Q: For a first-order reaction the half-life is 15\,min. The time required for the reactant concentration to fall to one-eighth of its initial value is:

45\,min. One-eighth remains after 3 half-lives, so t=3\times15=45\,min.

Q: For a first-order reaction, the ratio of the time for 99.9\% completion to the time for 90\% completion is:

3. t\propto\log\dfrac{100}{100-\%\text{completion}}; ratio =\dfrac{\log(1000)}{\log(10)}=\dfrac{3}{1}=3.

Q: For a zero-order reaction with k=0.05\,mol\,L^{-1}min^{-1} and initial concentration 0.80\,M, the concentration remaining after 6\,min is:

0.50\,M. [A]=[A]_0-kt=0.80-0.05\times6=0.50\,M.

Question 1

For a first-order reaction the half-life is 15\,min. The time required for the reactant concentration to fall to one-eighth of its initial value is:

Accepted Answer

45\,min. One-eighth remains after 3 half-lives, so t=3	imes15=45\,min.

Question 2

For a first-order reaction, the time required for 99.9\% completion is approximately how many times its half-life? (Take 2.303\log x=\ln x.)

Accepted Answer

10. t_{99.9\%}=\dfrac{2.303}{k}\log\dfrac{100}{0.1}=\dfrac{2.303	imes3}{k}=\dfrac{6.909}{k} and t_{1/2}=\dfrac{0.693}{k}, so the ratio =\dfrac{6.909}{0.693}\approx10.

Question 3

For a first-order reaction, the ratio of the time for 99.9\% completion to the time for 90\% completion is:

Accepted Answer

3. t\propto\log\dfrac{100}{100-\%	ext{completion}}; ratio =\dfrac{\log(1000)}{\log(10)}=\dfrac{3}{1}=3.

Question 4

The first-order gas-phase decomposition A(g)\rightarrow 2B(g)+C(g) starts with pure A at p_0=100\,mm\,Hg. After 10\,min the total pressure is p_t=200\,mm\,Hg. The rate constant is: (\log2=0.301)

Accepted Answer

0.0693\,min^{-1}. Here p_A=\dfrac{3p_0-p_t}{2}=\dfrac{300-200}{2}=50\,mm; k=\dfrac{2.303}{10}\log\dfrac{100}{50}=\dfrac{2.303	imes0.301}{10}=0.0693\,min^{-1}.

Question 5

For a zero-order reaction with k=0.010\,mol\,L^{-1}min^{-1} and initial concentration 0.50\,M, the time required for the concentration to fall to 0.20\,M is:

Accepted Answer

30\,min. t=\dfrac{[A]_0-[A]}{k}=\dfrac{0.50-0.20}{0.010}=30\,min.

Question 6

A zero-order reaction has k=0.030\,mol\,L^{-1}s^{-1} and initial concentration 0.24\,M. The time needed for the reactant to be completely consumed is:

Accepted Answer

8\,s. t=\dfrac{[A]_0}{k}=\dfrac{0.24}{0.030}=8\,s.

Question 7

For a zero-order reaction with k=0.05\,mol\,L^{-1}min^{-1} and initial concentration 0.80\,M, the concentration remaining after 6\,min is:

Accepted Answer

0.50\,M. [A]=[A]_0-kt=0.80-0.05	imes6=0.50\,M.

Question 8

For a first-order reaction the time constant is 	au=\dfrac{1}{k}. The fraction of the reactant that has decomposed after a time equal to two time constants (t=2	au) is: (e^{-2}=0.135)

Accepted Answer

0.865. Fraction remaining =e^{-kt}=e^{-2}=0.135, so fraction decomposed =1-0.135=0.865.

Question 9

A first-order reaction has k=2.0	imes10^{-3}\,s^{-1}. The time required for the concentration of the reactant to fall to \dfrac{1}{e} of its initial value is:

Accepted Answer

500\,s. [A]=[A]_0e^{-kt}=\dfrac{[A]_0}{e} when kt=1, so t=\dfrac{1}{k}=500\,s.

Question 10

For a first-order reaction, the fraction of the reactant remaining after 5 half-lives is:

Accepted Answer

\dfrac{1}{32}. Fraction remaining =\left(\dfrac12\right)^{5}=\dfrac{1}{32}.

