The rate constant of a chemical reaction increased from 0.100 s^-1 to 2.90s^-1 u

ID: 840824 • Letter: T

Question

The rate constant of a chemical reaction increased from 0.100 s^-1 to 2.90s^-1 upon raising the temperature from 25.0 C to 41.0 C .

Part A

Calculate the value of (1/T2-1/T1) where T1 is the initial temperature and T2 is the final temperature.

Express your answer numerically.

(1T/2-1/T1) =

Part B

Calculate the value of ln(k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

Express your answer numerically. ln(k1/k2) =

Part C

What is the activation energy of the reaction? Express your answer numerically in kilojoules per mole.

Ea =.......... kJ/mol

A) 1/T2 -1/T1 = (1/(273+41) - 1/(273+25))

= -1.71*10^-4

B) ln(k1/k2)

=ln(0.1/2.9)

= -3.367

C) -3.367 =(x/8.31) * (-1.71*10^-4)

or x=163.624 kJ/mol

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