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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Chemical Kinetics

For the reaction

A \to B

, the rate constant

K

(in

s^{- 1}

) is given by

lo g_{10} K = 20.35 - \frac{( 2.47 \times 1 0 ^{3} )}{T} .

The energy of activation in

kJ mol^{- 1}

is ______. (Nearest integer)
[Given :

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

]

Answer: 47

Solution

Comparing the given line with the Arrhenius form

lo g_{10} K = lo g_{10} A - \frac{E _{a}}{2.303 R T}

identifies

\frac{E _{a}}{2.303 R} = 2.47 \times 1 0^{3} K

. Hence

E_{a} = 2.303 \times 8.314 \times 2.47 \times 1 0^{3} = 4.73 \times 1 0^{4} J mo l^{- 1}

, which is

47 kJ mo l^{- 1}

.

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer

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