Integrated Rate Equations (Zero & First Order) — NEET Chemistry MCQs with Solutions
Free NEET 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
For a zero-order reaction, $[R] = [R]_0 - kt$. If $[R]_0 = 0.50$ M and $k = 0.05$ mol $L^{-1}s^{-1}$, the concentration of $R$ after 5 s is:
A. 0.30 M
B. 0.05 M
C. 0.20 M
D. 0.25 M ✓ Correct
Solution: $[R] = 0.50 - (0.05)(5) = 0.50 - 0.25 = 0.25$ M.
Q2 — Integrated Rate Equations (Zero & First Order) · easy · numerical
In a zero-order reaction the concentration of the reactant falls from 0.80 M to 0.40 M in 20 s. The rate constant is:
A. 0.01 mol $L^{-1}s^{-1}$
B. 0.02 mol $L^{-1}s^{-1}$ ✓ Correct
C. 0.20 mol $L^{-1}s^{-1}$
D. 0.04 mol $L^{-1}s^{-1}$
Solution: $k = \dfrac{[R]_0 - [R]}{t} = \dfrac{0.80 - 0.40}{20} = 0.02$ mol $L^{-1}s^{-1}$.
Q3 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction with $k = 0.02$ mol $L^{-1}s^{-1}$, the concentration of the reactant falls from $0.50$ M to $0.40$ M in:
A. 10 s
B. 5 s ✓ Correct
C. 25 s
D. 2 s
Solution: For zero order, $t = \dfrac{[R]_0 - [R]}{k} = \dfrac{0.50 - 0.40}{0.02} = 5$ s.
Q4 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a zero-order reaction, $[R]_0 = 0.02$ M and $k = 0.001$ mol $L^{-1}s^{-1}$. The half-life is:
A. 20 s
B. 10 s ✓ Correct
C. 5 s
D. 40 s
Solution: $t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.02}{2\times0.001} = 10$ s.
Q5 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A zero-order reaction has $[R]_0 = 0.10$ M and $k = 0.005$ mol $L^{-1}min^{-1}$. The time required for the reactant to be completely consumed is:
A. 50 min
B. 20 min ✓ Correct
C. 5 min
D. 10 min
Solution: For zero order, $[R] = 0$ when $t = \dfrac{[R]_0}{k} = \dfrac{0.10}{0.005} = 20$ min.
Q6 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a zero-order reaction with $[R]_0 = 1.0$ M and $k = 0.1$ mol $L^{-1}min^{-1}$, the concentration remaining after 6 min is:
A. 0.5 M
B. 0.4 M ✓ Correct
C. 0.1 M
D. 0.6 M
Solution: $[R] = 1.0 - (0.1)(6) = 1.0 - 0.6 = 0.4$ M.
Q7 — Integrated Rate Equations (Zero & First Order) · medium · numerical
Which of the following is a common example of a zero-order reaction?
A. The radioactive decay of a nucleus.
B. The decomposition of $N_2O_5$ in the gas phase.
C. The acid hydrolysis of an ester in dilute solution.
D. A gas reaction catalysed on a metal surface that is fully covered (saturated) with reactant. ✓ Correct
Solution: Surface-catalysed reactions on a saturated catalyst (and certain photochemical reactions) proceed at a rate independent of concentration, i.e. zero order.
Q8 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a zero-order reaction with $[R]_0 = 0.10$ M and $k = 0.02$ mol $L^{-1}min^{-1}$, the time taken for $[R]$ to fall to 0.02 M is:
A. 8 min
B. 2 min
C. 4 min ✓ Correct
D. 5 min
Solution: $t = \dfrac{[R]_0 - [R]}{k} = \dfrac{0.10 - 0.02}{0.02} = 4$ min.
Q9 — Integrated Rate Equations (Zero & First Order) · easy · numerical
A first-order reaction has a half-life of 138.6 s. Its rate constant is:
A. $5\times10^{-3}\,s^{-1}$ ✓ Correct
B. $1.4\times10^{-2}\,s^{-1}$
C. $5\times10^{-2}\,s^{-1}$
D. $2\times10^{-3}\,s^{-1}$
Solution: $k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{138.6} = 5\times10^{-3}\,s^{-1}$.
Q10 — Integrated Rate Equations (Zero & First Order) · easy · numerical
The rate constant of a first-order reaction is $0.0693\,min^{-1}$. Its half-life is:
A. 10 min ✓ Correct
B. 100 min
C. 5 min
D. 20 min
Solution: $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.0693} = 10$ min.
