Rate of a Chemical Reaction — NEET Chemistry MCQs with Solutions
Free NEET 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.
▶ Practise Rate of a Chemical Reaction online (free)
Questions with solutions
Q1 — Rate of a Chemical Reaction · easy · numerical
In a reaction the concentration of a reactant falls from $0.50\,M$ to $0.40\,M$ in $10\,s$. The average rate of disappearance of the reactant is:
A. $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1 \times 10^{-1}\,mol\,L^{-1}s^{-1}$
D. $1 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Average rate $= \dfrac{\Delta[R]}{\Delta t} = \dfrac{0.50-0.40}{10} = \dfrac{0.10}{10} = 1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q2 — Rate of a Chemical Reaction · easy · numerical
The rate of a chemical reaction (change of concentration with time) is expressed in the units:
A. $mol\,L^{-1}$
B. $mol\,L^{-1}s^{-1}$ ✓ Correct
C. $L\,mol^{-1}s^{-1}$
D. $mol\,L\,s^{-1}$
Solution: Rate = concentration per unit time $= \dfrac{mol\,L^{-1}}{s} = mol\,L^{-1}s^{-1}$.
Q3 — Rate of a Chemical Reaction · easy · numerical
For $2N_2O_5 \rightarrow 4NO_2 + O_2$, if oxygen is formed at a rate of $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$, the rate of formation of $NO_2$ is:
A. $5 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
B. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= \dfrac{1}{4}\dfrac{d[NO_2]}{dt} = \dfrac{d[O_2]}{dt}$, so $\dfrac{d[NO_2]}{dt} = 4 \times 2 \times 10^{-3} = 8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q4 — Rate of a Chemical Reaction · easy · numerical
The concentration of a reactant decreases from $2.0\,M$ to $1.6\,M$ in $20\,s$. The average rate of disappearance is:
A. $4 \times 10^{-1}\,mol\,L^{-1}s^{-1}$
B. $2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $8 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Average rate $= \dfrac{2.0-1.6}{20} = \dfrac{0.4}{20} = 2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q5 — Rate of a Chemical Reaction · easy · numerical
For $N_2 + 3H_2 \rightarrow 2NH_3$, if nitrogen disappears at $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$, the rate of disappearance of hydrogen is:
A. $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
B. $3 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $3.3 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $6 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: Hydrogen is consumed three times as fast as nitrogen: $-\dfrac{d[H_2]}{dt} = 3\left(-\dfrac{d[N_2]}{dt}\right) = 3 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q6 — Rate of a Chemical Reaction · easy · numerical
Which statement correctly distinguishes average rate from instantaneous rate?
A. Average rate is the change in concentration over a finite time interval, whereas instantaneous rate is the rate at a particular moment. ✓ Correct
B. Average rate and instantaneous rate are always numerically equal for every reaction.
C. Instantaneous rate is measured over a long time interval, while average rate is measured at a single instant.
D. The average rate of a reaction stays constant throughout the entire reaction.
Solution: Average rate uses a finite interval $\Delta t$; instantaneous rate is the limit as $\Delta t \to 0$ (the tangent slope at one instant).
Q7 — Rate of a Chemical Reaction · easy · numerical
In the reaction $A \rightarrow B$, the concentration of $B$ rises from $0$ to $0.6\,M$ in $30\,s$. The average rate of formation of $B$ is:
A. $2 \times 10^{-1}\,mol\,L^{-1}s^{-1}$
B. $1.8 \times 10^{1}\,mol\,L^{-1}s^{-1}$
C. $5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: Average rate $= \dfrac{\Delta[B]}{\Delta t} = \dfrac{0.6}{30} = 2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q8 — Rate of a Chemical Reaction · easy · numerical
For the reaction $2A \rightarrow B$, if $B$ is formed at $0.05\,mol\,L^{-1}s^{-1}$, the rate of disappearance of $A$ is:
A. $0.20\,mol\,L^{-1}s^{-1}$
B. $0.025\,mol\,L^{-1}s^{-1}$
C. $0.10\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $0.05\,mol\,L^{-1}s^{-1}$
Solution: $A$ is consumed twice as fast as $B$ is formed: $-\dfrac{d[A]}{dt} = 2\dfrac{d[B]}{dt} = 2 \times 0.05 = 0.10\,mol\,L^{-1}s^{-1}$.
