Rate of a Chemical Reaction — IISER Chemistry MCQs with Solutions
Free IISER 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
In a reaction the concentration of a reactant decreases from $0.75\,M$ to $0.60\,M$ in $30\,s$. The average rate of disappearance of the reactant is:
A. $5.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $1.5\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: Average rate $=\dfrac{0.75-0.60}{30}=\dfrac{0.15}{30}=5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q2 — Rate of a Chemical Reaction · easy · numerical
If concentration is measured in $mol\,L^{-1}$ and time in minutes, the average rate of a reaction has the units:
A. $min^{-1}$
B. $mol\,L^{-1}s^{-1}$
C. $mol\,L^{-1}min^{-1}$ ✓ Correct
D. $L\,mol^{-1}min^{-1}$
Solution: Rate $=\dfrac{\Delta[\,]}{\Delta t}$; with concentration in $mol\,L^{-1}$ and time in minutes the units are $mol\,L^{-1}min^{-1}$.
Q3 — Rate of a Chemical Reaction · easy · numerical
For $2SO_2 + O_2 \rightarrow 2SO_3$, oxygen is consumed at $3\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $SO_3$ is:
A. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $3\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}\dfrac{d[SO_3]}{dt}$, so $\dfrac{d[SO_3]}{dt}=2(3\times10^{-3})=6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q4 — Rate of a Chemical Reaction · easy · numerical
For $2SO_2 + O_2 \rightarrow 2SO_3$, $SO_2$ is consumed at $8\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ is:
A. $2\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $8\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.6\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{1}{2}\dfrac{d[SO_2]}{dt}=-\dfrac{d[O_2]}{dt}$, so $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}(8\times10^{-3})=4\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q5 — Rate of a Chemical Reaction · easy · numerical
For $4NO_2 + O_2 \rightarrow 2N_2O_5$, oxygen is consumed at $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $NO_2$ is:
A. $1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $6.25\times10^{-4}\,mol\,L^{-1}s^{-1}$
D. $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{4}\dfrac{d[NO_2]}{dt}=-\dfrac{d[O_2]}{dt}$, so $-\dfrac{d[NO_2]}{dt}=4(2.5\times10^{-3})=1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q6 — Rate of a Chemical Reaction · easy · numerical
For $4NO_2 + O_2 \rightarrow 2N_2O_5$, $N_2O_5$ is formed at $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ is:
A. $7.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $7.5\times10^{-4}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt}=-\dfrac{d[O_2]}{dt}$, so $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}(1.5\times10^{-3})=7.5\times10^{-4}\,mol\,L^{-1}s^{-1}$.
Q7 — Rate of a Chemical Reaction · easy · numerical
For the decomposition $2N_2O \rightarrow 2N_2 + O_2$, $N_2O$ disappears at $8\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $O_2$ is:
A. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $2\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $8\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $1.6\times10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{d[O_2]}{dt}$, so $\dfrac{d[O_2]}{dt}=\dfrac{1}{2}(8\times10^{-3})=4\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q8 — Rate of a Chemical Reaction · easy · numerical
For the decomposition $2NOBr \rightarrow 2NO + Br_2$, $Br_2$ is formed at $3\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $NOBr$ is:
A. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $3\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{2}\dfrac{d[NOBr]}{dt}=\dfrac{d[Br_2]}{dt}$, so $-\dfrac{d[NOBr]}{dt}=2(3\times10^{-3})=6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q9 — Rate of a Chemical Reaction · easy · numerical
For $N_2 + 3H_2 \rightarrow 2NH_3$, ammonia is formed at $4\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $N_2$ is:
A. $2\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $8\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{d[N_2]}{dt}=\dfrac{1}{2}\dfrac{d[NH_3]}{dt}=\dfrac{1}{2}(4\times10^{-3})=2\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q10 — Rate of a Chemical Reaction · easy · numerical
For $2SO_2 + O_2 \rightarrow 2SO_3$, the ratio of the rate of consumption of $SO_2$ to the rate of consumption of $O_2$ at any instant is:
A. $3 : 1$
B. $1 : 2$
C. $1 : 1$
D. $2 : 1$ ✓ Correct
Solution: From the stoichiometry $SO_2$ is consumed twice as fast as $O_2$, so the ratio is $2 : 1$.
Q11 — Rate of a Chemical Reaction · easy · numerical
For $4NO_2 + O_2 \rightarrow 2N_2O_5$, the ratio of the rate of consumption of $NO_2$ to the rate of formation of $N_2O_5$ is:
A. $4 : 1$
B. $2 : 1$ ✓ Correct
C. $1 : 4$
D. $1 : 2$
Solution: The coefficients are $4$ for $NO_2$ and $2$ for $N_2O_5$, so the ratio $=4:2=2:1$.
