation of

I2=−d[I2]dt{I_2} = {{ - d\left[ {{I_2}} \right]} \over {dt}}

hence

−d[H2]dt=−−d[I2]dt=12d[HI]dt{{ - d\left[ {{H_2}} \right]} \over {dt}} = - {{ - d\left[ {{I_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {HI} \right]} \over {dt}}

or

    \,\,\,\,
−2d[H2]dt=−2d[I2]dt=d[HI]dt- {{2d\left[ {{H_2}} \right]} \over {dt}} = - {{2d\left[ {{I_2}} \right]} \over {dt}} = {{d\left[ {HI} \right]} \over {dt}}
Q62
For a reaction of order n, the unit of the rate constant is :
A mol1−-n L1−-n s
B mol1−-n L2n s−-1
C mol1−-n Ln−-1 s−-1
D mol1−-n L1−-n s−-1
Correct Answer
Option C
Solution

Rate = k[A]n comparing units

(mol/l)sec⁡=k(moll)n{{(mol/l)} \over {\sec }} = k{\left( {{{mol} \over l}} \right)^n}
⇒k=mol(l−n)l(n−1)s−1\Rightarrow k = mo{l^{(l - n)}}{l^{(n - 1)}}{s^{ - 1}}
Q63
For a chemical reaction A+B→\mathrm{A}+\mathrm{B} \rightarrow Product, the order is 1 with respect to A\mathrm{A} and B\mathrm{B}. .tg .tg Rate\mathrm{Rate} mol L−1 S−1\mathrm{mol~L^{-1}~S^{-1}} [A]\mathrm{[A]} mol L−1\mathrm{mol~L^{-1}} [B]\mathrm{[B]} mol L−1\mathrm{mol~L^{-1}} 0.10 20 0.5 0.40 xx 0.5 0.80 40 yy What is the value of xx and yy ?
A 80 and 4
B 160 and 4
C 80 and 2
D 40 and 4
Correct Answer
Option C
Solution
r=K[A]1[ B]10.1=K(20)1(0.5)1........(i)0.40=K(x)1(0.5)1........(ii)0.80=K(40)1(y)1........(iii)\begin{aligned} & \mathrm{r}=\mathrm{K}[\mathrm{A}]^1[\mathrm{~B}]^1 \\\\ & 0.1=\mathrm{K}(20)^1(0.5)^1 ........(i) \\\\ & 0.40=\mathrm{K}(\mathrm{x})^1(0.5)^1 ........(ii) \\\\ & 0.80=\mathrm{K}(40)^1(\mathrm{y})^1 ........(iii) \end{aligned}

From (i) and (ii)

x=80x=80

From (i) and (iii)

y=2y=2
Q64
Which of the following statement is not true for radioactive decay?
A Half life is ln⁡2\ln 2 times of 1 rate constant \dfrac{1}{\text{ rate constant }}.
B Decay constant increases with increase in temperature.
C Decay constant does not depend upon temperature.
D Amount of radioactive substance remained after three half lives is 18\dfrac{1}{8} th of original amount.
Correct Answer
Option B
Solution

Answer: Option B is not true.

Explanation: Option A states that the half‐life t1/2t_{1/2} is ln⁡2\ln 2 times 1k\dfrac{1}{k}, where kk is the decay (rate) constant.

t1/2  =  ln⁡2k    =  ln⁡2×1k. t_{1/2} \;=\; \dfrac{\ln 2}{k} \;\;=\;\ln 2 \times \dfrac{1}{k}. This is correct.

Option B states that the decay constant kk increases with an increase in temperature.

However, for radioactive (nuclear) decay processes, the decay constant is essentially independent of external factors such as temperature and pressure.

Hence, saying that it increases with temperature is not true.

Option C states that the decay constant does not depend upon temperature, which is true for radioactive decay.

Option D states that the amount of radioactive substance left after three half‐lives is 18\tfrac{1}{8} of the original.

Since one half‐life leaves 12\tfrac{1}{2} of the sample, three half‐lives leave (12)3=18\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}.

This is correct.

Therefore, the statement that is not true is Option B.

Q65
A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will
A remain unchanged
B triple
C increase by a factor of 4
D double
Correct Answer
Option C
Solution

Since the reaction is

22

nd order w.r.t CO. Thus, rate law is given as.

r=k[CO]2r = k{\left[ {CO} \right]^2}

Let initial concentration of

COCO

is a i.e.

[CO]=a\left[ {CO} \right] = a

∴\therefore

    r1=k(a)2=ka2\,\,\,\,{r_1} = k{\left( a \right)^2} = k{a^2}

when concentration becomes doubled, i.e.

[CO]=2a\left[ {CO} \right] = 2a

∴\therefore

     r2=k(2a)2=4ka2\,\,\,\,\,{r_2} = k{\left( {2a} \right)^2} = 4k{a^2}

∴\therefore

     r2=4r1\,\,\,\,\,{r_2} = 4{r_1}

So, the rate of reaction becomes

44

times.

