Question

The rate constant of a chemical reaction increased from 0.100 s−1s−1 to 3.00 s−1s−1 upon raising the temperature from 25.0 ∘C∘C to 37.0 ∘C∘C .

Part A

Calculate the value of (1T2−1T1)(1T2−1T1) where T1T1 is the initial temperature and T2T2 is the final temperature.

Express your answer numerically.

View Available Hint(s)

(1T2−1T1)(1T2−1T1) =		K−1K−1

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Part B

Calculate the value of ln(k1k2)ln(k1k2) where k1k1 and k2k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.

Express your answer numerically.

View Available Hint(s)

ln(k1k2)ln(k1k2) =	nothing

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Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

View Available Hint(s)

EaEa =	nothing	kJ/molkJ/mol

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Answer 1

A)
Given:
T1 = 25 oC
=(25+273)K
= 298 K
T2 = 37 oC
=(37+273)K
= 310 K

Use:
(1/T1 - 1/T2)
= (1/298 - 1/310)
= 3.3557*10^-3 - 3.2258*10^-3
= 1.299*10^-4 K-1
Answer: 1.30*10^-4 K-1

B)
K1 = 0.1 s-1
K2 = 3 s-1

Use:
ln(K2/K1) = ln(3/0.1)
= ln(30)
= 3.4012
Answer: 3.40

C)
use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
3.4012 = (Ea/8.314)*(1.299*10^-4)
Ea = 217690 J/mol
Ea = 217.69 KJ/mol
Answer: 218 KJ/mol