Half-Life & Pseudo First Order Reactions — IISER Chemistry MCQs with Solutions
Free IISER Chemistry Half-Life & Pseudo First Order Reactions MCQs with step-by-step solutions (30 questions). Part of Chemical Kinetics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Half-Life & Pseudo First Order Reactions · easy · numerical
A first-order reaction has rate constant $k = 3.465 \times 10^{-3}\,s^{-1}$. Its half-life is:
A. $200\,s$ ✓ Correct
B. $100\,s$
C. $693\,s$
D. $346.5\,s$
Solution: For a first-order reaction $t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{3.465\times10^{-3}}=200\,s$.
Q2 — Half-Life & Pseudo First Order Reactions · easy · numerical
The half-life of a first-order reaction is $25\,min$. Its rate constant is:
A. $2.77 \times 10^{-2}\,min^{-1}$ ✓ Correct
B. $1.39 \times 10^{-2}\,min^{-1}$
C. $3.47 \times 10^{-2}\,min^{-1}$
D. $2.77 \times 10^{-2}\,s^{-1}$
Solution: $k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{25}=2.77\times10^{-2}\,min^{-1}$ (units $min^{-1}$, not $s^{-1}$).
Q3 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a first-order reaction with rate constant $k = 5 \times 10^{-4}\,s^{-1}$, the average (mean) life $\tau$ is:
A. $2000\,s$ ✓ Correct
B. $1386\,s$
C. $5000\,s$
D. $1000\,s$
Solution: Average life $\tau=\dfrac{1}{k}=\dfrac{1}{5\times10^{-4}}=2000\,s$. (The half-life would be $0.693/k=1386\,s$.)
Q4 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a first-order reaction, the fraction of the reactant remaining after 6 half-lives is:
A. $\dfrac{1}{128}$
B. $\dfrac{1}{64}$ ✓ Correct
C. $\dfrac{1}{12}$
D. $\dfrac{1}{32}$
Solution: After $n$ half-lives the fraction remaining is $\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^6=\dfrac{1}{64}$.
Q5 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a first-order reaction, how does the half-life change if the initial concentration of the reactant is doubled?
A. It becomes four times larger.
B. It remains unchanged, because for a first-order reaction $t_{1/2}=\dfrac{0.693}{k}$ is independent of $[A]_0$. ✓ Correct
C. It is halved, because the half-life is inversely proportional to the initial concentration.
D. It doubles, because the half-life is directly proportional to the initial concentration.
Solution: For a first-order reaction $t_{1/2}=\dfrac{0.693}{k}$ depends only on $k$; changing $[A]_0$ leaves the half-life unchanged.
Q6 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a zero-order reaction with $[R]_0 = 0.40\,mol\,L^{-1}$ and $k = 0.01\,mol\,L^{-1}min^{-1}$, the half-life is:
A. $40\,min$
B. $20\,min$ ✓ Correct
C. $80\,min$
D. $10\,min$
Solution: For a zero-order reaction $t_{1/2}=\dfrac{[R]_0}{2k}=\dfrac{0.40}{2\times0.01}=20\,min$.
Q7 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a zero-order reaction, if the initial concentration of the reactant is tripled, its half-life will:
A. remain unchanged, since half-life is independent of the initial concentration.
B. become nine times larger.
C. become three times larger, since $t_{1/2}=\dfrac{[R]_0}{2k}$ is directly proportional to $[R]_0$. ✓ Correct
D. become one-third as large.
Solution: For a zero-order reaction $t_{1/2}=\dfrac{[R]_0}{2k}\propto[R]_0$; tripling $[R]_0$ triples the half-life.
Q8 — Half-Life & Pseudo First Order Reactions · easy · numerical
The hydrolysis of methyl formate, $HCOOCH_3 + H_2O \rightarrow HCOOH + CH_3OH$, carried out in a large excess of water, is found to be first order. This is because:
A. the reaction is elementary and unimolecular.
B. water does not take part in the rate-determining step at all.
C. the molecularity of the reaction is genuinely one.
D. water is present in such large excess that $[H_2O]$ stays effectively constant, so the rate depends only on $[ester]$ (pseudo-first order). ✓ Correct
Solution: With water in vast excess $[H_2O]$ is nearly constant and merges into the rate constant, so $rate=k^{\prime}[ester]$ is first order though the molecularity is 2.