Q11 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction is 50% complete in 40 min. The rate constant is:
A. $3.47\times10^{-2}\,min^{-1}$
B. $0.693\,min^{-1}$
C. $1.73\times10^{-2}\,min^{-1}$ ✓ Correct
D. $6.93\times10^{-3}\,min^{-1}$
Solution: 50% completion means $t_{1/2} = 40$ min, so $k = \dfrac{0.693}{40} = 1.73\times10^{-2}\,min^{-1}$.
Q12 — Integrated Rate Equations (Zero & First Order) · medium · numerical
In a first-order reaction the concentration falls from 0.50 M to 0.125 M in 30 min. The rate constant is: (log 2 = 0.301, log 4 = 0.602)
A. $1.50\times10^{-2}\,min^{-1}$
B. $0.602\,min^{-1}$
C. $4.62\times10^{-2}\,min^{-1}$ ✓ Correct
D. $2.31\times10^{-2}\,min^{-1}$
Solution: $k = \dfrac{2.303}{30}\log\dfrac{0.50}{0.125} = \dfrac{2.303}{30}\times0.602 = 4.62\times10^{-2}\,min^{-1}$.
Q13 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a first-order reaction with $k = 0.0693\,min^{-1}$, the time required for 75% completion is: (log 2 = 0.301, log 4 = 0.602)
A. 10 min
B. 40 min
C. 20 min ✓ Correct
D. 30 min
Solution: $t = \dfrac{2.303}{k}\log\dfrac{100}{25} = \dfrac{2.303}{0.0693}\times0.602 = 20$ min.
Q14 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has $k = 2.303\times10^{-3}\,s^{-1}$. The time required for 90% completion is:
A. 1000 s ✓ Correct
B. 2000 s
C. 100 s
D. 500 s
Solution: $t = \dfrac{2.303}{k}\log\dfrac{100}{10} = \dfrac{2.303}{2.303\times10^{-3}}\times1 = 1000$ s.
Q15 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a first-order reaction, how is the time for 75% completion related to the time for 50% completion?
A. $t_{75\%} = 2\,t_{50\%}$ ✓ Correct
B. $t_{75\%} = 1.5\,t_{50\%}$
C. $t_{75\%} = 4\,t_{50\%}$
D. $t_{75\%} = 3\,t_{50\%}$
Solution: $t_{75\%} = \dfrac{2.303}{k}\log 4 = \dfrac{2.303}{k}(2\log 2) = 2\,t_{50\%}$.
Q16 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a first-order reaction, the time for 99% completion is how many times the time for 90% completion?
A. 3 times
B. 2 times ✓ Correct
C. 10 times
D. 1.5 times
Solution: $t_{99\%} \propto \log 100 = 2$ and $t_{90\%} \propto \log 10 = 1$, so $t_{99\%} = 2\,t_{90\%}$.
Q17 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a first-order reaction, the fraction of reactant remaining after 3 half-lives is:
A. 1/16
B. 1/8 ✓ Correct
C. 1/4
D. 1/6
Solution: After $n$ half-lives the fraction left is $(1/2)^n$; for $n = 3$ it is $(1/2)^3 = 1/8$.
Q18 — Integrated Rate Equations (Zero & First Order) · easy · numerical
For a first-order reaction, the fraction of reactant remaining after 2 half-lives is:
A. 1/8
B. 1/16
C. 1/2
D. 1/4 ✓ Correct
Solution: Fraction remaining $= (1/2)^2 = 1/4$.
Q19 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For the first-order gas reaction $A(g) \rightarrow B(g) + C(g)$, the initial pressure is 100 mm Hg and after 10 min the total pressure is 150 mm Hg. The rate constant is: (log 2 = 0.301)
A. $0.150\,min^{-1}$
B. $0.0693\,min^{-1}$ ✓ Correct
C. $0.301\,min^{-1}$
D. $0.0301\,min^{-1}$
Solution: Pressure of $A = 2P_0 - P_t = 200 - 150 = 50$ mm; $k = \dfrac{2.303}{10}\log\dfrac{100}{50} = \dfrac{2.303}{10}\times0.301 = 0.0693\,min^{-1}$.
Q20 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has $k = 4.606\times10^{-2}\,min^{-1}$. The time required for 90% of the reactant to react is:
A. 50 min ✓ Correct
B. 100 min
C. 25 min
D. 10 min
Solution: $t = \dfrac{2.303}{k}\log\dfrac{100}{10} = \dfrac{2.303}{4.606\times10^{-2}}\times1 = 50$ min.
Q21 — Integrated Rate Equations (Zero & First Order) · medium · numerical
In a first-order reaction the concentration drops from 0.20 M to 0.05 M in 60 min. The rate constant is: (log 2 = 0.301, log 4 = 0.602)
A. $4.62\times10^{-2}\,min^{-1}$
B. $0.602\,min^{-1}$
C. $2.31\times10^{-2}\,min^{-1}$ ✓ Correct
D. $1.16\times10^{-2}\,min^{-1}$
Solution: $k = \dfrac{2.303}{60}\log\dfrac{0.20}{0.05} = \dfrac{2.303}{60}\times0.602 = 2.31\times10^{-2}\,min^{-1}$.
