JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Chemical Kinetics

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It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ______

kJ mol^{- 1}

. [nearest integer]
(Given

ln 10 = 2.3

R = 8.3 JK^{- 1} mol^{- 1}

lo g 2 = 0.30

)

Answer: 59

Solution

Using the Arrhenius equation in logarithmic form:

lo g_{10} \frac{K _{2}}{K _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})

Given:

K_{2} = 2 K_{1}

T_{1} = 300

T_{2} = 309

lo g_{10} (2) = \frac{E _{a}}{2.303 \times 8.3} (\frac{9}{300 \times 309})

0.30 = \frac{E _{a} \times 9}{2.303 \times 8.3 \times 300 \times 309}

E_{a} = \frac{0.30 \times 2.303 \times 8.3 \times 300 \times 309}{9} = 59065

\approx 59

kJ mol

^{- 1}

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer

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