# The reaction N2O5 ---> 2NO2 + 1/2 O2 is first order with a half life of 1307 seconds. How long will it take to decompose 15% of the N2O5?

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### 1 Answer

N2O5 ------>2NO2+1/2O2

The reaction id frist order, then the rate of reaction is,

`r = k[N_2O_5]`

Therefore rate of decomposition of` N_2O_5` is,

`(d([N2O5]))/(dt)= -k[N_2O_5]`

Integrating wrt t,

`(d([N_2O_5]))/[N_2O_5] = -kdt`

`ln([N_2O_5]) = -kt + c`

If at `t = 0, [N_2O_5] = N,` then,

`ln(N) = 0 + c`

`ln([N_2O_5]) = -kt+ln(N)`

Therefore,

`ln([N_2O_5]/N) = -kt`

If the half life is 1307, then time to decompose 50% of N is 1307 seconds,

`ln(1/2) = -k xx 1307`

`k = 5.3 xx 10^(-4)`

Time to decompose 15% of N is,

`ln((100-15)/100) = -5.3 xx 10^(-4) t`

`t = 306.64` seconds

**Therefore time decompose 15% of N_2O_5 is 306.64 seconds.**

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