The equilibrium constant `K_(C)` for the reaction `N _(2) O _(4) hArr 2 NO _(2)` in chloroform at 291 Kis 1, 14. Calculate the free energy change of the reaction when the concentration of the two gases is 0.5 mol `dm^(-3)` each at the same temperature. (`R=0.082 L atm K^(-1)'mol^(-1)`)

Options

A. `-54.95 L atm `
B. `-38.94 L atm`
C. `-27.2 L atm`
D. `-19.67 L atm`

Answer

D

Explanation

From the given data, we have `T = 291K,R = 0.082 L atm K ^(-1) mol ^(-1) K _(C) = 1.14 ,C = 0.5 mol dm ^(-3)`
As `Q _(P) = Q _(C) (RT) ^( Delta n) and Delta n _(g) = 2 -1 = 1` in the case the reaction quotient `Q _(c)` for the reaction is
`Q _(P )= ([ NO _(2) ] ^(2))/( [N_(2) O _(4)]) xx 0.082 xx 291 = ((0.5))/(0.5 ) xx 0.082 xx 291= 11.93`
The equilibrium constatn is `K _(P) = K _(C) (RT) ^( Delta n _(g)) = 1. 14 xx (0.082 xx 291) = 27.2 atm`
Sumstituting these vlue in the equation `Delta G = Delta G ^(@) + RT ln Q _(P) = RT ln K _(P)+ RT ln Q _(P)`
`=-2.303 RT ( log K _(P) - log Q _(P)), ` we get `Delta G =- (0.082 xx 291 xx 2. 303)[log 27.2- log 11. 93]`
`=- 54.95 (1. 4346- 1. 0766 ) =- 1967 L atm`