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 = 291 K, 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_(g))` and `Delta n_(g) = 2-1 = 1` in this 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)^(2))/(0.5) xx 0.082 xx 291 = 11.93`
The equilibrium constant is `K_(P) = K_(C )(RT)^(Delta n_(g)) = 1.14 xx (0.082 xx 291) = 27.2` atm
Substituting these values 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) = -19.67` L atm