Q: The half-life of a first-order reaction is 20\,min. Its rate constant is:

3.465 \times 10^{-2}\,min^{-1}. k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{20} = 3.465\times10^{-2}\,min^{-1}.

Q: The average life \tau of a first-order reaction with k = 2 \times 10^{-3}\,s^{-1} is:

500\,s. Average (mean) life \tau = \dfrac{1}{k} = \dfrac{1}{2\times10^{-3}} = 500\,s.

Question 1

A first-order reaction has a rate constant k = 1.386 	imes 10^{-2}\,s^{-1}. Its half-life is:

Accepted Answer

50\,s. For a first-order reaction t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{1.386	imes10^{-2}} = 50\,s.

Question 2

The half-life of a first-order reaction is 20\,min. Its rate constant is:

Accepted Answer

3.465 	imes 10^{-2}\,min^{-1}. k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{20} = 3.465	imes10^{-2}\,min^{-1}.

Question 3

After 3 half-lives, the fraction of reactant remaining in a first-order reaction is:

Accepted Answer

\dfrac{1}{8}. Fraction left after n half-lives = \left(\dfrac{1}{2}\right)^n = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}.

Question 4

Which statement about the half-life of a first-order reaction is correct?

Accepted Answer

It is independent of the initial concentration of the reactant.. For a first-order reaction t_{1/2} = 0.693/k, which contains no concentration term, so it does not depend on [R]_0.

Question 5

A first-order reaction has a half-life of 10\,min. The time required for it to be 75\% complete is:

Accepted Answer

20\,min. 75\% complete leaves \dfrac{1}{4} = \left(\dfrac{1}{2}ight)^2, i.e. 2 half-lives = 2	imes10 = 20\,min.

Question 6

For a zero-order reaction with [R]_0 = 0.20\,mol\,L^{-1} and k = 0.02\,mol\,L^{-1}s^{-1}, the half-life is:

Accepted Answer

5\,s. For a zero-order reaction t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.20}{2	imes0.02} = 5\,s.

Question 7

The average life 	au of a first-order reaction with k = 2 	imes 10^{-3}\,s^{-1} is:

Accepted Answer

500\,s. Average (mean) life 	au = \dfrac{1}{k} = \dfrac{1}{2	imes10^{-3}} = 500\,s.

Question 8

A radioactive isotope has a half-life of 5\,days. The fraction of the sample remaining after 15\,days is:

Accepted Answer

\dfrac{1}{8}. 15\,days = 3 half-lives, so fraction left = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}.

Question 9

The acid hydrolysis of an ester in aqueous solution follows first-order kinetics (a pseudo-first-order reaction). This is because:

Accepted Answer

Water is present in large excess, so its concentration stays effectively constant and the rate depends only on the ester.. Although molecularity is 2, water is in huge excess so [H_2O] is nearly constant; the observed rate = k\,[ester], making it first order.

Question 10

A first-order reaction is 87.5\% complete in 90\,minutes. Its half-life is:

Accepted Answer

30\,min. 87.5\% complete leaves 12.5\% = \dfrac{1}{8} = \left(\dfrac{1}{2}\right)^3, i.e. 3 half-lives. So t_{1/2} = 90/3 = 30\,min.

Question 11

The concentration of a reactant in a first-order reaction falls from 0.80\,M to 0.10\,M in 30\,minutes. The half-life of the reaction is:

Accepted Answer

10\,min. 0.80 \rightarrow 0.40 \rightarrow 0.20 \rightarrow 0.10 is 3 half-lives in 30\,min, so t_{1/2} = 30/3 = 10\,min.

Question 12

A first-order reaction has a half-life of 69.3\,s. The fraction of the reactant remaining after 230.3\,s is: (use 2.303\log = \ln)

Accepted Answer

0.10. k = 0.693/69.3 = 0.01\,s^{-1}; \log\dfrac{[R]_0}{[R]} = \dfrac{kt}{2.303} = \dfrac{0.01	imes230.3}{2.303} = 1, so [R]_0/[R] = 10 and the fraction left = 0.10.

Question 13

A radioactive source has an activity of 8000 disintegrations per second and a half-life of 2\,hours. Its activity after 8\,hours is:

Accepted Answer

500\,dps. 8\,h = 4 half-lives, so activity = 8000	imes\left(\dfrac{1}{2}ight)^4 = 8000/16 = 500\,dps.

Question 14

A radioactive sample initially contains 6.4 	imes 10^{20} nuclei and has a half-life of 10\,min. The number of nuclei left after 40\,min is:

Accepted Answer

4 	imes 10^{19}. 40\,min = 4 half-lives, so N = 6.4	imes10^{20}	imes\left(\dfrac{1}{2}ight)^4 = \dfrac{6.4	imes10^{20}}{16} = 4	imes10^{19}.

Question 15

Three-fourths of a radioactive sample decays in 20\,minutes. Its half-life is:

Accepted Answer

10\,min. \dfrac{3}{4} decayed leaves \dfrac{1}{4} = \left(\dfrac{1}{2}\right)^2, i.e. 2 half-lives in 20\,min; t_{1/2} = 20/2 = 10\,min.

Question 16

For a zero-order reaction, if the initial concentration of the reactant is doubled, its half-life will:

Accepted Answer

become twice its original value. Zero-order t_{1/2} = \dfrac{[R]_0}{2k} \propto [R]_0; doubling [R]_0 doubles the half-life.

Question 17

Which statement correctly compares the half-lives of first- and zero-order reactions?

Accepted Answer

For a first-order reaction t_{1/2} is independent of [R]_0, whereas for a zero-order reaction t_{1/2} is directly proportional to [R]_0.. First order: t_{1/2} = 0.693/k (no [R]_0). Zero order: t_{1/2} = [R]_0/2k \propto [R]_0.

Question 18

The acid-catalysed hydrolysis of an ester (a pseudo-first-order reaction) is 50\% complete in 30\,min. The pseudo-first-order rate constant is:

Accepted Answer

2.31 	imes 10^{-2}\,min^{-1}. 50\% complete means t = t_{1/2} = 30\,min, so k = 0.693/30 = 2.31	imes10^{-2}\,min^{-1}.

Question 19

The inversion of cane sugar, C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6, is described as a pseudo-first-order reaction because:

Accepted Answer

water is in large excess so its concentration is essentially constant and the rate depends only on sucrose.. Rate = k\,[sucrose][H_2O], but [H_2O] is nearly constant (large excess), so rate = k^{\prime}[sucrose] appears first order though molecularity is 2.

Question 20

In a first-order reaction \dfrac{1}{16} of the reactant is left. If the half-life is 4\,min, the time elapsed is:

Accepted Answer

16\,min. \dfrac{1}{16} = \left(\dfrac{1}{2}ight)^4, i.e. 4 half-lives = 4	imes4 = 16\,min.

Half-Life & Pseudo First Order Reactions — NEET Chemistry MCQs with Solutions

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