Half-Life & Pseudo First Order Reactions — JEE Main Chemistry MCQs with Solutions
Free JEE Main 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
For a first-order reaction the mean (average) life is $\tau = 50\,s$. The time required to reach $75\%$ completion is: ($\log 2 = 0.301$)
A. $69.3\,s$ ✓ Correct
B. $138.6\,s$
C. $34.65\,s$
D. $50\,s$
Solution: Mean life $\tau = 1/k$, so $k = 0.02\,s^{-1}$. $t_{75\%} = 2\,t_{1/2} = 2\times\dfrac{0.693}{0.02} = 69.3\,s$ (also $\dfrac{2.303}{k}\log 4$).
Q2 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a reaction $A \rightarrow$ products, the half-life is $40\,min$ when $[A]_0 = 0.20\,M$ and $80\,min$ when $[A]_0 = 0.40\,M$. The order of the reaction is:
A. one-half
B. second
C. first
D. zero ✓ Correct
Solution: $t_{1/2} \propto [A]_0^{\,1-n}$. Here $\dfrac{80}{40} = 2 = 2^{1-n}$, so $1-n = 1 \Rightarrow n = 0$ (zero order).
Q3 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a reaction $A \rightarrow$ products, the half-life is $50\,s$ when $[A]_0 = 0.10\,M$ and $25\,s$ when $[A]_0 = 0.20\,M$. The order of the reaction is:
A. third
B. zero
C. second ✓ Correct
D. first
Solution: $t_{1/2} \propto [A]_0^{\,1-n}$: $\dfrac{25}{50} = \dfrac{1}{2} = 2^{1-n} \Rightarrow 1-n = -1 \Rightarrow n = 2$.
Q4 — Half-Life & Pseudo First Order Reactions · easy · numerical
A first-order reaction is $80\%$ complete in $60\,min$. Its half-life is: ($\log 2 = 0.301$, $\log 5 = 0.699$)
A. $\approx 25.8\,min$ ✓ Correct
B. $48\,min$
C. $30\,min$
D. $18\,min$
Solution: $t_{1/2} = t\cdot\dfrac{\log 2}{\log\frac{a}{a-x}} = 60\cdot\dfrac{0.301}{\log 5} = 60\cdot\dfrac{0.301}{0.699} = 25.8\,min$.
Q5 — Half-Life & Pseudo First Order Reactions · easy · numerical
A radioactive nuclide has a half-life of $30\,min$. The fraction of the sample remaining after $75\,min$ is: ($\sqrt{2} = 1.414$)
A. $0.250$
B. $0.354$
C. $0.125$
D. $\approx 0.177$ ✓ Correct
Solution: $n = \dfrac{75}{30} = 2.5$ half-lives; fraction $= \left(\tfrac{1}{2}\right)^{2.5} = \dfrac{1}{4\sqrt{2}} = \dfrac{1}{5.657} = 0.177$.
Q6 — Half-Life & Pseudo First Order Reactions · easy · numerical
The acid hydrolysis of an ester is pseudo-first-order and $25\%$ complete in $20\,min$. With $[H_2O] = 55\,mol\,L^{-1}$ (essentially constant), the true second-order rate constant is about: ($\log 2 = 0.301$, $\log 3 = 0.477$)
A. $\approx 7.9 \times 10^{-3}\,L\,mol^{-1}min^{-1}$
B. $\approx 5.2 \times 10^{-4}\,L\,mol^{-1}min^{-1}$
C. $\approx 1.4 \times 10^{-2}\,L\,mol^{-1}min^{-1}$
D. $\approx 2.6 \times 10^{-4}\,L\,mol^{-1}min^{-1}$ ✓ Correct
Solution: Pseudo $k^{\prime} = \dfrac{2.303}{20}\log\dfrac{100}{75} = \dfrac{2.303}{20}(0.125) = 1.44\times10^{-2}\,min^{-1}$; true $k = \dfrac{k^{\prime}}{[H_2O]} = \dfrac{0.0144}{55} = 2.6\times10^{-4}$.
