Following data are obtained for reaction :

N_{2}O_{5} → 2NO_{2} + 1/2O_{2}

t/s | 0 | 300 | 600 |

[N_{2}O_{5}]/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)

#### Solution

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

kt = `ln a_0/a_t`

Given

a_{0} = 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`