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Rate of a Chemical Reaction — JEE Main Chemistry MCQs with Solutions

Free JEE Main Chemistry Rate of a Chemical Reaction 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 — Rate of a Chemical Reaction · easy · numerical
For the roasting reaction $4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2$, at a certain instant $SO_2$ is being formed at $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ at that instant is:
A. $1.1\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $5.5\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
D. $2.9\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $\dfrac{1}{11}\left(-\dfrac{d[O_2]}{dt}\right)=\dfrac{1}{8}\dfrac{d[SO_2]}{dt}$, so $-\dfrac{d[O_2]}{dt}=\dfrac{11}{8}(4.0\times10^{-3})=5.5\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q2 — Rate of a Chemical Reaction · easy · numerical
For the combustion $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$, at an instant $CO_2$ is being formed at $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ at that instant is:
A. $1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$  ✓ Correct
B. $1.8\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $3.6\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $\dfrac{1}{3}\dfrac{d[CO_2]}{dt}=\dfrac{1}{5}\left(-\dfrac{d[O_2]}{dt}\right)$, so $-\dfrac{d[O_2]}{dt}=\dfrac{5}{3}(6.0\times10^{-3})=1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q3 — Rate of a Chemical Reaction · easy · numerical
For a reaction $A \rightarrow$ products the concentration of $A$ varies with time as $[A]=0.50-0.02\,t-0.003\,t^2$ (with $[A]$ in $mol\,L^{-1}$ and $t$ in $s$). The instantaneous rate of disappearance of $A$ at $t=10\,s$ is:
A. $0.02\,mol\,L^{-1}s^{-1}$
B. $0.06\,mol\,L^{-1}s^{-1}$
C. $0.14\,mol\,L^{-1}s^{-1}$
D. $0.08\,mol\,L^{-1}s^{-1}$  ✓ Correct
Solution: $-\dfrac{d[A]}{dt}=0.02+0.006\,t$; at $t=10\,s$ this is $0.02+0.06=0.08\,mol\,L^{-1}s^{-1}$.
Q4 — Rate of a Chemical Reaction · easy · numerical
For a reaction the concentration of the reactant $R$ is: $[R]=0.240\,M$ at $t=0$, $0.180\,M$ at $t=30\,s$ and $0.150\,M$ at $t=60\,s$. The average rate of disappearance of $R$ over the first $60\,s$ is:
A. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
Solution: Average rate $=\dfrac{\Delta[R]}{\Delta t}=\dfrac{0.240-0.150}{60}=\dfrac{0.090}{60}=1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q5 — Rate of a Chemical Reaction · easy · numerical
For a reaction $[A]$ falls uniformly (linearly) from $0.500\,M$ at $t=0$ to $0.300\,M$ at $t=50\,s$. The concentration of $A$ at $t=35\,s$ and the average rate of disappearance over the first $35\,s$ are, respectively:
A. $[A]=0.140\,M$ and rate $=4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $[A]=0.360\,M$ and rate $=4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
C. $[A]=0.360\,M$ and rate $=5.7\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $[A]=0.300\,M$ and rate $=4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Uniform rate $=\dfrac{0.500-0.300}{50}=4.0\times10^{-3}$; $[A]_{35}=0.500-(4.0\times10^{-3})(35)=0.360\,M$.
Q6 — Rate of a Chemical Reaction · easy · numerical
In a gaseous reaction a reactant is being consumed at $3.0\times10^{-4}\,mol\,L^{-1}s^{-1}$ at a temperature where $RT=20\,L\,atm\,mol^{-1}$. Using $p=CRT$, the rate of decrease of its partial pressure is:
A. $1.5\times10^{-5}\,atm\,s^{-1}$
B. $6.0\times10^{-2}\,atm\,s^{-1}$
C. $6.0\times10^{-3}\,atm\,s^{-1}$  ✓ Correct
D. $3.0\times10^{-4}\,atm\,s^{-1}$
Solution: $p=CRT\Rightarrow \dfrac{dp}{dt}=RT\dfrac{dC}{dt}=20\times3.0\times10^{-4}=6.0\times10^{-3}\,atm\,s^{-1}$.
