Integrated Rate Equations (Zero & First Order) — IISER Chemistry MCQs with Solutions
Free IISER Chemistry Integrated Rate Equations (Zero & First Order) 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 — Integrated Rate Equations (Zero & First Order) · easy · numerical
A first-order reaction is $20\%$ complete in $10\,min$. The rate constant is: (take $\log 1.25 = 0.097$)
A. $2.30\times10^{-2}\,min^{-1}$
B. $6.9\times10^{-2}\,min^{-1}$
C. $2.23\times10^{-2}\,min^{-1}$ ✓ Correct
D. $1.0\times10^{-2}\,min^{-1}$
Solution: $k = \dfrac{2.303}{t}\log\dfrac{a}{a-x} = \dfrac{2.303}{10}\log\dfrac{100}{80} = 0.2303\times0.097 = 2.23\times10^{-2}\,min^{-1}$.
Q2 — Integrated Rate Equations (Zero & First Order) · easy · numerical
A first-order reaction has a half-life of $20\,min$. Starting from $0.80\,M$, the concentration of the reactant after $40\,min$ is:
A. $0.60\,M$
B. $0.40\,M$
C. $0.10\,M$
D. $0.20\,M$ ✓ Correct
Solution: $40\,min = 2$ half-lives, so $[R] = \dfrac{0.80}{2^2} = \dfrac{0.80}{4} = 0.20\,M$.
Q3 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction, $[R] = [R]_0 - kt$. If $[R]_0 = 0.55\,M$ and $k = 0.03\,mol\,L^{-1}min^{-1}$, the concentration of $R$ after $10\,min$ is:
A. $0.25\,M$ ✓ Correct
B. $0.30\,M$
C. $0.35\,M$
D. $0.20\,M$
Solution: $[R] = 0.55 - (0.03)(10) = 0.55 - 0.30 = 0.25\,M$.
Q4 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction with $k = 0.05\,mol\,L^{-1}s^{-1}$, the time required for the concentration to fall from $0.60\,M$ to $0.15\,M$ is:
A. $6\,s$
B. $12\,s$
C. $15\,s$
D. $9\,s$ ✓ Correct
Solution: $t = \dfrac{[R]_0 - [R]}{k} = \dfrac{0.60 - 0.15}{0.05} = \dfrac{0.45}{0.05} = 9\,s$.
Q5 — Integrated Rate Equations (Zero & First Order) · easy · numerical
A zero-order reaction has $[R]_0 = 0.40\,M$ and $k = 0.008\,mol\,L^{-1}s^{-1}$. The time required for the reactant to be completely consumed is:
A. $100\,s$
B. $40\,s$
C. $25\,s$
D. $50\,s$ ✓ Correct
Solution: Complete consumption occurs when $[R]=0$: $t = \dfrac{[R]_0}{k} = \dfrac{0.40}{0.008} = 50\,s$.
Q6 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a first-order reaction, a straight line is obtained when one plots:
A. $\dfrac{1}{[R]}$ against time $t$
B. $[R]^2$ against time $t$
C. $[R]$ against time $t$
D. $\log[R]$ against time $t$ ✓ Correct
Solution: The integrated first-order law $\log[R] = \log[R]_0 - \dfrac{k}{2.303}t$ is linear in $t$, so a plot of $\log[R]$ against $t$ is a straight line.
Q7 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a first-order reaction, a plot of $\log[R]$ against time $t$ (in minutes) is a straight line of slope $-0.010\,min^{-1}$. The rate constant $k$ is:
A. $1.0\times10^{-2}\,min^{-1}$
B. $4.6\times10^{-2}\,min^{-1}$
C. $6.93\times10^{-3}\,min^{-1}$
D. $2.30\times10^{-2}\,min^{-1}$ ✓ Correct
Solution: For a first-order reaction the slope of $\log[R]$ vs $t$ is $-\dfrac{k}{2.303}$. Hence $k = 2.303\times0.010 = 2.30\times10^{-2}\,min^{-1}$.