Q9 — Half-Life & Pseudo First Order Reactions · easy · numerical
A first-order reaction has a half-life of $12\,min$. The time required for the reactant concentration to fall to one-sixteenth of its initial value is:
A. $36\,min$
B. $24\,min$
C. $60\,min$
D. $48\,min$ ✓ Correct
Solution: $\dfrac{1}{16}=\left(\dfrac{1}{2}\right)^4$, so 4 half-lives are needed: $4\times12=48\,min$.
Q10 — Half-Life & Pseudo First Order Reactions · easy · numerical
A radioactive isotope has a half-life of $6\,hours$. The fraction of the sample remaining after $1\,day$ is:
A. $\dfrac{1}{4}$
B. $\dfrac{1}{8}$
C. $\dfrac{1}{16}$ ✓ Correct
D. $\dfrac{1}{32}$
Solution: $1\,day = 24\,h = 4$ half-lives, so the fraction remaining is $\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}$.
Q11 — Half-Life & Pseudo First Order Reactions · easy · numerical
A $32\,g$ sample of a radioactive nuclide has a half-life of $15\,min$. The mass remaining after $45\,min$ is:
A. $16\,g$
B. $2\,g$
C. $4\,g$ ✓ Correct
D. $8\,g$
Solution: $45\,min = 3$ half-lives, so mass $= 32\times\left(\dfrac{1}{2}\right)^3=\dfrac{32}{8}=4\,g$.
Q12 — Half-Life & Pseudo First Order Reactions · easy · numerical
For the pseudo-first-order hydrolysis of an ester, $rate = k[ester][H_2O]$ with true rate constant $k = 2.0 \times 10^{-4}\,L\,mol^{-1}s^{-1}$ and $[H_2O] = 55\,mol\,L^{-1}$ (constant). The observed pseudo-first-order rate constant $k^{\prime}$ is:
A. $3.6 \times 10^{-6}\,s^{-1}$
B. $1.1 \times 10^{-2}\,s^{-1}$ ✓ Correct
C. $2.0 \times 10^{-4}\,s^{-1}$
D. $1.1 \times 10^{-2}\,L\,mol^{-1}s^{-1}$
Solution: $k^{\prime}=k[H_2O]=2.0\times10^{-4}\times55=1.1\times10^{-2}\,s^{-1}$ (first-order units $s^{-1}$).
Q13 — Half-Life & Pseudo First Order Reactions · medium · numerical
In a first-order reaction the concentration of the reactant falls from $0.50\,M$ to $0.0625\,M$ in $36\,min$. The half-life of the reaction is:
A. $9\,min$
B. $6\,min$
C. $18\,min$
D. $12\,min$ ✓ Correct
Solution: $\dfrac{0.50}{0.0625}=8=2^3$, a fall to $\dfrac{1}{8}$ i.e. 3 half-lives $=36\,min$, so $t_{1/2}=12\,min$.
Q14 — Half-Life & Pseudo First Order Reactions · medium · numerical
A first-order reaction is $90\%$ complete in $30\,min$. Its half-life is: ($\log 2 = 0.301$)
A. $15\,min$
B. $9.03\,min$ ✓ Correct
C. $3.01\,min$
D. $10\,min$
Solution: $k=\dfrac{2.303}{30}\log\dfrac{100}{10}=0.0768\,min^{-1}$; $t_{1/2}=\dfrac{0.693}{k}=9.03\,min$ (also $t_{1/2}=0.301\times t_{90\%}$).
Q15 — Half-Life & Pseudo First Order Reactions · medium · numerical
For a reaction $A \rightarrow$ products, the half-life is $100\,s$ when $[A]_0 = 0.20\,M$ and also $100\,s$ when $[A]_0 = 0.40\,M$. The order of the reaction is:
A. order $1.5$.
B. second order, because half-life is inversely proportional to the initial concentration.
C. zero order, because half-life is directly proportional to the initial concentration.
D. first order, because the half-life is independent of the initial concentration. ✓ Correct
Solution: Only for a first-order reaction is $t_{1/2}$ independent of $[A]_0$; since the half-life is unchanged when $[A]_0$ changes, the order is 1.
Q16 — Half-Life & Pseudo First Order Reactions · medium · numerical
For $A \rightarrow$ products the half-life is $30\,min$ when $[A]_0 = 0.10\,M$ and $45\,min$ when $[A]_0 = 0.15\,M$. The order of the reaction is:
A. zero order, since $t_{1/2}\propto[A]_0$. ✓ Correct
B. order $1.5$.
C. second order, since $t_{1/2}\propto\dfrac{1}{[A]_0}$.
D. first order, since $t_{1/2}$ is independent of $[A]_0$.
Solution: $[A]_0$ rises by a factor $1.5$ and $t_{1/2}$ also rises by $1.5$, so $t_{1/2}\propto[A]_0$, which is a zero-order reaction.