Q7 — Half-Life & Pseudo First Order Reactions · easy · numerical
The inversion of cane sugar in dilute acid is pseudo-first-order with $k^{\prime} = 4.6 \times 10^{-3}\,min^{-1}$. The time for the solution to lose $90\%$ of its cane sugar is:
A. $\approx 150\,min$
B. $\approx 1000\,min$
C. $\approx 500\,min$ ✓ Correct
D. $\approx 230\,min$
Solution: $t_{90\%} = \dfrac{2.303}{k^{\prime}}\log\dfrac{100}{10} = \dfrac{2.303}{4.6\times10^{-3}}\times 1 = 500\,min$.
Q8 — Half-Life & Pseudo First Order Reactions · easy · numerical
A radioactive source has an initial activity of $12800$ disintegrations per second and a half-life of $20\,min$. Its activity after $50\,min$ is: ($\sqrt{2} = 1.414$)
A. $1600\,dps$
B. $4525\,dps$
C. $3200\,dps$
D. $\approx 2263\,dps$ ✓ Correct
Solution: $n = \dfrac{50}{20} = 2.5$; $A = 12800\left(\tfrac12\right)^{2.5} = \dfrac{12800}{5.657} = 2263\,dps$.
Q9 — Half-Life & Pseudo First Order Reactions · easy · numerical
A zero-order reaction has a half-life of $25\,min$ when $[A]_0 = 0.50\,M$. Keeping $k$ the same, if the initial concentration is changed to $0.80\,M$, the half-life becomes:
A. $40\,min$ ✓ Correct
B. $64\,min$
C. $15.6\,min$
D. $25\,min$
Solution: Zero order: $t_{1/2} = \dfrac{[A]_0}{2k} \propto [A]_0$. $t_{1/2} = 25\times\dfrac{0.80}{0.50} = 40\,min$.
Q10 — Half-Life & Pseudo First Order Reactions · easy · numerical
A wooden artefact shows a $^{14}C$ activity that is $30\%$ of that in living wood. If the half-life of $^{14}C$ is $5730\,yr$, the age of the artefact is: ($\log 2 = 0.301$, $\log 3 = 0.477$)
A. $\approx 6870\,yr$
B. $\approx 9960\,yr$ ✓ Correct
C. $\approx 11460\,yr$
D. $\approx 5730\,yr$
Solution: $t = \dfrac{t_{1/2}}{0.693}\times 2.303\log\dfrac{100}{30} = \dfrac{5730}{0.693}(2.303)(0.523) = 9960\,yr$ (since $\log\tfrac{10}{3} = 0.523$).
Q11 — Half-Life & Pseudo First Order Reactions · easy · numerical
A reactant $A$ decomposes by two parallel first-order paths: $A \rightarrow B$ ($k_1 = 3 \times 10^{-3}\,s^{-1}$) and $A \rightarrow C$ ($k_2 = 2 \times 10^{-3}\,s^{-1}$). The half-life for the disappearance of $A$ is:
A. $99\,s$
B. $346.5\,s$
C. $231\,s$
D. $138.6\,s$ ✓ Correct
Solution: For parallel first-order decay $k_{eff} = k_1 + k_2 = 5\times10^{-3}\,s^{-1}$; $t_{1/2} = \dfrac{0.693}{5\times10^{-3}} = 138.6\,s$.
Q12 — Half-Life & Pseudo First Order Reactions · easy · numerical
$A$ decomposes by two parallel first-order paths with $k_1 = 6 \times 10^{-4}\,s^{-1}$ (giving $B$) and $k_2 = 2 \times 10^{-4}\,s^{-1}$ (giving $C$). The percentage of $A$ that ends up as $B$ is:
A. $75\%$ ✓ Correct
B. $25\%$
C. $50\%$
D. $60\%$
Solution: Branching fraction to $B = \dfrac{k_1}{k_1+k_2} = \dfrac{6}{6+2} = 0.75 = 75\%$.
Q13 — Half-Life & Pseudo First Order Reactions · easy · numerical
For a reaction $A \rightarrow$ products, doubling the initial concentration of $A$ reduces the half-life to one-fourth of its original value. The order of the reaction is:
A. third ✓ Correct
B. zero
C. first
D. second
Solution: $t_{1/2} \propto [A]_0^{\,1-n}$: $\dfrac{1}{4} = 2^{1-n} \Rightarrow 1-n = -2 \Rightarrow n = 3$.