Q7 — Rate of a Chemical Reaction · easy · numerical
The partial pressure of a gaseous reactant falls at $0.030\,atm\,s^{-1}$ at a temperature where $RT=24\,L\,atm\,mol^{-1}$. The rate of disappearance of the reactant expressed in concentration units is:
A. $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.25\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $0.72\,mol\,L^{-1}s^{-1}$
D. $1.25\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
Solution: $C=\dfrac{p}{RT}\Rightarrow \dfrac{dC}{dt}=\dfrac{1}{RT}\dfrac{dp}{dt}=\dfrac{0.030}{24}=1.25\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q8 — Rate of a Chemical Reaction · easy · numerical
For $2A + 3B \rightarrow C + 2D$, the rate of the reaction defined as $r=-\dfrac{1}{2}\dfrac{d[A]}{dt}=-\dfrac{1}{3}\dfrac{d[B]}{dt}=\dfrac{d[C]}{dt}$ equals $1.2\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $B$ is:
A. $1.2\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $2.4\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $4.0\times10^{-4}\,mol\,L^{-1}s^{-1}$
D. $3.6\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
Solution: $-\dfrac{d[B]}{dt}=3r=3(1.2\times10^{-3})=3.6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q9 — Rate of a Chemical Reaction · easy · numerical
For the gaseous decomposition $A(g) \rightarrow 2B(g) + C(g)$ carried out in a rigid vessel, the partial pressure of $A$ is falling at $0.010\,atm\,s^{-1}$ at an instant. The rate of increase of the total pressure at that instant is:
A. $0.020\,atm\,s^{-1}$  ✓ Correct
B. $0.010\,atm\,s^{-1}$
C. $0.040\,atm\,s^{-1}$
D. $0.030\,atm\,s^{-1}$
Solution: $p_A$ falls at $0.010$, $p_B$ rises at $0.020$, $p_C$ rises at $0.010$; $\dfrac{dp_{tot}}{dt}=-0.010+0.020+0.010=+0.020\,atm\,s^{-1}$.
Q10 — Rate of a Chemical Reaction · easy · numerical
For the redox reaction $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$, $CO_2$ is evolved at $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $MnO_4^-$ is:
A. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
D. $2.5\times10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $\dfrac{1}{2}\left(-\dfrac{d[MnO_4^-]}{dt}\right)=\dfrac{1}{10}\dfrac{d[CO_2]}{dt}$, so $-\dfrac{d[MnO_4^-]}{dt}=\dfrac{2}{10}(5.0\times10^{-3})=1.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q11 — Rate of a Chemical Reaction · easy · numerical
In a reaction the amount of product $x$ (in $mol\,L^{-1}$) formed up to time $t$ (in $s$) is $x=0.4\,t-0.01\,t^2$. The rate of formation of the product at $t=15\,s$ is:
A. $0.10\,mol\,L^{-1}s^{-1}$  ✓ Correct
B. $0.02\,mol\,L^{-1}s^{-1}$
C. $0.25\,mol\,L^{-1}s^{-1}$
D. $0.40\,mol\,L^{-1}s^{-1}$
Solution: $\dfrac{dx}{dt}=0.4-0.02\,t$; at $t=15\,s$ this is $0.4-0.30=0.10\,mol\,L^{-1}s^{-1}$.
