Question

The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

T/°C	0	20	40	60	80
105 × k/s−1	0.0787	1.70	25.7	178	2140

Draw a graph between ln k and `1/T` and calculate the values of A and E_a. Predict the rate constant at 30º and 50ºC.

Answer 1

The rate constants for the decomposition of N₂O₅ at different temperatures are shown below.

T(°C)	T(K)	`1/T`	k(s−1)	ln k (= 2.303 log k)
0	273	3.6 × 10−3	7.87 × 10−7	−14.06
20	293	3.4 × 10−3	1.70 × 10−5	−10.98
40	313	3.19 × 10−3	25.7 × 10−5	−8.266
60	333	3.00 × 10−3	178 × 10−5	−6.332
80	353	2.8 × 10−3	2140 × 10−5	−3.844

Slope of the line = tan θ

= `(y_2 - y_1)/(x_2 - x_1)`

= `(-10.98 - (-14.06))/(3.4 - 3.6) xx 10^3`

= `3.08/-0.2 xx 10^3`

= −15.4 × 10³

E_a = −slope × R

= −(−15.4 × 10³ × 8.314)

= 128.035 kJ K⁻¹ mol⁻¹

Again,

ln A = `ln k + E_a/(RT)`

= `-14.06 + (128.035 xx 10^3)/(8.314 xx 273)` 

= `-14.06 + 128035/2269.722`

= −14.06 + 56.40

ln A = 42.34

⇒ A = antilog 42.34

= 0.3388 × 10¹⁹ 

Values of the rate constant k at 303 K and 323 K can be obtained from the graph.

First, k is obtained corresponding to `1/(303 K) and 1/(323 K)`, and then k is calculated.