Question

The slope of the line in the graph of log k versus `1/T` for the reaction 

\[\ce{N2O -> 2NO2 + \frac{1}{2} O2_{(g)}}\]

is −5000 K. Calculate the energy of activation for the reaction. (R = 8.314 J K⁻¹ mol⁻¹).

Answer 1

Given:

Slope of the graph of log k vs `1T` = –5000 K

R = 8.314 J K⁻¹ mol⁻¹

Use the logarithmic form of the Arrhenius equation:

`log k = log A - E_a/(2.303 R) * 1/T`

So, comparing with the equation of a straight line y = mx + c, the slope is

Slope = `-E_a/(2.303 R)`

∴ `5000 = E_a/(2.303 xx 8.314)`

⇒ E_a ​= 5000 × 2.303 × 8.314

E_a ​= 5000 × 19.147

= 95,735 J/mol

E_a ≈ 95.7 kJ/mol