Q12 — Rate of a Chemical Reaction · easy · numerical
For $A \rightarrow$ products with $[A]_0=0.80\,M$, the fraction of $A$ reacted is $0.20$ at $t=100\,s$ and $0.50$ at $t=250\,s$. The average rate of disappearance of $A$ between $100\,s$ and $250\,s$ is:
A. $2.67\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.6\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
C. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $9.6\times10^{-4}\,mol\,L^{-1}s^{-1}$
Solution: $[A]_{100}=0.80(0.80)=0.64$, $[A]_{250}=0.80(0.50)=0.40$; average rate $=\dfrac{0.64-0.40}{150}=1.6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q13 — Rate of a Chemical Reaction · easy · numerical
For $2H_2S + SO_2 \rightarrow 3S + 2H_2O$, at an instant $H_2S$ is being consumed at $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of sulfur ($S$) at that instant is:
A. $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
D. $2.67\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $\dfrac{1}{2}\left(-\dfrac{d[H_2S]}{dt}\right)=\dfrac{1}{3}\dfrac{d[S]}{dt}$, so $\dfrac{d[S]}{dt}=\dfrac{3}{2}(4.0\times10^{-3})=6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q14 — Rate of a Chemical Reaction · easy · numerical
For $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$ in a closed rigid vessel, at an instant $-\dfrac{dp_{NO_2}}{dt}=8.0\times10^{-3}\,atm\,s^{-1}$. The rate of increase of the total pressure at that instant is:
A. $1.2\times10^{-2}\,atm\,s^{-1}$
B. $2.0\times10^{-3}\,atm\,s^{-1}$
C. $8.0\times10^{-3}\,atm\,s^{-1}$
D. $4.0\times10^{-3}\,atm\,s^{-1}$  ✓ Correct
Solution: $p_{NO}$ rises at $8.0\times10^{-3}$, $p_{O_2}$ rises at $4.0\times10^{-3}$, $p_{NO_2}$ falls at $8.0\times10^{-3}$; net $\dfrac{dp_{tot}}{dt}=+4.0\times10^{-3}\,atm\,s^{-1}$.
Q15 — Rate of a Chemical Reaction · easy · numerical
For $A + 3B \rightarrow 2C$ carried out in a $2\,L$ closed vessel, $0.60\,mol$ of $C$ is produced in $50\,s$. The average rate of consumption of $B$ over this period is:
A. $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.8\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $9.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
D. $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $[C]$ rises at $\dfrac{0.60/2}{50}=6.0\times10^{-3}$; $-\dfrac{d[B]}{dt}=\dfrac{3}{2}\dfrac{d[C]}{dt}=\dfrac{3}{2}(6.0\times10^{-3})=9.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q16 — Rate of a Chemical Reaction · medium · numerical
The gaseous reaction $A(g) \rightarrow 3B(g)$ is studied in a rigid vessel starting from pure $A$ at $0.80\,atm$. At an instant when the total pressure is $1.60\,atm$, the total pressure is rising at $0.040\,atm\,s^{-1}$. The rate of disappearance of $A$ (in $atm\,s^{-1}$) at that instant is:
A. $0.020\,atm\,s^{-1}$  ✓ Correct
B. $0.013\,atm\,s^{-1}$
C. $0.060\,atm\,s^{-1}$
D. $0.040\,atm\,s^{-1}$
Solution: With $A\rightarrow3B$, $p_{tot}=p_{A,0}+2x$ where $x$ is the pressure of $A$ reacted, so $\dfrac{dp_{tot}}{dt}=2\left(-\dfrac{dp_A}{dt}\right)$; $-\dfrac{dp_A}{dt}=\dfrac{0.040}{2}=0.020\,atm\,s^{-1}$ (the $1.60\,atm$ value is not needed).
Q17 — Rate of a Chemical Reaction · medium · numerical
For the gas-phase reaction $2A(g) \rightarrow B(g)$ at $500\,K$ ($R=0.0821\,L\,atm\,K^{-1}mol^{-1}$), the concentration of $A$ is decreasing at $4.0\times10^{-4}\,mol\,L^{-1}s^{-1}$. The rate of increase of the partial pressure of $B$ is:
A. $1.64\times10^{-2}\,atm\,s^{-1}$
B. $2.0\times10^{-4}\,atm\,s^{-1}$
C. $4.1\times10^{-3}\,atm\,s^{-1}$
D. $8.2\times10^{-3}\,atm\,s^{-1}$  ✓ Correct
Solution: $\dfrac{d[B]}{dt}=\dfrac{1}{2}(4.0\times10^{-4})=2.0\times10^{-4}$; $\dfrac{dp_B}{dt}=RT\dfrac{d[B]}{dt}=(41.05)(2.0\times10^{-4})=8.2\times10^{-3}\,atm\,s^{-1}$.