Q14 — Half-Life & Pseudo First Order Reactions · easy · numerical
A first-order reaction $A \rightarrow P$ (1:1) begins with $1.6\,mol$ of $A$. After a time equal to $2.5$ half-lives, the amount of product $P$ formed is: ($\sqrt{2} = 1.414$)
A. $\approx 1.32\,mol$ ✓ Correct
B. $\approx 0.28\,mol$
C. $0.40\,mol$
D. $1.20\,mol$
Solution: $A$ left $= 1.6\left(\tfrac12\right)^{2.5} = \dfrac{1.6}{5.657} = 0.283\,mol$; $P = 1.6 - 0.283 = 1.32\,mol$.
Q15 — Half-Life & Pseudo First Order Reactions · easy · numerical
An acid-catalysed reaction is pseudo-first-order in the substrate with $k^{\prime} = k[H^+]$. When $[H^+] = 0.10\,M$, $k^{\prime} = 2.0 \times 10^{-3}\,s^{-1}$. If $[H^+]$ is raised to $0.25\,M$, the half-life of the reaction becomes:
A. $138.6\,s$ ✓ Correct
B. $277\,s$
C. $86.6\,s$
D. $346.5\,s$
Solution: $k^{\prime} \propto [H^+]$, so $k^{\prime} = 2.0\times10^{-3}\times\dfrac{0.25}{0.10} = 5.0\times10^{-3}\,s^{-1}$; $t_{1/2} = \dfrac{0.693}{5\times10^{-3}} = 138.6\,s$.
Q16 — Half-Life & Pseudo First Order Reactions · medium · numerical
The half-life of a reaction is $200\,s$ when $[A]_0 = 0.10\,M$ and $141.4\,s$ when $[A]_0 = 0.20\,M$. The order of the reaction is: ($\sqrt{2} = 1.414$)
A. $\dfrac{1}{2}$
B. $\dfrac{3}{2}$ ✓ Correct
C. $2$
D. $1$
Solution: $t_{1/2} \propto [A]_0^{\,1-n}$: $\dfrac{141.4}{200} = 0.707 = 2^{-1/2} = 2^{1-n} \Rightarrow 1-n = -\tfrac12 \Rightarrow n = \tfrac32$.
Q17 — Half-Life & Pseudo First Order Reactions · medium · numerical
The acid hydrolysis of an ester is pseudo-first-order and $60\%$ is hydrolysed in $50\,min$. If water is present at $55\,mol\,L^{-1}$, the true bimolecular rate constant is: ($\log 2 = 0.301$, $\log 5 = 0.699$)
A. $\approx 6.7 \times 10^{-4}\,L\,mol^{-1}min^{-1}$
B. $\approx 1.0 \times 10^{-2}\,L\,mol^{-1}min^{-1}$
C. $\approx 1.8 \times 10^{-2}\,L\,mol^{-1}min^{-1}$
D. $\approx 3.3 \times 10^{-4}\,L\,mol^{-1}min^{-1}$ ✓ Correct
Solution: $k^{\prime} = \dfrac{2.303}{50}\log\dfrac{100}{40} = \dfrac{2.303}{50}(0.398) = 1.83\times10^{-2}\,min^{-1}$; true $k = \dfrac{k^{\prime}}{55} = 3.3\times10^{-4}$.
Q18 — Half-Life & Pseudo First Order Reactions · medium · numerical
$A$ decomposes by two parallel first-order paths: $A \rightarrow B$ ($k_1 = 1.5 \times 10^{-3}\,s^{-1}$) and $A \rightarrow C$ ($k_2 = 0.5 \times 10^{-3}\,s^{-1}$). Starting from pure $A$, the time for $75\%$ of $A$ to react and the ratio $[B]:[C]$ at that instant are: ($\log 2 = 0.301$)
A. $1386\,s$ and $3:1$
B. $346\,s$ and $3:1$
C. $693\,s$ and $3:1$ ✓ Correct
D. $693\,s$ and $1:1$
Solution: $k_{eff} = 2\times10^{-3}$; $t_{75\%} = \dfrac{2.303}{2\times10^{-3}}\log 4 = 693\,s$. Products form in ratio $k_1:k_2 = 3:1$ at all times.
