For the re action of \(\mathrm{H}_{2}\) with \(\mathrm{I}_{2}\), the rate constant is \(2.5 \times 10^{-4} \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at \(327^{\circ} \mathrm{C}\) and \(1.0 \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\) at \(527^{\circ} \mathrm{C}\). The activation energy for the reaction, in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is: ( \(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) )

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Chemical Kinetics

For the re action of

H_{2}

with

I_{2}

, the rate constant is

2.5 \times 1 0^{- 4} dm^{3} mol^{- 1} s^{- 1}

32 7^{\circ} C

and

1.0 dm^{3} mol^{- 1} s^{- 1}

52 7^{\circ} C

. The activation energy for the reaction, in

kJ mol^{- 1}

is: (

R = 8.314 J K^{- 1} mol^{- 1}

)

Answer: (B)

Two rate constants at two temperatures feed straight into the Arrhenius relation

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} \cdot \frac{T _{2} - T _{1}}{T _{1} T _{2}}

, with

T_{1} = 600 K

and

T_{2} = 800 K

. The left side is

lo g \frac{1.0}{2.5 \times 1 0 ^{- 4}} = lo g 4000 = 3.60

, and the temperature factor is

\frac{200}{600 \times 800}

. Solving for the activation energy,

E_{a} = 3.60 \times 2.303 \times 8.314 \times \frac{600 \times 800}{200} \approx 166 kJ mol^{- 1}

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option