Q8 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction, a plot of $[R]$ against time $t$ is a straight line of slope $-4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate constant $k$ is:
A. $1.7\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $9.2\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $-4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: For a zero-order reaction $[R] = [R]_0 - kt$, so the slope equals $-k$. Thus $k = 4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q9 — Integrated Rate Equations (Zero & First Order) · easy · numerical
A first-order reaction has a half-life of $30\,min$. The time required for the reaction to be $75\%$ complete is:
A. $120\,min$
B. $90\,min$
C. $45\,min$
D. $60\,min$ ✓ Correct
Solution: For a first-order reaction $t_{3/4} = 2\,t_{1/2}$ (75% complete means the concentration falls to $\tfrac14$, i.e. two half-lives). So $t_{3/4} = 2\times30 = 60\,min$.
Q10 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a first-order reaction, the fraction of reactant remaining after 4 half-lives is:
A. $\dfrac{1}{16}$ ✓ Correct
B. $\dfrac{1}{4}$
C. $\dfrac{1}{8}$
D. $\dfrac{1}{32}$
Solution: After $n$ half-lives the fraction remaining is $\left(\dfrac{1}{2}\right)^n$. For $n=4$ this is $\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}$.
Q11 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction with $[R]_0 = 0.40\,M$ and $k = 0.05\,mol\,L^{-1}min^{-1}$, the half-life is:
A. $8\,min$
B. $2\,min$
C. $5\,min$
D. $4\,min$ ✓ Correct
Solution: For a zero-order reaction $t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.40}{2\times0.05} = \dfrac{0.40}{0.10} = 4\,min$.
Q12 — Integrated Rate Equations (Zero & First Order) · easy · numerical
A first-order reaction has a rate constant $k = 6.93\times10^{-3}\,s^{-1}$. Its half-life is:
A. $10\,s$
B. $200\,s$
C. $50\,s$
D. $100\,s$ ✓ Correct
Solution: $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{6.93\times10^{-3}} = 100\,s$.
Q13 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has $k = 2.303\times10^{-2}\,s^{-1}$. The time required for the reactant concentration to fall to $25\%$ of its initial value is: ($\log 2 = 0.301$)
A. $120\,s$
B. $30.1\,s$
C. $100\,s$
D. $60.2\,s$ ✓ Correct
Solution: $25\%$ remaining means $\dfrac{a}{a-x} = 4$. $t = \dfrac{2.303}{k}\log 4 = \dfrac{2.303}{2.303\times10^{-2}}\times0.602 = 100\times0.602 = 60.2\,s$.
Q14 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has a half-life of $10\,min$. The time required for the concentration to fall to $6.25\%$ of its initial value is:
A. $50\,min$
B. $20\,min$
C. $30\,min$
D. $40\,min$ ✓ Correct
Solution: $6.25\% = \dfrac{1}{16} = \left(\dfrac{1}{2}\right)^4$, i.e. 4 half-lives. Time $= 4\times10 = 40\,min$.
Q15 — Integrated Rate Equations (Zero & First Order) · medium · numerical
In a first-order reaction the concentration falls from $0.48\,M$ to $0.06\,M$ in $24\,min$. The rate constant is: ($\log 2 = 0.301$)
A. $8.67\times10^{-2}\,min^{-1}$ ✓ Correct
B. $0.903\,min^{-1}$
C. $4.34\times10^{-2}\,min^{-1}$
D. $2.89\times10^{-2}\,min^{-1}$
Solution: $\dfrac{a}{a-x} = \dfrac{0.48}{0.06} = 8$, so $\log 8 = 3\log 2 = 0.903$. $k = \dfrac{2.303}{24}\times0.903 = 0.09596\times0.903 = 8.67\times10^{-2}\,min^{-1}$.
Q16 — Integrated Rate Equations (Zero & First Order) · medium · numerical
The first-order gaseous decomposition $A(g) \rightarrow 2B(g)$ starts with pure $A$ at a pressure of $400\,mm\,Hg$. After $15\,min$ the total pressure is $500\,mm\,Hg$. The rate constant is: ($\log 2 = 0.301$, $\log 3 = 0.477$)
A. $3.84\times10^{-2}\,min^{-1}$
B. $9.6\times10^{-3}\,min^{-1}$
C. $4.6\times10^{-2}\,min^{-1}$
D. $1.92\times10^{-2}\,min^{-1}$ ✓ Correct
Solution: Let $x$ be the drop in $p_A$. Total $= (400-x) + 2x = 400 + x = 500$, so $x = 100$ and $p_A = 300\,mm$. $\dfrac{a}{a-x} = \dfrac{400}{300} = \dfrac{4}{3}$, $\log\dfrac{4}{3} = 0.602 - 0.477 = 0.125$. $k = \dfrac{2.303}{15}\times0.125 = 1.92\times10^{-2}\,min^{-1}$.