Q9 — Rate of a Chemical Reaction · easy · numerical
The instantaneous rate of a reaction is best described as:
A. The slope of the tangent to the concentration versus time curve at that instant. ✓ Correct
B. The same as the average rate taken over the whole reaction.
C. The slope of the straight line joining the initial and final points of the curve.
D. Always zero at the very start of a reaction.
Solution: The instantaneous rate is $\lim_{\Delta t \to 0}\dfrac{\Delta[X]}{\Delta t}$, i.e. the slope of the tangent to the concentration-time curve at that moment.
Q10 — Rate of a Chemical Reaction · medium · numerical
For $2N_2O_5 \rightarrow 4NO_2 + O_2$, the rate of disappearance of $N_2O_5$ is $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of the reaction is:
A. $1 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
B. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= -\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt} = \dfrac{1}{2}(4 \times 10^{-3}) = 2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q11 — Rate of a Chemical Reaction · medium · numerical
For $2N_2O_5 \rightarrow 4NO_2 + O_2$, if $N_2O_5$ disappears at $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$, the rate of formation of $NO_2$ is:
A. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.6 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
C. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $-\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt} = \dfrac{1}{4}\dfrac{d[NO_2]}{dt}$, so $\dfrac{d[NO_2]}{dt} = 2\left(-\dfrac{d[N_2O_5]}{dt}\right) = 2 \times 4 \times 10^{-3} = 8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q12 — Rate of a Chemical Reaction · medium · numerical
For $N_2 + 3H_2 \rightarrow 2NH_3$, hydrogen disappears at $6 \times 10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $NH_3$ is:
A. $6 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
B. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $9 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{3}\dfrac{d[H_2]}{dt} = \dfrac{1}{2}\dfrac{d[NH_3]}{dt}$, so $\dfrac{d[NH_3]}{dt} = \dfrac{2}{3}(6 \times 10^{-3}) = 4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q13 — Rate of a Chemical Reaction · medium · numerical
For $N_2 + 3H_2 \rightarrow 2NH_3$, $NH_3$ is formed at $2 \times 10^{-4}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $H_2$ is:
A. $2 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
B. $1.33 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
C. $3 \times 10^{-4}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $6 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{3}\dfrac{d[H_2]}{dt} = \dfrac{1}{2}\dfrac{d[NH_3]}{dt}$, so $-\dfrac{d[H_2]}{dt} = \dfrac{3}{2}(2 \times 10^{-4}) = 3 \times 10^{-4}\,mol\,L^{-1}s^{-1}$.
Q14 — Rate of a Chemical Reaction · medium · numerical
For the reaction $3A + B \rightarrow 2C + D$, $C$ is formed at $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $A$ is:
A. $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
B. $3 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1.5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $6.7 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= -\dfrac{1}{3}\dfrac{d[A]}{dt} = \dfrac{1}{2}\dfrac{d[C]}{dt}$, so $-\dfrac{d[A]}{dt} = \dfrac{3}{2}(1 \times 10^{-2}) = 1.5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q15 — Rate of a Chemical Reaction · medium · numerical
The concentration of a reactant drops from $0.80\,M$ to $0.60\,M$ in $40\,s$. The average rate of disappearance is:
A. $5 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $2 \times 10^{-1}\,mol\,L^{-1}s^{-1}$
C. $5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: Average rate $= \dfrac{0.80-0.60}{40} = \dfrac{0.20}{40} = 5 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q16 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products, $[A]$ is $0.100\,M$ at $t=0$, $0.080\,M$ at $t=20\,s$ and $0.065\,M$ at $t=40\,s$. The average rate of reaction during the first $20\,s$ is:
A. $7.5 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
B. $2.0 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1.0 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $1.0 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= \dfrac{0.100-0.080}{20} = \dfrac{0.020}{20} = 1.0 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q17 — Rate of a Chemical Reaction · medium · numerical
For $A \rightarrow$ products, $[A]$ is $0.100\,M$ at $t=0$, $0.080\,M$ at $t=20\,s$ and $0.065\,M$ at $t=40\,s$. The average rate during the interval $20\,s$ to $40\,s$ is:
A. $7.5 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.0 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $3.25 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $7.5 \times 10^{-4}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: Rate $= \dfrac{0.080-0.065}{40-20} = \dfrac{0.015}{20} = 7.5 \times 10^{-4}\,mol\,L^{-1}s^{-1}$ (smaller than in the first interval, as the rate falls with time).