Q66
The reaction A2+B2→2ABA_2 + B_2 \rightarrow 2AB follows the mechanism:A2⇌k−1k1A+AA_2 \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} A + A (fast)A+B2→k2AB+BA + B_2 \xrightarrow{k_2} AB + B (slow)A+B→ABA + B \rightarrow AB (fast)The overall order of the reaction is:
A 3
B 2.5
C 1.5
D 2
Correct Answer
Option C
Solution

Reaction given:

A2+B2→2ABA_2+B_2 \rightarrow 2 A B

Mechanism:

A2⇌k−1k1A+AA_2\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{k_{-1}}^{k_1}} A+A

(fast)

A+B2→k2AB+B (slow) A+B⟶AB (fast) \begin{aligned} & A+B_2 \xrightarrow{k_2} A B+B \text{ (slow) } \\ & A+B \longrightarrow A B \text{ (fast) } \end{aligned}

The rate determining step in a mechanism is the slow step.

So, late =k2[A][B2]=k_2[A]\left[B_2\right] AA is the intermediate; to determine the concentration of AA, the equilibrium is considered.

A2⇌k−1k1A+AA_2\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{k_{-1}}^{k_1}} A+A
k1k−1=[A][A][A2]=[A]2[A2]\begin{aligned} \frac{k_1}{k_{-1}} & =\frac{[A][A]}{\left[A_2\right]} \\ & =\frac{[A]^2}{\left[A_2\right]} \end{aligned}
S0,[A]2=k1k−1[A2][A]={k1k−1[A2]}1/2\begin{aligned} S_{0,}[A]^2 & =\frac{k_1}{k_{-1}}\left[A_2\right] \\ {[A] } & =\left\{\frac{k_1}{k_{-1}}\left[A_2\right]\right\}^{1 / 2} \end{aligned}

Substitute [A][A] in the late, late =k2[A][B2]=k_2[A]\left[B_2\right]

=k2(k1k−1)1/2[A2]1/2[B2]=k2(k1k−1)1/2[A2]1/2[B2]1\begin{aligned} & =k_2\left(\frac{k_1}{k_{-1}}\right)^{1 / 2}\left[A_2\right]^{1 / 2}\left[B_2\right] \\ & =k_2\left(\frac{k_1}{k_{-1}}\right)^{1 / 2}\left[A_2\right]^{1 / 2}\left[B_2\right]^1 \end{aligned}
 Overall rate =12+1=32=1.5\begin{aligned} \text{ Overall rate } & =\frac{1}{2}+1 \\ & =\frac{3}{2} \\ & =1.5 \end{aligned}
Q67
The rate law for a reaction between the substances A and B is given by Rate = k[A]n [B]m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as
A (m + n)
B (n - m)
C 2( n - m)
D 12(m+n){1 \over {{2^{(m + n)}}}}
Correct Answer
Option C
Solution
Rate1=k[A]n[B]m;Rat{e_1} = k{\left[ A \right]^n}{\left[ B \right]^m};
Rate2=k[2A]n[1/2B]mRat{e_2} = k{\left[ {2A} \right]^n}{\left[ {{{1}/{2}}B} \right]^m}

∴\therefore

    Rate2Rate1=k[2A]n[1/2B]mk[A]n[B]m\,\,\,\,{{Rat{e_2}} \over {Rat{e_1}}} = {{k{{\left[ {2A} \right]}^n}{{\left[ {{{1}/{2}}B} \right]}^m}} \over {k{{\left[ A \right]}^n}{{\left[ B \right]}^m}}}
=[2]n[1/2]m=2n.2−m=2n−m= {\left[ 2 \right]^n}{\left[ {{{1}/{2}}} \right]^m} = {2^n}{.2^{ - m}} = {2^{n - m}}
Q68
In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is
A 30 minutes
B 60 minutes
C 7.5 minutes
D 15 minutes
Correct Answer
Option A
Solution

As the concentration of reactant decreases from

0.80.8

to

0.40.4

in

1515

minutes hence the

t1/2{t_{1/2}}

is

1515

minutes. To fall the concentration from

0.10.1

to

0.0250.025

we need two half lives i.e.,

3030

minutes.

Q69
At 30∘C30^{\circ} \mathrm{C}, the half life for the decomposition of AB2\mathrm{AB}_{2} is 200 s200 \mathrm{~s} and is independent of the initial concentration of AB2\mathrm{AB}_{2}. The time required for 80%80 \% of the AB2\mathrm{AB}_{2} to decompose is Given: log⁡2=0.30\log 2=0.30 log⁡3=0.48\quad \log 3=0.48
A 200 s
B 323 s
C 467 s
D 532 s
Correct Answer
Option C
Solution

Since, half-life is independent of the initial concentration of

AB2A B_{2}

. Hence, the reaction is "First Order".

k=2.303log⁡2t1/2k=\frac{2.303 \log 2}{t_{1 / 2}}
2.303log⁡2t1/2=2.303tlog⁡100(100−80)\frac{2.303 \log 2}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{100}{(100-80)}
2.303×0.3200=2.303tlog⁡5\frac{2.303 \times 0.3}{200}=\frac{2.303}{t} \log 5
t=467 st=467 \mathrm{~s}
Q70
The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2=0.301) :
A 230.3 minutes
B 23.03 minutes
C 46.06 minutes
D 460.6 minutes
Correct Answer
Option C
Solution

For first order reaction

k=2.303tlog⁡100100−99k = {{2.303} \over t}\log {{100} \over {100 - 99}}
0.6936.93=2.303tlog⁡1001{{0.693} \over {6.93}} = {{2.303} \over t}\log {{100} \over 1}
0.6936.93=2.303×2t{{0.693} \over {6.93}} = {{2.303 \times 2} \over t}
⇒t=46.06min⁡\Rightarrow t = 46.06\min
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