PLEASE ANSWER ALL PARTS The rate constant of a chemical reaction increased from

Question

PLEASE ANSWER ALL PARTS

The rate constant of a chemical reaction increased from 0.100 s1 to 3.20 s1 upon raising the temperature from 25.0 C to 53.0 C .

Part A

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

Express your answer numerically.

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.

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

Explanation / Answer

A) (1/T2 - 1/T1)

(1/326-1/298) = -2.95x10-4

B) i hope this is ln K2/K1

ln (3.2/0.1)

=3.46

C) activation energy =

ln(k2/k1) = -Ea/R(1/T2 - 1/T1)

3.46= -Ea/8.314 (-2.95x10-4)

-1.1748x104 = -Ea/8.314

Ea= 9.7675 x 104 J

Ea= 97.675 K J

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.