#### Question

The rate constant for the first order decomposition of H_{2}O_{2} is given by the following equation:

log *k *= 14.34 − 1.25 × 10^{4 }K/*T. *Calculate *E*_{a} for this reaction and at what temperature will its half-period be 256 minutes?

#### Solution

Arrhenius equation is given by,

`k = Ae^((-E_a)/"RT")`

`=>In k = In A - E_a/"RT"`

`=>Ink = lof A - E_a/"RT"`

`=> log k = log A - E_a/(2.303 RT)` ... i

The given equation is

log k = 14.34 - 21.25 x 10^{4} K/T ... ii

From equation (i) and (ii), we obtain

`E_a/(2.303RT) = (1.25xx10^4 K)/T`

`=> E_a = 1.25 xx 10^4K xx 2.303 xxR`

= 1.25 × 10^{4} K × 2.303 × 8.314 J K^{−1} mol^{−1}

= 239339.3 J mol^{−}1 (approximately)

= 239.34 kJ mol^{−1}

Also, when *t*_{1/2} = 256 minutes,

`k = 0.693/t_(1/2)`

= 0.693/256

= 2.707 × 10^{−3} min^{−1}

= 4.51 × 10^{−5} s^{−1}

It is also given that, log *k* = 14.34 − 1.25 × 10^{4} K/*T*

`=>log(4.51xx10^(-5))` = `14.34 - (1.25xx10^(4)K)/T`

`=>log(0.654 - 0.5) = 14.34 - (1.25 xx 10^4K)/T`

`=> (1.25xx10^4K)/T = 18.686`

`=> T = (1.25xx10^4K)/18.686`

= 668.95 K

= 669 K (approximately)