Q12 — Rate of a Chemical Reaction · easy · numerical
The instantaneous rate of a reaction is best defined as:
A. the rate constant multiplied by the elapsed time
B. the rate at a particular instant, equal to the slope of the tangent to the concentration–time curve at that instant ✓ Correct
C. the change in concentration divided by the total time of the reaction
D. the average rate taken over the whole course of the reaction
Solution: The instantaneous rate is $-\dfrac{d[R]}{dt}$ at a given moment, i.e. the slope of the tangent to the concentration–time curve at that point.
Q13 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products the concentration of $A$ is $0.500\,M$ at $t=0$, $0.440\,M$ at $t=20\,s$ and $0.380\,M$ at $t=50\,s$. The average rate of reaction during the first $20\,s$ is:
A. $2.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: Average rate $=\dfrac{0.500-0.440}{20}=\dfrac{0.060}{20}=3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q14 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products the concentration of $A$ is $0.500\,M$ at $t=0$, $0.440\,M$ at $t=20\,s$ and $0.380\,M$ at $t=50\,s$. The average rate of reaction during the interval $20\,s$ to $50\,s$ is:
A. $1.2\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $3.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $6.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Average rate $=\dfrac{0.440-0.380}{50-20}=\dfrac{0.060}{30}=2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q15 — Rate of a Chemical Reaction · medium · numerical
For a gas-phase reaction the partial pressure of the reactant falls at $0.060\,atm\,s^{-1}$ at a temperature where $RT=30\,L\,atm\,mol^{-1}$. Using $[\,]=\dfrac{p}{RT}$, the rate of disappearance in concentration units is:
A. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $1.8\,mol\,L^{-1}s^{-1}$
C. $5.0\times10^{-4}\,mol\,L^{-1}s^{-1}$
D. $2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{d[\,]}{dt}=\dfrac{1}{RT}\left(-\dfrac{dp}{dt}\right)=\dfrac{0.060}{30}=2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q16 — Rate of a Chemical Reaction · medium · numerical
A gaseous reactant is consumed at $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ at a temperature where $RT=40\,L\,atm\,mol^{-1}$. The rate of fall of its partial pressure is:
A. $1.25\times10^{-4}\,atm\,s^{-1}$
B. $8.0\times10^{-3}\,atm\,s^{-1}$
C. $5.0\times10^{-3}\,atm\,s^{-1}$
D. $0.20\,atm\,s^{-1}$ ✓ Correct
Solution: $p=[\,]RT\Rightarrow -\dfrac{dp}{dt}=RT\left(-\dfrac{d[\,]}{dt}\right)=40\times5.0\times10^{-3}=0.20\,atm\,s^{-1}$.
Q17 — Rate of a Chemical Reaction · medium · numerical
In a reaction the concentration of product $P$ follows $[P]=0.02\,t+0.001\,t^2$ (with $[P]$ in $mol\,L^{-1}$ and $t$ in $s$). The rate of formation of $P$ at $t=10\,s$ is:
A. $2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $1.2\times10^{-1}\,mol\,L^{-1}s^{-1}$
C. $3.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $4.0\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $\dfrac{d[P]}{dt}=0.02+0.002\,t$; at $t=10\,s$, $=0.02+0.02=0.04=4.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q18 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products the concentration follows $[A]=1.0-0.05\,t+0.001\,t^2$ (with $[A]$ in $mol\,L^{-1}$ and $t$ in $s$). The instantaneous rate of disappearance of $A$ at $t=10\,s$ is:
A. $5.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $7.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $3.0\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{d[A]}{dt}=0.05-0.002\,t$; at $t=10\,s$, $=0.05-0.02=0.03=3.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q19 — Rate of a Chemical Reaction · medium · numerical
$0.8\,mol$ of a reactant contained in a $4\,L$ vessel is completely consumed in $200\,s$. The average rate of the reaction is:
A. $4.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Initial concentration $=\dfrac{0.8}{4}=0.20\,M$; average rate $=\dfrac{0.20}{200}=1.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q20 — Rate of a Chemical Reaction · medium · numerical
For $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$, oxygen is consumed at $6\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $CO_2$ is:
A. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $4\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $9\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{1}{2}\dfrac{d[CO_2]}{dt}$, so $\dfrac{d[CO_2]}{dt}=\dfrac{2}{3}(6\times10^{-3})=4\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q21 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products the concentration follows $[A]=0.50\,e^{-0.04\,t}$ (with $[A]$ in $mol\,L^{-1}$ and $t$ in $s$). The instantaneous rate of disappearance of $A$ at $t=0$ is:
A. $5.0\times10^{-1}\,mol\,L^{-1}s^{-1}$
B. $1.25\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $4.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{d[A]}{dt}=0.04\times0.50\,e^{-0.04t}$; at $t=0$, $=0.04\times0.50=2.0\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q22 — Rate of a Chemical Reaction · medium · numerical
For $2SO_2 + O_2 \rightarrow 2SO_3$ the rate of the reaction (defined as $-\dfrac{1}{2}\dfrac{d[SO_2]}{dt}$) is $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $SO_3$ is:
A. $1.25\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $=\dfrac{1}{2}\dfrac{d[SO_3]}{dt}$, so $\dfrac{d[SO_3]}{dt}=2(2.5\times10^{-3})=5.0\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q23 — Rate of a Chemical Reaction · medium · numerical
For a reaction of positive order, as the reaction proceeds with time the instantaneous rate of disappearance of the reactant:
A. decreases, because the concentration of the reactant falls ✓ Correct
B. is always equal to the average rate over the whole reaction
C. remains constant throughout the reaction
D. increases, because more product is formed
Solution: For a positive-order reaction the rate depends on reactant concentration; as $[R]$ falls with time, the instantaneous rate decreases.