Q17 — Half-Life & Pseudo First Order Reactions · medium · numerical
A radioactive source has an activity of $6400$ disintegrations per second and a half-life of $15\,min$. Its activity after $45\,min$ is:
A. $2133\,dps$
B. $400\,dps$
C. $800\,dps$ ✓ Correct
D. $1600\,dps$
Solution: $45\,min = 3$ half-lives, so activity $=6400\times\left(\dfrac{1}{2}\right)^3=800\,dps$.
Q18 — Half-Life & Pseudo First Order Reactions · medium · numerical
A radioactive nuclide decays to $\dfrac{1}{64}$ of its initial amount in $30\,min$. Its half-life is:
A. $3\,min$
B. $5\,min$ ✓ Correct
C. $6\,min$
D. $10\,min$
Solution: $\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^6$, so 6 half-lives $=30\,min$, giving $t_{1/2}=5\,min$.
Q19 — Half-Life & Pseudo First Order Reactions · medium · numerical
The acid-catalysed hydrolysis of an ester (pseudo-first order) is $75\%$ complete in $40\,min$. The pseudo-first-order rate constant is: ($\log 2 = 0.301$)
A. $1.73 \times 10^{-2}\,min^{-1}$
B. $3.47 \times 10^{-2}\,s^{-1}$
C. $1.73 \times 10^{-2}\,s^{-1}$
D. $3.47 \times 10^{-2}\,min^{-1}$ ✓ Correct
Solution: $k^{\prime}=\dfrac{2.303}{40}\log\dfrac{100}{25}=\dfrac{2.303}{40}(0.602)=3.47\times10^{-2}\,min^{-1}$.
Q20 — Half-Life & Pseudo First Order Reactions · medium · numerical
The inversion of cane sugar is pseudo-first order with observed rate constant $k^{\prime} = 5.5 \times 10^{-3}\,min^{-1}$. If $[H_2O] = 55\,mol\,L^{-1}$ stays constant, the true second-order rate constant is:
A. $3.0 \times 10^{-1}\,L\,mol^{-1}min^{-1}$
B. $1.0 \times 10^{-4}\,min^{-1}$
C. $5.5 \times 10^{-3}\,L\,mol^{-1}min^{-1}$
D. $1.0 \times 10^{-4}\,L\,mol^{-1}min^{-1}$ ✓ Correct
Solution: $k^{\prime}=k[H_2O]\Rightarrow k=\dfrac{k^{\prime}}{[H_2O]}=\dfrac{5.5\times10^{-3}}{55}=1.0\times10^{-4}\,L\,mol^{-1}min^{-1}$.
Q21 — Half-Life & Pseudo First Order Reactions · medium · numerical
A first-order reaction has $k = 0.02\,min^{-1}$. The time required for the reactant to fall to $20\%$ of its initial concentration is: ($\log 5 = 0.699$)
A. $34.6\,min$
B. $80.5\,min$ ✓ Correct
C. $115\,min$
D. $161\,min$
Solution: $t=\dfrac{2.303}{k}\log\dfrac{100}{20}=\dfrac{2.303}{0.02}(0.699)=80.5\,min$.
Q22 — Half-Life & Pseudo First Order Reactions · medium · numerical
A first-order reaction $A \rightarrow P$ (1 : 1 stoichiometry) starts with $2.0\,mol$ of $A$. After $3$ half-lives, the amount of product $P$ formed is:
A. $0.75\,mol$
B. $1.50\,mol$
C. $1.75\,mol$ ✓ Correct
D. $0.25\,mol$
Solution: After 3 half-lives $A$ remaining $=2.0\times\left(\dfrac{1}{2}\right)^3=0.25\,mol$, so $P=2.0-0.25=1.75\,mol$.
Q23 — Half-Life & Pseudo First Order Reactions · medium · numerical
A first-order reaction is $50\%$ complete in $20\,min$. The time required for it to be $87.5\%$ complete is:
A. $30\,min$
B. $40\,min$
C. $60\,min$ ✓ Correct
D. $80\,min$
Solution: $t_{1/2}=20\,min$; $87.5\%$ complete leaves $\dfrac{1}{8}=\left(\dfrac{1}{2}\right)^3$, i.e. 3 half-lives $=60\,min$.
