JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Chemical Kinetics

Consider

A k_{1} B

and

C k_{2} D

are two reactions. If the rate constant

(k_{1})

of the

A \to B

reaction can be expressed by the following equation

lo g_{10} k = 14.34 - \frac{1.5 \times 1 0 ^{4}}{T / K}

and activation energy of

C \to D

reaction

(Ea_{2})

\frac{1}{5}

th of the

A \to B

reaction

(Ea_{1})

, then the value of

(Ea_{2})

is _____

kJ mol^{- 1}

. (Nearest Integer)

Answer: 57

Solution

Writing the Arrhenius equation as

lo g k = lo g A - \frac{E _{a}}{2.303 R T}

and comparing with

lo g k = 14.34 - \frac{1.5 \times 1 0 ^{4}}{T}

gives

\frac{E _{a 1}}{2.303 R} = 1.5 \times 1 0^{4}

. Hence

E_{a 1} = 1.5 \times 1 0^{4} \times 2.303 \times 8.314 = 2.872 \times 1 0^{5} J mo l^{- 1} \approx 287.2 kJ mo l^{- 1}

. Since

E_{a 2} = \frac{1}{5} E_{a 1}

E_{a 2} = \frac{287.2}{5} = 57.4 kJ mo l^{- 1}

. The activation energy of the second reaction is

57 kJ mo l^{- 1}

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer