Question

A first order reaction is 50% complete in 30 minutes at 27°C and in 10 minutes at 47°C. Calculate the reaction rate constant at 27°C and the energy of activation of the reaction in kJ/mol.

Answer 1

From the given data, it is clear that

Half-life of the reaction at 27°C `(t_(1//2)^(27^circ C))` = 30 min.

Half-life of the reaction at 47°C `(t_(1//2)^(47^circ C))` = 10 min.

For a first order reaction, k = `0.693/t_(1//2)`

∴ Rate constant of the reaction at 27°C (k^27°C)

= `0.693/(t_(1//2)^(27^circ C))`

= `0.693/30`

= 0.0231 min⁻¹

Rate constant of the reaction at 47°C (k^47°C)

= `0.693/(t_(1//2)^(47^circ C))`

= `0.693/10`

= 0.0693 min⁻¹

According to the Arrhenius equation,

`log_10  k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`

In the present case,

k₁ = 0.0231 min⁻¹, 

k₂ = 0.0693 min⁻¹, 

T₁ = 27 + 273 = 300 K;

T₂ = 47 + 273 = 320 K and 

R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹.

Substituting the values, we have

`log_10  0.0693/0.0231 = E_a/(2.303 xx 8.314 xx 10^-3) [1/300 - 1/320]`

or, log₁₀ 3 = `E_a/0.01915 xx 20/(300 xx 320)`

or, E_a = `(0.01915 xx 300 xx 320 xx log_10 3)/20`

= `877.1397/20`

= 43.856

= 43.86 kJ mol⁻¹