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← JEE Main 2022 · 28 Jul · Shift 2
ChemistryInteger+4 / -1Question 58 of 90

JEE Main 2022 · 28 Jul · Shift 2, Chemistry question 58

For a reaction, given below is the graph of ln k vs 1T\dfrac1T. The activation energy for the reaction is equal to ______ cal mol−1^{-1}. (nearest integer)

(Given : R = 2 cal K−1^{-1} mol−1^{-1})

ln k against 1/T (K⁻¹): a straight line from ln k = 20 at 1/T = 0 down to 0 at 1/T = 5.
Answer and worked solution

Official answer: 8

Slope −205=−4=−EaR-\frac{20}5=-4=-\frac{E_a}R, so Ea=8E_a=8 cal mol−1^{-1}.

Answer: 8

Topics:Arrhenius equation
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