Q18 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products: $[A]=1.00\,M$ at $t=0$, $0.80\,M$ at $t=40\,s$, $0.65\,M$ at $t=80\,s$. Assuming $[A]$ changes linearly within each $40\,s$ interval, the estimated average rate of disappearance of $A$ between $t=20\,s$ and $t=60\,s$ is:
A. $3.75\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $4.375\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
C. $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: First interval rate $=\dfrac{0.20}{40}=5.0\times10^{-3}$ so $[A]_{20}=0.90$; second interval rate $=\dfrac{0.15}{40}=3.75\times10^{-3}$ so $[A]_{60}=0.725$; average $=\dfrac{0.90-0.725}{40}=4.375\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q19 — Rate of a Chemical Reaction · medium · numerical
For a reaction $A \rightarrow$ products the concentration follows $[A]=\dfrac{1}{1+0.5\,t}$ (with $[A]$ in $mol\,L^{-1}$ and $t$ in $s$). The rate of disappearance of $A$ at $t=2\,s$ is:
A. $0.0625\,mol\,L^{-1}s^{-1}$
B. $0.125\,mol\,L^{-1}s^{-1}$  ✓ Correct
C. $0.25\,mol\,L^{-1}s^{-1}$
D. $0.50\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{d[A]}{dt}=\dfrac{0.5}{(1+0.5\,t)^2}$; at $t=2\,s$ this is $\dfrac{0.5}{(2)^2}=\dfrac{0.5}{4}=0.125\,mol\,L^{-1}s^{-1}$.
Q20 — Rate of a Chemical Reaction · medium · numerical
For $3A \rightarrow 2B$, $[B]$ increases from $0$ to $0.12\,M$ during the first $30\,s$, and the instantaneous rate of formation of $B$ at $t=30\,s$ is $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The instantaneous rate of disappearance of $A$ at $t=30\,s$ is:
A. $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.33\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
D. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Use the instantaneous rate: $-\dfrac{d[A]}{dt}=\dfrac{3}{2}\dfrac{d[B]}{dt}=\dfrac{3}{2}(2.0\times10^{-3})=3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ (the average $\dfrac{0.12}{30}=4.0\times10^{-3}$ must not be used).
Q21 — Rate of a Chemical Reaction · medium · numerical
For $2A(g) \rightarrow 4B(g) + C(g)$ in a rigid vessel starting from pure $A$ at $p_0=200\,mm\,Hg$, the total pressure at an instant is $350\,mm\,Hg$, and at that instant $p_A$ is falling at $2.0\,mm\,Hg\,s^{-1}$. The fraction of $A$ decomposed and $\dfrac{dp_{tot}}{dt}$ are, respectively:
A. $75\%$ and $3.0\,mm\,Hg\,s^{-1}$
B. $50\%$ and $2.0\,mm\,Hg\,s^{-1}$
C. $25\%$ and $1.5\,mm\,Hg\,s^{-1}$
D. $50\%$ and $3.0\,mm\,Hg\,s^{-1}$  ✓ Correct
Solution: If $a$ is the pressure of $A$ reacted, $p_{tot}=p_0+\dfrac{3}{2}a$, so $350=200+\dfrac{3}{2}a\Rightarrow a=100$ ($50\%$ decomposed); $\dfrac{dp_{tot}}{dt}=\dfrac{3}{2}\dfrac{da}{dt}=\dfrac{3}{2}(2.0)=3.0\,mm\,Hg\,s^{-1}$.
