ering /assignment/6073077 Secure I ht Course Home UNIT 2 HOMEWORK-Cheomium com/m

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ering /assignment/6073077 Secure I ht Course Home UNIT 2 HOMEWORK-Cheomium com/myct/itemView?assignment ProblemlD-95885403 i Secure https UNIT 2: HOMEWORK Exercise 13.62 Constants I Periodic Table The rate constant (k) for a reaction was measured as a function of temperature. A plot of In k versus 1/Tin K) is linear and has a slope of -9.70x 103 K Part A Calculate the activation energy for the reaction. Express your answer with the appropriate units. E. Value Units Submit Provide Feedback here to search

Explanation / Answer

Solution:- ln k vs 1/T is a Arrhenius plot to determine activation energy.

lnk = - (1/T)Ea/R + lnA

on comparing this equation with y = mx + c

Slope = -Ea/R

Ea = -slope(R)

Where R is the universal gas conatant and it's value is 8.314 x 10-3 kJ/mol.K and Ea is activation energy.

Ea = -(-9.70 x 103 K )(8.314 x 10-3 kJ/mol.K)

Ea = 80.6 kJ/mol

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