Q17 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A zero-order reaction is $40\%$ complete in $20\,min$. The time required for the reactant to be completely consumed is:
A. $25\,min$
B. $50\,min$ ✓ Correct
C. $40\,min$
D. $100\,min$
Solution: For a zero-order reaction the amount reacted is proportional to time. $40\%$ in $20\,min$ corresponds to $2\%\,min^{-1}$, so $100\%$ takes $\dfrac{20}{0.40} = 50\,min$.
Q18 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A zero-order reaction has $k = 2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ and $[R]_0 = 0.10\,M$. The time required for $[R]$ to fall to $0.04\,M$ is:
A. $20\,s$
B. $50\,s$
C. $60\,s$
D. $30\,s$ ✓ Correct
Solution: $t = \dfrac{[R]_0 - [R]}{k} = \dfrac{0.10 - 0.04}{2.0\times10^{-3}} = \dfrac{0.06}{2.0\times10^{-3}} = 30\,s$.
Q19 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A zero-order reaction has a half-life of $20\,min$ when $[R]_0 = 0.40\,M$. Keeping the rate constant unchanged, if the initial concentration is raised to $0.60\,M$ the half-life becomes:
A. $30\,min$ ✓ Correct
B. $20\,min$
C. $15\,min$
D. $40\,min$
Solution: For a zero-order reaction $t_{1/2} = \dfrac{[R]_0}{2k} \propto [R]_0$. So $t_{1/2}$ scales with $[R]_0$: $20\times\dfrac{0.60}{0.40} = 30\,min$.
Q20 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction $A \rightarrow 2B$ begins with $[A]_0 = 0.60\,M$. After two half-lives, the concentration of $B$ is:
A. $0.45\,M$
B. $0.90\,M$ ✓ Correct
C. $1.20\,M$
D. $0.30\,M$
Solution: After two half-lives $[A] = \dfrac{0.60}{4} = 0.15\,M$, so $0.45\,M$ of $A$ has reacted. Each $A$ gives $2B$, so $[B] = 2\times0.45 = 0.90\,M$.
Q21 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has $k = 0.0347\,min^{-1}$. The percentage of the reactant remaining after $40\,min$ is: ($\log 2 = 0.301$)
A. $12.5\%$
B. $50\%$
C. $20\%$
D. $25\%$ ✓ Correct
Solution: $t_{1/2} = \dfrac{0.693}{0.0347} \approx 20\,min$, so $40\,min$ is 2 half-lives and $\dfrac{1}{2^2} = \dfrac{1}{4} = 25\%$ remains.
Q22 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has $k = 1.386\times10^{-2}\,min^{-1}$. Starting from $0.80\,M$, the concentration remaining after $150\,min$ is:
A. $0.20\,M$
B. $0.05\,M$
C. $0.10\,M$ ✓ Correct
D. $0.40\,M$
Solution: $t_{1/2} = \dfrac{0.693}{1.386\times10^{-2}} = 50\,min$, so $150\,min = 3$ half-lives. $[R] = \dfrac{0.80}{2^3} = \dfrac{0.80}{8} = 0.10\,M$.
Q23 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a reaction $A \rightarrow$ products, a plot of $[A]$ against $t$ is a straight line while a plot of $\log[A]$ against $t$ is curved. If the straight-line slope is $-0.010\,mol\,L^{-1}s^{-1}$ and $[A]_0 = 0.10\,M$, the order and half-life are:
A. first order, $t_{1/2} = 69.3\,s$
B. first order, $t_{1/2} = 5\,s$
C. zero order, $t_{1/2} = 10\,s$
D. zero order, $t_{1/2} = 5\,s$ ✓ Correct
Solution: A linear $[A]$-vs-$t$ plot means zero order, with $k = 0.010\,mol\,L^{-1}s^{-1}$. Then $t_{1/2} = \dfrac{[A]_0}{2k} = \dfrac{0.10}{2\times0.010} = 5\,s$.