Q22 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a zero-order reaction, a plot of $[R]$ against time $t$ is a straight line. Its slope is equal to:
A. $k$
B. $-\dfrac{k}{2.303}$
C. $-k$ ✓ Correct
D. $-2.303k$
Solution: From $[R] = [R]_0 - kt$, the graph of $[R]$ vs $t$ has slope $-k$ and intercept $[R]_0$.
Q23 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a first-order reaction, a plot of $\log[R]$ against time $t$ is a straight line whose slope equals:
A. $-\dfrac{k}{2.303}$ ✓ Correct
B. $-k$
C. $-2.303k$
D. $\dfrac{k}{2.303}$
Solution: $\log[R] = \log[R]_0 - \dfrac{k}{2.303}t$, so the slope of $\log[R]$ vs $t$ is $-\dfrac{k}{2.303}$.
Q24 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has $k = 1\times10^{-3}\,s^{-1}$. The time needed for the reactant concentration to fall to one-eighth of its initial value is:
A. 693 s
B. 1386 s
C. 2079 s ✓ Correct
D. 3000 s
Solution: One-eighth remains after 3 half-lives; $t_{1/2} = \dfrac{0.693}{10^{-3}} = 693$ s, so $t = 3\times693 = 2079$ s.
Q25 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction is 75% complete in 40 min. The rate constant is:
A. $1.73\times10^{-2}\,min^{-1}$
B. $0.693\,min^{-1}$
C. $3.47\times10^{-2}\,min^{-1}$ ✓ Correct
D. $6.93\times10^{-2}\,min^{-1}$
Solution: 75% complete means 2 half-lives, so $t_{1/2} = 20$ min and $k = \dfrac{0.693}{20} = 3.47\times10^{-2}\,min^{-1}$.
Q26 — Integrated Rate Equations (Zero & First Order) · medium · numerical
In a first-order reaction the concentration decreases from 0.10 M to 0.025 M in 200 s. The rate constant is: (log 2 = 0.301, log 4 = 0.602)
A. $3.47\times10^{-3}\,s^{-1}$
B. $6.93\times10^{-3}\,s^{-1}$ ✓ Correct
C. $0.602\,s^{-1}$
D. $1.39\times10^{-2}\,s^{-1}$
Solution: $k = \dfrac{2.303}{200}\log\dfrac{0.10}{0.025} = \dfrac{2.303}{200}\times0.602 = 6.93\times10^{-3}\,s^{-1}$.
Q27 — Integrated Rate Equations (Zero & First Order) · medium · numerical
For a zero-order reaction with $[R]_0 = 0.06$ M and $k = 0.01$ mol $L^{-1}min^{-1}$, the time in which half of the reactant is consumed is:
A. 1.5 min
B. 6 min
C. 3 min ✓ Correct
D. 12 min
Solution: $t_{1/2} = \dfrac{[R]_0}{2k} = \dfrac{0.06}{2\times0.01} = 3$ min.
Q28 — Integrated Rate Equations (Zero & First Order) · medium · numerical
A first-order reaction has $k = 2.303\times10^{-2}\,min^{-1}$. The percentage of reactant remaining after 100 min is:
A. 25%
B. 50%
C. 1%
D. 10% ✓ Correct
Solution: $\log\dfrac{[R]_0}{[R]} = \dfrac{kt}{2.303} = \dfrac{2.303\times10^{-2}\times100}{2.303} = 1$, so $\dfrac{[R]_0}{[R]} = 10$ and 10% remains.
Q29 — Integrated Rate Equations (Zero & First Order) · easy · numerical
Which statement about a zero-order reaction is correct?
A. The half-life is independent of the initial concentration.
B. The rate is constant and independent of the reactant concentration. ✓ Correct
C. The rate doubles every time the concentration is doubled.
D. A plot of $\log[R]$ against time is a straight line.
Solution: For a zero-order reaction the rate equals $k$ and does not depend on concentration; $[R]$ vs $t$ (not $\log[R]$) is linear.
Q30 — Integrated Rate Equations (Zero & First Order) · easy · numerical
The rate constant of a first-order reaction:
A. has units of $mol\,L^{-1}s^{-1}$.
B. has units of $s^{-1}$ and is independent of the concentration units used. ✓ Correct
C. has units of $L\,mol^{-1}s^{-1}$.
D. depends on the initial concentration of the reactant.
Solution: For a first-order reaction $k$ has units of time$^{-1}$ (e.g. $s^{-1}$), independent of how concentration is expressed.