Q19 — Half-Life & Pseudo First Order Reactions · medium · numerical
A first-order reaction has $k = 2.5 \times 10^{-3}\,s^{-1}$. The percentage of reactant left after $10\,min$ is: ($\ln x = 2.303\log x$, $\log e = 0.4343$)
A. $\approx 30\%$
B. $\approx 77.7\%$
C. $\approx 22.3\%$ ✓ Correct
D. $\approx 15\%$
Solution: $kt = 2.5\times10^{-3}\times 600 = 1.5$; $\dfrac{[A]}{[A]_0} = e^{-1.5} = 0.223 = 22.3\%$ (the $77.7\%$ is the amount reacted).
Q20 — Half-Life & Pseudo First Order Reactions · medium · numerical
For a first-order reaction, the time to reduce the reactant from $100\%$ to $10\%$ of its initial amount, expressed as a multiple of the half-life $t_{1/2}$, is: ($\log 2 = 0.301$)
A. $2.30\,t_{1/2}$
B. $10\,t_{1/2}$
C. $3.00\,t_{1/2}$
D. $3.32\,t_{1/2}$ ✓ Correct
Solution: $t = \dfrac{2.303}{k}\log 10$ and $t_{1/2} = \dfrac{2.303}{k}\log 2$, so $\dfrac{t}{t_{1/2}} = \dfrac{\log 10}{\log 2} = \dfrac{1}{0.301} = 3.32$.
Q21 — Half-Life & Pseudo First Order Reactions · medium · numerical
A radioactive nuclide decays simultaneously by $\alpha$-emission ($t_{1/2} = 1620\,yr$) and $\beta$-emission ($t_{1/2} = 405\,yr$). Its effective (overall) half-life is:
A. $2025\,yr$
B. $405\,yr$
C. $1012.5\,yr$
D. $324\,yr$ ✓ Correct
Solution: Rate constants add: $\dfrac{1}{t_{eff}} = \dfrac{1}{1620} + \dfrac{1}{405} = \dfrac{1+4}{1620} = \dfrac{1}{324}$, so $t_{eff} = 324\,yr$.
Q22 — Half-Life & Pseudo First Order Reactions · medium · numerical
For $A \rightarrow$ products the half-lives are: $[A]_0 = 0.10\,M \Rightarrow 200\,s$; $0.20\,M \Rightarrow 100\,s$; $0.40\,M \Rightarrow 50\,s$. The rate constant of the reaction (with proper units) is:
A. $0.02\,L\,mol^{-1}s^{-1}$
B. $0.05\,L\,mol^{-1}s^{-1}$ ✓ Correct
C. $3.47 \times 10^{-3}\,s^{-1}$
D. $5.0\,L\,mol^{-1}s^{-1}$
Solution: $t_{1/2}$ halves as $[A]_0$ doubles $\Rightarrow$ second order, $t_{1/2} = \dfrac{1}{k[A]_0}$. $k = \dfrac{1}{200\times0.10} = 0.05\,L\,mol^{-1}s^{-1}$.
Q23 — Half-Life & Pseudo First Order Reactions · medium · numerical
The count rate of a radioactive sample falls from $10000$ to $3000$ counts per minute in $40\,min$. Its half-life is: ($\log 2 = 0.301$, $\log 3 = 0.477$)
A. $\approx 30\,min$
B. $\approx 23\,min$ ✓ Correct
C. $\approx 40\,min$
D. $\approx 12\,min$
Solution: $t_{1/2} = t\cdot\dfrac{\log 2}{\log\frac{N_0}{N}} = 40\cdot\dfrac{0.301}{\log 3.33} = 40\cdot\dfrac{0.301}{0.523} = 23\,min$.
Q24 — Half-Life & Pseudo First Order Reactions · medium · numerical
A zero-order reaction with $[A]_0 = 0.10\,M$ has a half-life of $25\,min$. Measured from the start, the time at which the reaction is $90\%$ complete is:
A. $50\,min$
B. $45\,min$ ✓ Correct
C. $90\,min$
D. $22.5\,min$
Solution: $k = \dfrac{[A]_0}{2t_{1/2}} = \dfrac{0.10}{50} = 2\times10^{-3}\,mol\,L^{-1}min^{-1}$; for $90\%$, $x = 0.09$, $t = \dfrac{x}{k} = \dfrac{0.09}{2\times10^{-3}} = 45\,min$.