Q24 — Rate of a Chemical Reaction · medium · numerical
For the reaction $4NO_2 + O_2 \rightarrow 2N_2O_5$, the rate of reaction is correctly written as:
A. $-\dfrac{1}{2}\dfrac{d[NO_2]}{dt}=-\dfrac{d[O_2]}{dt}=+\dfrac{1}{4}\dfrac{d[N_2O_5]}{dt}$
B. $-4\dfrac{d[NO_2]}{dt}=-\dfrac{d[O_2]}{dt}=+2\dfrac{d[N_2O_5]}{dt}$
C. $-\dfrac{d[NO_2]}{dt}=-\dfrac{1}{4}\dfrac{d[O_2]}{dt}=+\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt}$
D. $-\dfrac{1}{4}\dfrac{d[NO_2]}{dt}=-\dfrac{d[O_2]}{dt}=+\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt}$ ✓ Correct
Solution: Divide each species rate by its stoichiometric coefficient (negative for reactants): $-\dfrac{1}{4}\dfrac{d[NO_2]}{dt}=-\dfrac{d[O_2]}{dt}=+\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt}$.
Q25 — Rate of a Chemical Reaction · medium · numerical
For $3A + 2B \rightarrow 4C + D$, product $C$ is formed at $8\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $A$ is:
A. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $1.07\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $8\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{1}{4}\dfrac{d[C]}{dt}$, so $-\dfrac{d[A]}{dt}=\dfrac{3}{4}(8\times10^{-3})=6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q26 — Rate of a Chemical Reaction · medium · numerical
For $3A + 2B \rightarrow 4C + D$, $A$ is consumed at $9\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $C$ is:
A. $9\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $6.75\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $1.8\times10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{1}{4}\dfrac{d[C]}{dt}$, so $\dfrac{d[C]}{dt}=\dfrac{4}{3}(9\times10^{-3})=1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$.
Q27 — Rate of a Chemical Reaction · medium · numerical
For $2H_2 + O_2 \rightarrow 2H_2O$, hydrogen is consumed at $5\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ is:
A. $1.0\times10^{-2}\,mol\,L^{-1}s^{-1}$
B. $1.0\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $5\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{1}{2}\dfrac{d[H_2]}{dt}=-\dfrac{d[O_2]}{dt}$, so $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}(5\times10^{-3})=2.5\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q28 — Rate of a Chemical Reaction · medium · numerical
For the reaction $N_2O_4 \rightarrow 2NO_2$, the rate of the reaction (equal to $-\dfrac{d[N_2O_4]}{dt}$) is $3\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $NO_2$ is:
A. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $1.2\times10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1.5\times10^{-3}\,mol\,L^{-1}s^{-1}$
D. $3\times10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{d[N_2O_4]}{dt}=\dfrac{1}{2}\dfrac{d[NO_2]}{dt}$, so $\dfrac{d[NO_2]}{dt}=2(3\times10^{-3})=6\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q29 — Rate of a Chemical Reaction · medium · numerical
For $2O_3 \rightarrow 3O_2$, ozone is consumed at $6\times10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $O_2$ is:
A. $4\times10^{-3}\,mol\,L^{-1}s^{-1}$
B. $6\times10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.8\times10^{-2}\,mol\,L^{-1}s^{-1}$
D. $9\times10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{1}{2}\dfrac{d[O_3]}{dt}=\dfrac{1}{3}\dfrac{d[O_2]}{dt}$, so $\dfrac{d[O_2]}{dt}=\dfrac{3}{2}(6\times10^{-3})=9\times10^{-3}\,mol\,L^{-1}s^{-1}$.
Q30 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products the average rate of disappearance of $A$ over the first $100\,s$ is $2.0\times10^{-3}\,mol\,L^{-1}s^{-1}$. If $[A]_0=0.50\,M$, the concentration of $A$ at $t=100\,s$ is:
A. $0.30\,M$ ✓ Correct
B. $0.40\,M$
C. $0.48\,M$
D. $0.20\,M$
Solution: $[A]$ falls by (average rate)$\times t=2.0\times10^{-3}\times100=0.20\,M$; so $[A]=0.50-0.20=0.30\,M$.