Q24 — Integrated Rate Equations (Zero & First Order) · medium · numerical
Comparing the half-life expressions $t_{1/2} = \dfrac{[R]_0}{2k}$ (zero order) and $t_{1/2} = \dfrac{0.693}{k}$ (first order), which statement is correct?
A. The half-lives of both reactions are independent of the initial concentration
B. The half-lives of both reactions depend on the initial concentration
C. The half-life of a zero-order reaction depends on the initial concentration, whereas that of a first-order reaction does not ✓ Correct
D. The half-life of a first-order reaction depends on the initial concentration, whereas that of a zero-order reaction does not
Solution: For zero order $t_{1/2} \propto [R]_0$, so it varies with initial concentration; for first order $t_{1/2} = \dfrac{0.693}{k}$ is constant, independent of $[R]_0$.
Q25 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has a half-life of $15\,min$. The time required for the reaction to be $87.5\%$ complete is:
A. $30\,min$
B. $15\,min$
C. $45\,min$ ✓ Correct
D. $60\,min$
Solution: $87.5\%$ complete leaves $12.5\% = \dfrac{1}{8} = \left(\dfrac{1}{2}\right)^3$, i.e. 3 half-lives. Time $= 3\times15 = 45\,min$.
Q26 — Integrated Rate Equations (Zero & First Order) · medium · numerical
In a first-order reaction the concentration is $0.50\,M$ at $t = 10\,min$ and $0.125\,M$ at $t = 25\,min$. The rate constant is: ($\log 2 = 0.301$)
A. $9.24\times10^{-2}\,min^{-1}$ ✓ Correct
B. $4.62\times10^{-2}\,min^{-1}$
C. $0.602\,min^{-1}$
D. $6.93\times10^{-2}\,min^{-1}$
Solution: Over $\Delta t = 15\,min$ the concentration falls by a factor $\dfrac{0.50}{0.125} = 4$. $k = \dfrac{2.303}{15}\log 4 = 0.15353\times0.602 = 9.24\times10^{-2}\,min^{-1}$.
Q27 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has a half-life of $20\,min$. The time required for it to be $99\%$ complete is:
A. $100\,min$
B. $66\,min$
C. $200\,min$
D. $133\,min$ ✓ Correct
Solution: $k = \dfrac{0.693}{20} = 0.03465\,min^{-1}$. For $99\%$ completion $\dfrac{a}{a-x} = 100$, so $t = \dfrac{2.303}{0.03465}\log 100 = 66.46\times2 = 132.9 \approx 133\,min$.
Q28 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A zero-order reaction has $[R]_0 = 0.50\,M$ and is complete in $25\,min$. The concentration of $R$ remaining after $10\,min$ is:
A. $0.20\,M$
B. $0.25\,M$
C. $0.30\,M$ ✓ Correct
D. $0.10\,M$
Solution: $k = \dfrac{[R]_0}{t_{\text{complete}}} = \dfrac{0.50}{25} = 0.02\,mol\,L^{-1}min^{-1}$. Then $[R] = 0.50 - (0.02)(10) = 0.30\,M$.
Q29 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction is $60\%$ complete in $30\,min$. The rate constant is: ($\log 2.5 = 0.398$)
A. $1.53\times10^{-2}\,min^{-1}$
B. $0.398\,min^{-1}$
C. $6.1\times10^{-2}\,min^{-1}$
D. $3.06\times10^{-2}\,min^{-1}$ ✓ Correct
Solution: $60\%$ complete gives $\dfrac{a}{a-x} = \dfrac{100}{40} = 2.5$. $k = \dfrac{2.303}{30}\log 2.5 = 0.076767\times0.398 = 3.06\times10^{-2}\,min^{-1}$.
Q30 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For the zero-order reaction $A \rightarrow 2P$ (with $k = -\dfrac{d[A]}{dt} = 0.010\,mol\,L^{-1}min^{-1}$) and $[A]_0 = 0.30\,M$, the concentration of product $P$ formed after $10\,min$ is:
A. $0.30\,M$
B. $0.10\,M$
C. $0.20\,M$ ✓ Correct
D. $0.40\,M$
Solution: In $10\,min$, $A$ falls by $kt = 0.010\times10 = 0.10\,M$. Each mole of $A$ gives 2 of $P$, so $[P] = 2\times0.10 = 0.20\,M$.