Q22 — Rate of a Chemical Reaction · medium · numerical
For $A + 2B \rightarrow 3C$, the rate of the reaction is $r$. At an instant $C$ is being formed at $9.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The values of $r$ and $-\dfrac{d[B]}{dt}$ at that instant are, respectively:
A. $3.0\times10^{-3}$ and $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
B. $4.5\times10^{-3}$ and $9.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $9.0\times10^{-3}$ and $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $3.0\times10^{-3}$ and $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $r=\dfrac{1}{3}\dfrac{d[C]}{dt}=3.0\times10^{-3}$; $-\dfrac{d[B]}{dt}=2r=6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q23 — Rate of a Chemical Reaction · medium · numerical
For $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$ at $500\,K$ ($R=0.0821\,L\,atm\,K^{-1}mol^{-1}$), $NO_2$ is being formed at $4.0\times10^{-5}\,mol\,L^{-1}s^{-1}$. The rate of decrease of the partial pressure of $O_2$ is:
A. $4.1\times10^{-4}\,atm\,s^{-1}$
B. $2.0\times10^{-5}\,atm\,s^{-1}$
C. $1.64\times10^{-3}\,atm\,s^{-1}$
D. $8.2\times10^{-4}\,atm\,s^{-1}$  ✓ Correct
Solution: $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}\dfrac{d[NO_2]}{dt}=2.0\times10^{-5}$; $\dfrac{dp}{dt}=RT(2.0\times10^{-5})=(41.05)(2.0\times10^{-5})=8.2\times10^{-4}\,atm\,s^{-1}$.
Q24 — Rate of a Chemical Reaction · medium · numerical
For a reaction $A \rightarrow$ products, $[A]$ (in $M$) follows $[A]=0.60\,e^{-0.05\,t}$ with $t$ in $s$. Using $e^{-0.5}=0.607$, the average rate of disappearance of $A$ over the first $10\,s$ and the instantaneous rate at $t=0$ are, respectively:
A. $3.64\times10^{-2}$ and $3.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $3.0\times10^{-2}$ and $2.36\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $2.36\times10^{-2}$ and $2.36\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2.36\times10^{-2}$ and $3.0\times10^{-2}\,mol\,L^{-1}s^{-1}$  ✓ Correct
Solution: $[A]_{10}=0.60(0.607)=0.364$, so average $=\dfrac{0.60-0.364}{10}=2.36\times10^{-2}$; instantaneous at $t=0$ is $0.05\times0.60=3.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q25 — Rate of a Chemical Reaction · medium · numerical
In a reaction $aA \rightarrow bB$ it is found that $-\dfrac{d[A]}{dt}=3\left(\dfrac{d[B]}{dt}\right)$ at every instant. If $A$ disappears at $9.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ and the reaction rate (per $r=-\dfrac{1}{a}\dfrac{d[A]}{dt}=\dfrac{1}{b}\dfrac{d[B]}{dt}$) is $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$, the simplest ratio $a:b$ and $\dfrac{d[B]}{dt}$ are:
A. $a:b=1:3$ and $\dfrac{d[B]}{dt}=3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $a:b=3:1$ and $\dfrac{d[B]}{dt}=2.7\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $a:b=3:1$ and $\dfrac{d[B]}{dt}=3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
D. $a:b=2:6$ and $\dfrac{d[B]}{dt}=1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{d[A]}{dt}=\dfrac{a}{b}\dfrac{d[B]}{dt}=3\dfrac{d[B]}{dt}\Rightarrow a:b=3:1$; $\dfrac{d[B]}{dt}=\dfrac{1}{3}(9.0\times10^{-3})=3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ (indeed $a=6,\,b=2$).
Q26 — Rate of a Chemical Reaction · medium · numerical
For $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$ carried out in a rigid vessel, the total pressure is falling at $4.0\times10^{-3}\,atm\,s^{-1}$ at an instant. The rate of consumption of $SO_2$ (partial pressure) at that instant is:
A. $2.0\times10^{-3}\,atm\,s^{-1}$
B. $1.2\times10^{-2}\,atm\,s^{-1}$
C. $4.0\times10^{-3}\,atm\,s^{-1}$
D. $8.0\times10^{-3}\,atm\,s^{-1}$  ✓ Correct
Solution: If $x$ is the extent (pressure of $O_2$ reacted), $p_{tot}$ changes by $-2x-x+2x=-x$, so $\dfrac{dp_{tot}}{dt}=-\dfrac{dx}{dt}\Rightarrow \dfrac{dx}{dt}=4.0\times10^{-3}$; $-\dfrac{dp_{SO_2}}{dt}=2\dfrac{dx}{dt}=8.0\times10^{-3}\,atm\,s^{-1}$.
