Question

Following data are obtained for reaction :

N₂O₅ → 2NO₂ + 1/2O₂

t/s	0	300	600
[N2O5]/mol L–1	1.6 × 10-2	0.8 × 10–2	0.4 × 10–2

1) Show that it follows first order reaction.

2) Calculate the half-life.

(Given log 2 = 0.3010, log 4 = 0.6021)

Answer 1

1) For 1st order reaction the integral rate law is :

kt = `ln  a_0/a_t`

Given

a₀ = 1.6×10⁻² mol L⁻¹

For t = 300 s, a_t = 0.8×10⁻² mol L⁻¹

For t = 600 s, a_t = 0.4×10⁻² mol L⁻¹

Using first set of data in the rate law,

`k xx 300 = ln (1.6 xx 10^(-2))/(0.8xx 10^(-2))`

k = `0.00231 s^(-1)`

Using second set of data in the rate law, 

`k xx 600 = ln  (1.6 xx 10^(-2))/(0.4 xx 10^(-2))`

k = 0.00231 s^-1

The value of k is consistent, therefore it follows first order reaction.

2) The half life of first order reaction is given by the following equation:

`t_(1/2) = (ln 2)/k = 2.303 xx (log 2)/k`

`:. t_(1/2) = 2.303 xx (log 2)/0.00231 = 300.08 s`