The rate constant of a chemical reaction increased from 0.100 s-1 to 3.10 s-1 upon raising the temperature from 25.0 ∘C to 51.0 ∘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.

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.

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

Question

Fred

Physics

28 January, 00:57

The rate constant of a chemical reaction increased from 0.100 s-1 to 3.10 s-1 upon raising the temperature from 25.0 ∘C to 51.0 ∘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.

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.

Part C

What is the activation energy of the reaction?

Express your answer numerically in kilojoules per mole.

+5

Answers (1)

Know the Answer?

Velez · Answer 1

A. (1/T₂ - 1/T₁) = - 2.69 x 10⁻⁴ K⁻¹

B. ln (k₁/k₂) = - 3.434

C. E = 106.13 KJ/mol

Explanation:

Part A:

we have:

T₁ = 25°C = 298 K

T₂ = 51°C = 324 K

(1/T₂ - 1/T₁) = (1/324 k - 1/298 k)

(1/T₂ - 1/T₁) = - 2.69 x 10⁻⁴ K⁻¹

Part B:

ln (k₁/k₂) = ln (0.1/3.1)

ln (k₁/k₂) = - 3.434

Part C:

The activation energy can be found out by using Arrhenius Equation:

ln (k₁/k₂) = E/R (1/T₂ - 1/T₁)

where,

E = Activation Energy

R = General Gas Constant = 8.314 J/mol. k

Therefore,

-3.434 = (E/8.314 J/mol. k) (-2.69 x 10⁻⁴ K⁻¹)

E = 106134.8 J/mol = 106.13 KJ/mol