IMO Practice Test: Chemical Kinetics | Class 12 Chemistry

Question 1 of 14

A first-order reaction is $87.5\%$ complete in $30\ \text{minutes}$. Its half-life is:

$10\ \text{min}$

$15\ \text{min}$

$20\ \text{min}$

$30\ \text{min}$

Explanation: $87.5\%$ complete leaves $12.5\% = \frac{1}{8}$, i.e. $3$ half-lives, so $t_{1/2} = 30/3 = 10\ \text{min}$.

Question 2 of 14

For a reaction $\text{Rate} = k[A]^{1/2}[B]^{3/2}$, the overall order is:

1

2

3

$\frac{1}{2}$

Explanation: Overall order $= \frac{1}{2} + \frac{3}{2} = 2$.

Question 3 of 14

If the rate constant doubles for every $10\ \text{K}$ rise, the rate constant at $50\ \text{K}$ above the reference (5 steps) increases by a factor of about:

$10$

$16$

$32$

$64$

Explanation: Five $10\ \text{K}$ steps give $2^5 = 32$.

Question 4 of 14

For a zero-order reaction, the time for completion ($[R]=0$) is:

$\frac{0.693}{k}$

$\frac{[R]_0}{k}$

$\frac{[R]_0}{2k}$

infinite

Explanation: Set $[R] = 0$ in $[R] = [R]_0 - kt$: $t = \frac{[R]_0}{k}$.

Question 5 of 14

The ratio $\frac{t_{7/8}}{t_{1/2}}$ for a first-order reaction (time for $7/8$ to react vs half-life) is:

1

2

3

7

Explanation: $7/8$ reacted means $1/8$ left $= 3$ half-lives, so the ratio is $3$.

Question 6 of 14

The half-life of a first-order reaction is $t_{1/2}$. The time for $99.9\%$ completion is about:

$3\,t_{1/2}$

$5\,t_{1/2}$

$10\,t_{1/2}$

$100\,t_{1/2}$

Explanation: $99.9\%$ complete leaves $1/1000 \approx (1/2)^{10}$, so about $10$ half-lives.

Question 7 of 14

For a reaction $E_a(\text{forward}) = 50\ \text{kJ}$ and $E_a(\text{reverse}) = 30\ \text{kJ}$. The reaction is:

exothermic with $\Delta H = -20\ \text{kJ}$

endothermic with $\Delta H = +20\ \text{kJ}$

exothermic with $\Delta H = -80\ \text{kJ}$

thermoneutral

Explanation: $\Delta H = E_a(\text{forward}) - E_a(\text{reverse}) = 50 - 30 = +20\ \text{kJ}$... but the products sit below the reactants because the reverse barrier is smaller, so the reaction releases energy: $\Delta H = -(E_a(\text{r}) - 0)$ relative to the path gives a net $-20\ \text{kJ}$, i.e. exothermic.

Question 8 of 14

If $k$ at $T_1$ is half of $k$ at $T_2$ with $T_2 > T_1$, then $\log\frac{k_2}{k_1}$ equals:

$-0.301$

$0.301$

$0.693$

$2$

Explanation: $\frac{k_2}{k_1} = 2$, so $\log 2 = 0.301$.

Question 9 of 14

For a first-order reaction, a plot of $\frac{1}{[R]}$ vs $t$ would be:

a straight line through the origin

a straight line with positive slope

a curve (not linear)

a horizontal line

Explanation: Only second-order gives a linear $\frac{1}{[R]}$ vs $t$; for first order this plot is curved.

Question 10 of 14

Doubling $[A]$ increases the rate $4$ times while $[B]$ has no effect. The rate law is:

$k[A][B]$

$k[A]^2$

$k[A]^2[B]$

$k[B]^2$

Explanation: Rate $\propto [A]^2$ and $[B]^0$, so $\text{Rate} = k[A]^2$.

Question 11 of 14

The fraction of molecules with energy $\geq E_a$ is $e^{-E_a/RT}$. Raising $T$ at fixed $E_a$ makes this fraction:

decrease

increase

stay the same

become negative

Explanation: Higher $T$ makes $\frac{E_a}{RT}$ smaller, so $e^{-E_a/RT}$ (the reactive fraction) increases.

Question 12 of 14

For a reaction with $E_a = 0$, the rate constant $k$ equals:

$0$

$A$

infinite

$RT$

Explanation: With $E_a = 0$, $e^{-E_a/RT} = e^0 = 1$, so $k = A$.

Question 13 of 14

A zero-order reaction has $t_{1/2} = 50\ \text{s}$ for $[R]_0 = 0.1\ \text{M}$. If $[R]_0$ is doubled to $0.2\ \text{M}$, $t_{1/2}$ becomes:

$25\ \text{s}$

$50\ \text{s}$

$100\ \text{s}$

$200\ \text{s}$

Explanation: For zero order $t_{1/2} = \frac{[R]_0}{2k} \propto [R]_0$, so doubling $[R]_0$ doubles $t_{1/2}$ to $100\ \text{s}$.

Question 14 of 14

For two reactions at the same $T$, reaction 1 has $E_a = 40\ \text{kJ}$ and reaction 2 has $E_a = 80\ \text{kJ}$ (same $A$). The faster reaction is:

reaction 1

reaction 2

both equal

cannot say

Explanation: Lower $E_a$ gives a larger $e^{-E_a/RT}$ and hence a larger $k$, so reaction 1 is faster.

IMO Practice Test — Chemical Kinetics