The decomposition of A into product has value of k as `4.5 × 10^(3) s^(–1) at 10°C`
and energy of activation `60 kJ mol^(–1)`. At what temperature would k be
`1.5 × 10^(4)s^(–1)`?

Answer

`24^(@)C`

Explanation

`k_(1)=4.5xx10^(3)s^(-1),k_(2)=1.5xx10^(4)s^(-1)`
`T_(1)=10+273=283K,T_(2)=?`
`E_(a)=60kJ mol^(-1)`
Using Arrhenius equation `:`
`log(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`log``(1.5xx10^(4))/(4.5xx10^(3))=(60xx1000J mol^(-1))/(2.303xx8.314 J K^(-1)mol^(-1))((T_(2)=283)/(283T_(2)))`
`or log 3.333=3133.63((T_(2)-283)/(283T_(2)))`
`or (0.5228)/(3133.63)=((T_(2)-283)/(293T_(2)))`
`or 0.0472T_(2)=T_(2)-283`
`or 0.9528T_(2)=283`
`orT_(2)=(283)/(0.9528)=297K=297-273^(@)C=24^(@)C`