Q24 — Half-Life & Pseudo First Order Reactions · medium · numerical
A first-order reaction has $k = 0.0693\,min^{-1}$. The percentage of the reactant remaining after $40\,min$ is: ($\ln 2 = 0.693$)
A. $3.125\%$
B. $6.25\%$ ✓ Correct
C. $25\%$
D. $12.5\%$
Solution: $t_{1/2}=\dfrac{0.693}{0.0693}=10\,min$; $40\,min = 4$ half-lives, so remaining $=\left(\dfrac{1}{2}\right)^4=6.25\%$.
Q25 — Half-Life & Pseudo First Order Reactions · medium · numerical
For $A \rightarrow$ products the half-life is $80\,s$ when $[A]_0 = 0.05\,M$ and $40\,s$ when $[A]_0 = 0.10\,M$. The order of the reaction is:
A. order $1.5$.
B. first order, since $t_{1/2}$ is independent of $[A]_0$.
C. second order, since $t_{1/2}\propto\dfrac{1}{[A]_0}$. ✓ Correct
D. zero order, since $t_{1/2}\propto[A]_0$.
Solution: $[A]_0$ doubles while $t_{1/2}$ halves, so $t_{1/2}\propto\dfrac{1}{[A]_0}$, which is a second-order reaction.
Q26 — Half-Life & Pseudo First Order Reactions · medium · numerical
For a first-order reaction the half-life is $138.6\,s$. The average (mean) life $\tau$ is:
A. $96.1\,s$
B. $138.6\,s$
C. $200\,s$ ✓ Correct
D. $277.2\,s$
Solution: $k=\dfrac{0.693}{138.6}=5\times10^{-3}\,s^{-1}$; $\tau=\dfrac{1}{k}=200\,s$ (i.e. $\tau=1.44\,t_{1/2}$).
Q27 — Half-Life & Pseudo First Order Reactions · medium · numerical
An acid-catalysed reaction is pseudo-first order with $k^{\prime} = k[H^+]$. At $[H^+] = 0.05\,M$ its half-life is $40\,min$. When $[H^+]$ is raised to $0.20\,M$, the half-life becomes:
A. $160\,min$
B. $10\,min$ ✓ Correct
C. $40\,min$
D. $20\,min$
Solution: $k^{\prime}\propto[H^+]$; raising $[H^+]$ by a factor $4$ makes $k^{\prime}$ four times larger, so $t_{1/2}=\dfrac{0.693}{k^{\prime}}$ falls to $\dfrac{40}{4}=10\,min$.
Q28 — Half-Life & Pseudo First Order Reactions · medium · numerical
A radioactive sample initially contains $4.8 \times 10^{20}$ nuclei and has a half-life of $12\,min$. The number of nuclei remaining after $36\,min$ is:
A. $3.0 \times 10^{19}$
B. $2.4 \times 10^{20}$
C. $1.2 \times 10^{20}$
D. $6.0 \times 10^{19}$ ✓ Correct
Solution: $36\,min = 3$ half-lives, so $N=4.8\times10^{20}\times\left(\dfrac{1}{2}\right)^3=6.0\times10^{19}$.
Q29 — Half-Life & Pseudo First Order Reactions · medium · numerical
A first-order reaction is $60\%$ complete in $20\,min$. Its rate constant is: ($\log 2 = 0.301$, $\log 5 = 0.699$)
A. $1.15 \times 10^{-2}\,min^{-1}$
B. $2.29 \times 10^{-2}\,min^{-1}$
C. $4.58 \times 10^{-2}\,s^{-1}$
D. $4.58 \times 10^{-2}\,min^{-1}$ ✓ Correct
Solution: $k=\dfrac{2.303}{20}\log\dfrac{100}{40}=\dfrac{2.303}{20}(0.398)=4.58\times10^{-2}\,min^{-1}$ (here $\log2.5=0.699-0.301=0.398$).
Q30 — Half-Life & Pseudo First Order Reactions · medium · numerical
The acid-catalysed inversion of cane sugar is pseudo-first order with $k^{\prime} = 2.31 \times 10^{-3}\,min^{-1}$. The time for the solution to become half-inverted is:
A. $433\,min$
B. $300\,min$ ✓ Correct
C. $150\,min$
D. $693\,min$
Solution: $t_{1/2}=\dfrac{0.693}{k^{\prime}}=\dfrac{0.693}{2.31\times10^{-3}}=300\,min$.