Question 11

In a first-order reaction the concentration falls from 0.60\,M to 0.15\,M in 40\,min. The rate constant is: (\log2=0.301)

Accepted Answer

0.0347\,min^{-1}. k=\dfrac{2.303}{40}\log\dfrac{0.60}{0.15}=\dfrac{2.303}{40}\log4=\dfrac{2.303	imes0.602}{40}=0.0347\,min^{-1}.

Question 12

For a first-order reaction a plot of \log[A] against time t (in minutes) is a straight line of slope -2.0	imes10^{-3}\,min^{-1}. The rate constant is:

Accepted Answer

4.606	imes10^{-3}\,min^{-1}. Slope =-\dfrac{k}{2.303}, so k=2.303	imes2.0	imes10^{-3}=4.606	imes10^{-3}\,min^{-1}.

Question 13

For a zero-order reaction a plot of [A] against time is a straight line of slope -0.02\,mol\,L^{-1}s^{-1}. If the initial concentration is 0.10\,M, the half-life of the reaction is:

Accepted Answer

2.5\,s. For zero order slope =-k, so k=0.02; t_{1/2}=\dfrac{[A]_0}{2k}=\dfrac{0.10}{2	imes0.02}=2.5\,s.

Question 14

A zero-order reaction takes 10\,min for the concentration to fall from 0.20\,M to 0.10\,M. The time it then takes to fall further from 0.10\,M to 0.05\,M is:

Accepted Answer

5\,min. For zero order t=\dfrac{\Delta[A]}{k}; k=\dfrac{0.10}{10}=0.010, so the next drop needs \dfrac{0.05}{0.010}=5\,min (successive half-lives halve).

Question 15

A first-order reaction A \rightarrow 2B has an initial concentration of A equal to 0.50\,M. After one half-life, the concentration of B is:

Accepted Answer

0.50\,M. After one half-life [A]=0.25\,M, so 0.25\,M of A has reacted, giving [B]=2	imes0.25=0.50\,M.

Question 16

The first-order decomposition A(g)\rightarrow 2B(g)+C(g) begins with pure A at p_0=100\,mm\,Hg. After 50\,min the total pressure is 280\,mm\,Hg. The rate constant is:

Accepted Answer

4.606	imes10^{-2}\,min^{-1}. p_A=\dfrac{3p_0-p_t}{2}=\dfrac{300-280}{2}=10\,mm; k=\dfrac{2.303}{50}\log\dfrac{100}{10}=\dfrac{2.303	imes1}{50}=0.04606\,min^{-1}.

Question 17

A first-order reaction has k=4.606	imes10^{-3}\,s^{-1}. The time required for 99.9\% of the reactant to decompose is: (\log2=0.301)

Accepted Answer

1500\,s. t=\dfrac{2.303}{k}\log\dfrac{100}{0.1}=\dfrac{2.303	imes3}{4.606	imes10^{-3}}=1500\,s.

Question 18

For a first-order reaction, the ratio t_{7/8}:t_{3/4}:t_{1/2} (the times for 87.5\%, 75\% and 50\% completion) is:

Accepted Answer

3:2:1. t\propto\log\dfrac{1}{	ext{fraction left}}; \log8:\log4:\log2=3\log2:2\log2:\log2=3:2:1.

Question 19

A first-order reaction has k=4.0	imes10^{-3}\,s^{-1} and initial concentration 0.20\,M. The concentration remaining after a time equal to three time constants (t=3/k) is: (e^{-3}=0.0498)

Accepted Answer

9.96	imes10^{-3}\,M. [A]=[A]_0e^{-kt}=0.20	imes e^{-3}=0.20	imes0.0498=9.96	imes10^{-3}\,M.

Question 20

For the zero-order reaction 2A \rightarrow products the rate \left(=-\dfrac12\dfrac{d[A]}{dt}\right) equals k=0.02\,mol\,L^{-1}min^{-1}. Starting from [A]_0=0.50\,M, the time for [A] to fall to 0.10\,M is:

Accepted Answer

10\,min. -\dfrac{d[A]}{dt}=2k=0.04, so t=\dfrac{[A]_0-[A]}{2k}=\dfrac{0.40}{0.04}=10\,min (the stoichiometric factor of 2 must be included).

Integrated Rate Equations (Zero & First Order) — JEE Main Chemistry MCQs with Solutions

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