Q25 — Half-Life & Pseudo First Order Reactions · medium · numerical
In a first-order reaction $30\%$ of the reactant decomposes in time $t$. The percentage that has decomposed after time $2t$ is:
A. $60\%$
B. $70\%$
C. $49\%$
D. $51\%$ ✓ Correct
Solution: The fraction remaining is multiplicative: after $t$ it is $0.70$; after $2t$ it is $(0.70)^2 = 0.49$. Decomposed $= 1 - 0.49 = 51\%$.
Q26 — Half-Life & Pseudo First Order Reactions · medium · numerical
For a first-order reaction, the ratio of the time taken for $\dfrac{2}{3}$ of the reactant to react to the time taken for $\dfrac{1}{3}$ to react is: ($\log 2 = 0.301$, $\log 3 = 0.477$)
A. $3.00$
B. $2.00$
C. $\approx 2.71$ ✓ Correct
D. $1.50$
Solution: Ratio $= \dfrac{\log(1/\frac13)}{\log(1/\frac23)} = \dfrac{\log 3}{\log(3/2)} = \dfrac{0.477}{0.477-0.301} = \dfrac{0.477}{0.176} = 2.71$.
Q27 — Half-Life & Pseudo First Order Reactions · medium · numerical
A radioactive sample contains $8.0 \times 10^{18}$ atoms and has a half-life of $10\,hr$. Its initial activity (disintegrations per second) is:
A. $4.28 \times 10^{10}\,dps$
B. $1.54 \times 10^{14}\,dps$ ✓ Correct
C. $1.925 \times 10^{-5}\,dps$
D. $5.54 \times 10^{17}\,dps$
Solution: $k = \dfrac{0.693}{10\times3600} = 1.925\times10^{-5}\,s^{-1}$; activity $= kN = 1.925\times10^{-5}\times 8.0\times10^{18} = 1.54\times10^{14}\,dps$.
Q28 — Half-Life & Pseudo First Order Reactions · medium · numerical
A reaction $A + B \rightarrow$ products is first order in each (second order overall). Run with $[B]_0 = 1.0\,M \gg [A]_0 = 0.01\,M$, it gives an observed first-order constant $k_{obs} = 3.0 \times 10^{-2}\,s^{-1}$. If instead $[B]_0 = 2.0\,M$, the observed half-life of $A$ is:
A. $46.2\,s$
B. $23.1\,s$
C. $5.78\,s$
D. $11.55\,s$ ✓ Correct
Solution: Pseudo-first-order $k_{obs} = k[B] \propto [B]$; doubling $[B]$ gives $k_{obs} = 6.0\times10^{-2}\,s^{-1}$, so $t_{1/2} = \dfrac{0.693}{0.06} = 11.55\,s$.
Q29 — Half-Life & Pseudo First Order Reactions · medium · numerical
For a first-order reaction, the fraction of the initial reactant remaining after a time equal to the mean life $\tau = 1/k$ is: ($e = 2.718$)
A. $0.632$
B. $0.693$
C. $0.500$
D. $\approx 0.368$ ✓ Correct
Solution: At $t = \tau = 1/k$, $\dfrac{[A]}{[A]_0} = e^{-k\tau} = e^{-1} = 0.368$ (the $0.632$ is the fraction reacted; $0.5$ occurs at $t_{1/2}$).
Q30 — Half-Life & Pseudo First Order Reactions · medium · numerical
For an $n$th-order reaction ($n \neq 1$), $t_{1/2} \propto [A]_0^{\,1-n}$. When the initial concentration is reduced to one-quarter, the half-life increases eightfold. The order of the reaction is:
A. $2$
B. $\dfrac{3}{2}$
C. $3$
D. $\dfrac{5}{2}$ ✓ Correct
Solution: $\left(\tfrac14\right)^{1-n} = 8 \Rightarrow 2^{-2(1-n)} = 2^{3} \Rightarrow -2(1-n) = 3 \Rightarrow n = \tfrac52$.