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. `- 523.2 kJ mol^(-1)`
B. `- 221.1 kj mol^(-1)`
C. `- 339.3 kJ mol^(-1)`
D. `- 439.3 kJ mol^(-1)`

Answer

C

Explanation

Thermochemical equation for the oxidation of ammonia is give as
`NH_(3) (g) + (3)/(4) O_(2) (g) rarr (1)/(2) N_(2) (g) + (3)/(2) H_(2) O (1)`
we are given `Delta H = - 382.64 kJ mol^(-1)`
`T = 298 K`
`Delta S = - 145.6 J K mol^(-1)`
`= - 0.1456 kJ K^(-1) mol^(-1)`
According to thermodynamics,
`Delta G = Delta H - T Delta S`
`= (- 382. 64 kJ mol^(-1)) - (298 K)`
`(- 0.1456 kJ l^(-1) mol^(-1)`
`= - (382.64 kJ mol^(-1)) + (43.39 kJ mol^(-1))`
`= - 339.25 kJ mol^(-1))`