Standard enthalpy and standard entropy change for the oxidation of `NH_(3)` at `298K` are `-382.64KJ mol^(-1)` and `145.6Jmol^(-1)` respectively. Standard free energy change for the same reaction at `298K` is

Options

A. `-221.1 KJ "mol"^(-1)`
B. `-339.3 KJ "mol"^(-1)`
C. `-439.3 KJ "mol"^(-1)`
D. `-523.2 KJ "mol"^(-1)`

Answer

B

Explanation

`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)" "…(i)`
Given that , `DeltaH^(@)=-382.64 KJ "mol"^(-1)`
`DeltaS^(@)=-145.6 JK^(-1) "mol"^(-1)`
`=-145.6xx10^(-3) KJ K^(-1)`
T=298 K
On putting the given values in eq. (i) we get , or `DeltaG^(@)=-382.64-[298xx(-145.6xx10^(-3))]`
`=-339.3 KJ "mol"^(-1)`