Chemical Kinetics and Nuclear Chemistry — JEE Main practice

53 questions

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Sample questions with solutions

Q1 · 2026

Decomposition of A is a first order reaction at T(K) and is given by A(g) → B(g) + C(g).

In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min $^{-1}$ ) of the reaction? (log 2 = 0.3)

A.
$6.9\times 10^{-4}$
B.
$6.9\times 10^{-1}$
C.
$6.9\times 10^{-2}$
D.
$6.9\times 10^{-3}$

Answer: D

Set up the pressure relationships. Given A decomposes to B and C in equal proportions, if $x$ bar of A decomposes,

P_{total} = (1-x)+x+x = 1+x

Solve for x using the given total pressure. Given after 100 min, $P_{total}=1.5$ bar,

1+x=1.5 \quad \Rightarrow \quad x=0.5

Find the remaining pressure of A. Given

P_A = 1-x = 0.5\ \mathrm{bar}

Apply the first-order rate law using pressure as a proxy for concentration. Given

k = \frac{2.303}{t}\log\left(\frac{P_{A,0}}{P_{A,t}}\right) = \frac{2.303}{100}\log\left(\frac{1}{0.5}\right) = \frac{2.303}{100}\times 0.3 = 6.9\times 10^{-3}\ \mathrm{min}^{-1}

Hence, the answer is option D.

Q2 · 2026

$\mathrm{A}\rightarrow$ product (First order reaction).

Three sets of experiment were performed for a reaction under similar experimental conditions:

Run 1 ⇒ 100 mL of 10 M solution of reactant A

Run 2 ⇒ 200 mL of 10 M solution of reactant A

Run 3 ⇒ 100 mL of 10 M solution of reactant A + 100 mL of $\mathrm{H}_2\mathrm{O}$ added.

The correct variation of rate of reaction is

A.
Run 1 = Run 2 = Run 3
B.
Run 3 < Run 1 < Run 2
C.
Run 1 < Run 2 < Run 3
D.
Run 3 < Run 1 = Run 2

Answer: D

Recall that first-order rate depends only on concentration, not volume. Since Rate $=k[A]$ ,
Determine [A] in each run. Given Run 1 and Run 2 both use 10 M solution (just different volumes, concentration unchanged), while Run 3 is diluted with equal volume of water,

[A]_{Run 3} = \frac{10\times 100}{100+100} = 5\ \mathrm{M}

Compare the rates. Given

\text{Rate}_1 = k(10),\quad \text{Rate}_2 = k(10),\quad \text{Rate}_3 = k(5)

so Run 1 = Run 2 > Run 3.

Hence, the answer is option D.

Q3 · 2026

Correct statements regarding Arrhenius equation among the following are:

A. Factor $e^{-E_a/RT}$ corresponds to fraction of molecules having kinetic energy less than Ea.

B. At a given temperature, lower the Ea, faster is the reaction.

C. Increase in temperature by about $10^{\circ}\mathrm{C}$ doubles the rate of reaction.

D. Plot of $\log k$ vs $\frac{1}{T}$ gives a straight line with slope $=-\frac{E_a}{R}$ .

Choose the correct answer from the options given below:

A.
B and C Only
B.
A and B Only
C.
B and D Only
D.
A and C Only

Answer: A

Check statement A. Since $e^{-E_a/RT}$ actually represents the fraction of molecules with energy equal to or greater than $E_a$ (able to overcome the barrier), not less than, statement A is incorrect.
Check statement B. Given $k = Ae^{-E_a/RT}$ , a smaller $E_a$ gives a larger $e^{-E_a/RT}$ and thus a larger $k$ (faster reaction). Statement B is correct.
Check statement C. Given this is a well-known empirical rule (rate roughly doubles for every 10°C rise for many reactions), statement C is correct.
Check statement D. Since for base-10 logarithm,

\log k = \log A - \frac{E_a}{2.303R}\left(\frac{1}{T}\right)

the slope is $-\frac{E_a}{2.303R}$ , not $-\frac{E_a}{R}$ (that slope applies to $\ln k$ vs $1/T$ ). Statement D is incorrect.

Hence, the correct statements are B and C only — the answer is option A.

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