Q18 — Rate of a Chemical Reaction · medium · numerical
For a gas-phase reaction, how are the rate expressed in concentration units and the rate expressed in partial-pressure units related (at constant temperature)?
A. The rate in concentration units equals the rate in pressure units divided by $RT^2$.
B. The rate in concentration units equals the rate in pressure units divided by RT, because concentration = pressure/RT. ✓ Correct
C. The rate in concentration units equals the rate in pressure units multiplied by RT.
D. The rate in pressure units and in concentration units are always numerically equal.
Solution: Since $p = CRT$, $C = \dfrac{p}{RT}$ and $\dfrac{dC}{dt} = \dfrac{1}{RT}\dfrac{dp}{dt}$; the concentration rate is the pressure rate divided by $RT$.
Q19 — Rate of a Chemical Reaction · medium · numerical
For $4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$, water is formed at $3 \times 10^{-2}\,mol\,L^{-1}s^{-1}$. The rate of consumption of $O_2$ is:
A. $2.5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $3.6 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
C. $1.5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $3 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: $-\dfrac{1}{5}\dfrac{d[O_2]}{dt} = \dfrac{1}{6}\dfrac{d[H_2O]}{dt}$, so $-\dfrac{d[O_2]}{dt} = \dfrac{5}{6}(3 \times 10^{-2}) = 2.5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$.
Q20 — Rate of a Chemical Reaction · medium · numerical
For $4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$, $NH_3$ disappears at $8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of the reaction is:
A. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $3.2 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= -\dfrac{1}{4}\dfrac{d[NH_3]}{dt} = \dfrac{1}{4}(8 \times 10^{-3}) = 2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q21 — Rate of a Chemical Reaction · medium · numerical
For the reaction $2A + 3B \rightarrow$ products, $B$ is consumed at $9 \times 10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $A$ is:
A. $9 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
B. $1.35 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
C. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $6 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: $\dfrac{-d[A]/dt}{-d[B]/dt} = \dfrac{2}{3}$, so $-\dfrac{d[A]}{dt} = \dfrac{2}{3}(9 \times 10^{-3}) = 6 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q22 — Rate of a Chemical Reaction · medium · numerical
$0.5\,mol$ of a reactant contained in a $2\,L$ vessel is completely consumed in $100\,s$. The average rate of the reaction is:
A. $2.5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
B. $5 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $5 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
D. $2.5 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
Solution: Concentration change $= \dfrac{0.5}{2} = 0.25\,M$; average rate $= \dfrac{0.25}{100} = 2.5 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q23 — Rate of a Chemical Reaction · medium · numerical
For $H_2 + I_2 \rightarrow 2HI$, $HI$ is formed at $1 \times 10^{-4}\,mol\,L^{-1}s^{-1}$. The rate of disappearance of $H_2$ is:
A. $1 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
B. $2 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
C. $5 \times 10^{-5}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $2.5 \times 10^{-5}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= -\dfrac{d[H_2]}{dt} = \dfrac{1}{2}\dfrac{d[HI]}{dt} = \dfrac{1}{2}(1 \times 10^{-4}) = 5 \times 10^{-5}\,mol\,L^{-1}s^{-1}$.
Q24 — Rate of a Chemical Reaction · medium · numerical
Which expression correctly represents the rate of the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$?
A. $Rate = -\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt} = \dfrac{1}{4}\dfrac{d[NO_2]}{dt} = \dfrac{d[O_2]}{dt}$ ✓ Correct
B. $Rate = -\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt} = \dfrac{d[NO_2]}{dt} = \dfrac{1}{4}\dfrac{d[O_2]}{dt}$
C. $Rate = -\dfrac{d[N_2O_5]}{dt} = \dfrac{1}{4}\dfrac{d[NO_2]}{dt} = \dfrac{d[O_2]}{dt}$
D. $Rate = -2\dfrac{d[N_2O_5]}{dt} = 4\dfrac{d[NO_2]}{dt} = \dfrac{d[O_2]}{dt}$
Solution: Each term is the species rate divided by its stoichiometric coefficient, giving $-\dfrac{1}{2}\dfrac{d[N_2O_5]}{dt} = \dfrac{1}{4}\dfrac{d[NO_2]}{dt} = \dfrac{d[O_2]}{dt}$.