Q27 — Rate of a Chemical Reaction · medium · numerical
A reaction $2A \rightarrow 3B$ proceeds at a constant rate of reaction $r=5.0\times10^{-4}\,mol\,L^{-1}s^{-1}$ (zero order). Starting with $[A]_0=0.100\,M$ and $[B]_0=0$, the concentration of $B$ after $100\,s$ is:
A. $0.050\,M$
B. $0.075\,M$
C. $0.150\,M$  ✓ Correct
D. $0.100\,M$
Solution: $\dfrac{d[B]}{dt}=3r=1.5\times10^{-3}$, so after $100\,s$, $[B]=(1.5\times10^{-3})(100)=0.150\,M$ (meanwhile $-\dfrac{d[A]}{dt}=2r=1.0\times10^{-3}$ exhausts the $0.100\,M$ of $A$ at $t=100\,s$).
Q28 — Rate of a Chemical Reaction · medium · numerical
A gaseous reactant is being consumed at $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$, which corresponds to a fall in its partial pressure of $0.050\,atm\,s^{-1}$. Taking $R=0.0821\,L\,atm\,K^{-1}mol^{-1}$, the temperature of the reaction is closest to:
A. $273\,K$
B. $610\,K$
C. $152\,K$
D. $305\,K$  ✓ Correct
Solution: $\dfrac{dp}{dt}=RT\dfrac{dC}{dt}\Rightarrow RT=\dfrac{0.050}{2.0\times10^{-3}}=25\,L\,atm\,mol^{-1}$; $T=\dfrac{25}{0.0821}\approx305\,K$.
Q29 — Rate of a Chemical Reaction · medium · numerical
For $4A + B \rightarrow 2C + 3D$, at a certain instant $A$ is disappearing at $8.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. The values of $-\dfrac{d[B]}{dt}$, $\dfrac{d[C]}{dt}$ and $\dfrac{d[D]}{dt}$ at that instant are, respectively:
A. $3.2\times10^{-2}$, $1.6\times10^{-2}$ and $2.4\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $8.0\times10^{-3}$, $4.0\times10^{-3}$ and $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $2.0\times10^{-3}$, $1.6\times10^{-2}$ and $2.4\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2.0\times10^{-3}$, $4.0\times10^{-3}$ and $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
Solution: The reaction rate $r=\dfrac{1}{4}(8.0\times10^{-3})=2.0\times10^{-3}$; then $-\dfrac{d[B]}{dt}=r=2.0\times10^{-3}$, $\dfrac{d[C]}{dt}=2r=4.0\times10^{-3}$, $\dfrac{d[D]}{dt}=3r=6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q30 — Rate of a Chemical Reaction · medium · numerical
The decomposition $2A(g) \rightarrow B(g) + 3C(g)$ occurs in a rigid $10\,L$ flask at $300\,K$ ($R=0.0821\,L\,atm\,K^{-1}mol^{-1}$). At an instant the total pressure is increasing at $0.090\,atm\,s^{-1}$. The rate of formation of $C$ in concentration units at that instant is:
A. $1.83\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $5.48\times10^{-3}\,mol\,L^{-1}s^{-1}$  ✓ Correct
C. $0.135\,mol\,L^{-1}s^{-1}$
D. $3.65\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: If $x$ is the pressure of $A$ reacted, $p_{tot}$ changes by $-x+0.5x+1.5x=+x$, so $\dfrac{dp_{tot}}{dt}=\dfrac{dx}{dt}=0.090$; $\dfrac{dp_C}{dt}=1.5(0.090)=0.135$, and $\dfrac{d[C]}{dt}=\dfrac{0.135}{RT}=\dfrac{0.135}{24.63}=5.48\times10^{-3}\,mol\,L^{-1}s^{-1}$ (the $10\,L$ is not needed).