Question

The rate constant of a reaction at 500°C is 1.6 × 10³ M⁻¹ s⁻¹. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol?

Answer 1

Given:

Rate constant (k) = 1.6 × 10³ M⁻¹s⁻¹,

Temperature (T) = 500 + 273 = 773 K,

Activation energy (E_a) = 56 kJ mol⁻¹ = 56 × 10³ J mol⁻¹

To find:

Frequency factor (A)

Formula:

K = `"Ae"^(-E_a//RT)`

Calculation:

Substituting the given values

`1.6 xx 10^3 M^-1 s^-1 = A xx e^(((-56 xx 10^3 J  mol^-1)/(8.314  J K^-1   mol^-1  xx  773  K)))`

∴ `(1.6 xx 10^3 M^-1 s^-1)/A = "e"^(((-56000)/(8.314 xx 773)))`

∴ `log ((1.6 xx 10^3  M^-1 s^-1)/A) = (-56000)/(8.314 xx 773 xx 2.303)`

∴ `(1.6 xx 10^3 M^-1 s^-1)/A` = antilog(−3.7836)

∴ `(1.6 xx 10^3 M^-1 s^-1)/"A" = 1.646 xx 10^-4`

∴ A = `((1.6 xx 10^3 M^-1 s^-1)/(1.646 xx 10^-4)) = 9.72 xx 10^6 M^-1 s^-1`

The frequency factor of reaction is `9.72 × 10^6  M^-1 s^-1`.