Q25 — Rate of a Chemical Reaction · medium · numerical
A reactant with initial concentration $1.0\,M$ is found to be $0.7\,M$ after $50\,s$. The average rate of disappearance over this period is:
A. $6 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
B. $6 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $3 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $1.4 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
Solution: Average rate $= \dfrac{1.0-0.7}{50} = \dfrac{0.3}{50} = 6 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q26 — Rate of a Chemical Reaction · medium · numerical
For a gas-phase reaction $A(g) \rightarrow B(g)$ at a temperature where $RT = 25\,L\,atm\,mol^{-1}$, the partial pressure of $A$ falls at $0.50\,atm\,min^{-1}$. The rate of disappearance of $A$ in concentration units is:
A. $0.50\,mol\,L^{-1}min^{-1}$
B. $0.04\,mol\,L^{-1}min^{-1}$
C. $12.5\,mol\,L^{-1}min^{-1}$
D. $0.02\,mol\,L^{-1}min^{-1}$ ✓ Correct
Solution: $[A] = \dfrac{p_A}{RT}$, so $-\dfrac{d[A]}{dt} = \dfrac{1}{RT}\left(-\dfrac{dp_A}{dt}\right) = \dfrac{0.50}{25} = 0.02\,mol\,L^{-1}min^{-1}$.
Q27 — Rate of a Chemical Reaction · medium · numerical
A gaseous reactant is consumed at $0.004\,mol\,L^{-1}s^{-1}$ at a temperature where $RT = 25\,L\,atm\,mol^{-1}$. The rate of fall of its partial pressure is:
A. $1.6 \times 10^{-4}\,atm\,s^{-1}$
B. $0.1\,atm\,s^{-1}$ ✓ Correct
C. $0.05\,atm\,s^{-1}$
D. $0.004\,atm\,s^{-1}$
Solution: Since $p = CRT$, $-\dfrac{dp}{dt} = RT\left(-\dfrac{dC}{dt}\right) = 25 \times 0.004 = 0.1\,atm\,s^{-1}$.
Q28 — Rate of a Chemical Reaction · medium · numerical
For $2A \rightarrow B + C$, the concentration of $B$ increases from $0$ to $0.020\,M$ in $100\,s$. The rate of disappearance of $A$ is:
A. $1 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
B. $4 \times 10^{-4}\,mol\,L^{-1}s^{-1}$ ✓ Correct
C. $2 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
D. $8 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
Solution: $\dfrac{d[B]}{dt} = \dfrac{0.020}{100} = 2 \times 10^{-4}$; since $-\dfrac{1}{2}\dfrac{d[A]}{dt} = \dfrac{d[B]}{dt}$, $-\dfrac{d[A]}{dt} = 2 \times 2 \times 10^{-4} = 4 \times 10^{-4}\,mol\,L^{-1}s^{-1}$.
Q29 — Rate of a Chemical Reaction · medium · numerical
The concentration of a reactant decreases linearly from $0.50\,M$ to $0.20\,M$ between $t=100\,s$ and $t=400\,s$. The rate of disappearance during this interval is:
A. $1 \times 10^{-2}\,mol\,L^{-1}s^{-1}$
B. $7 \times 10^{-4}\,mol\,L^{-1}s^{-1}$
C. $1 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
D. $3 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= \dfrac{0.50-0.20}{400-100} = \dfrac{0.30}{300} = 1 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.
Q30 — Rate of a Chemical Reaction · medium · numerical
For $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, oxygen is consumed at $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$. The rate of formation of $CO_2$ is:
A. $2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$ ✓ Correct
B. $8 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
C. $1 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
D. $4 \times 10^{-3}\,mol\,L^{-1}s^{-1}$
Solution: Rate $= -\dfrac{1}{2}\dfrac{d[O_2]}{dt} = \dfrac{d[CO_2]}{dt}$, so $\dfrac{d[CO_2]}{dt} = \dfrac{1}{2}(4 \times 10^{-3}) = 2 \times 10^{-3}\,mol\,L^{-1}s^{-1}$.