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← JEE Main 2022 · 27 Jun · Shift 2
ChemistryInteger+4 / -1Question 88 of 90

JEE Main 2022 · 27 Jun · Shift 2, Chemistry question 88

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^{-1}. [nearest integer]

(Given ln⁡10=2.3\ln10=2.3, R=8.3R=8.3 J K−1^{-1} mol−1^{-1}, log⁡2=0.30\log2=0.30)

Answer and worked solution

Official answer: 59

Arrhenius. ln⁡2=EaR(1300−1309)\ln2=\dfrac{E_a}R\left(\dfrac1{300}-\dfrac1{309}\right), so Ea=0.69×8.3×300×3099≈59E_a=\dfrac{0.69\times8.3\times300\times309}9\approx59 kJ mol−1^{-1}.

Answer: 59