The rate constant of a chemical reaction increased from 0.100 s1 to 2.70 s1 upon

ID: 573157 • Letter: T

Question

The rate constant of a chemical reaction increased from 0.100 s1 to 2.70 s1 upon raising the temperature from 25.0 C to 41.0 C .

Part A

Part complete

Calculate the value of (1T21T1) where T1 is the initial temperature and T2 is the final temperature.

Express your answer numerically.

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1.71×104

Correct

Part B

Calculate the value of ln(k1k2) 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.

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Submit

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

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(1T21T1) =

1.71×104

Given:
T1 = 25.0 oC
=(25.0+273)K
= 298 K
T2 = 41.0 oC
=(41.0+273)K
= 314 K
K1 = 0.1 s-1
K2 = 2.7 s-1

A)
1/T2 - 1/T1
=1/314 - 1/298
=(3.185*10^-3)-(3.356*10^-3)
= -1.71*10^-4

B)
ln(K1/K2)
= ln(0.1/2.7)
= ln(0.0370)
= -3.296

C)
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
3.2958 = (Ea/8.314)*(1.71*10^-4)
Ea = 160251 J/mol
Ea = 160.2513 KJ/mol
Answer: 160 KJ/mol

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