For the equilibrium at 298 K: `N_2O_(4(g)) hArr 2NO_(2(g)), G_(N_2O_4)^@ = 100 "kJ mol"^(-1)` and `G_(NO_2)^@ = 50 "kJ mol"^(-1)` . If 5 moles of `N_2O_4` and 2 moles of `NO_2` are taken initially in one litre container then which statements are correct?

Options

A. Reaction proceeds in forward direction
B. `K_c=1`
C. `DeltaG`= - 0.55 kJ , `DeltaG^@=0`
D. At equilibrium `[N_2O_4]` = 4.894 M and `[NO_2]`= 2.212 M

Answer

A, B, C, D

Explanation

`DeltaG=DeltaG^@ + 2.303 RT log Q`
`DeltaG^@ = 2xxG_(NO_2)^@ - G_(N_2O_4)^@ = 2 xx 50 -100 =0`
`therefore DeltaG=0+ 2.303 xx 8.314 xx 10^(-3) xx 298 "log" 2^2/5`
= 0-0.55 kJ
`therefore DeltaG`=- 0.55 kJ, i.e., reaction proceeds in forward direction .
Also, `DeltaG^@=0=-2.303 RT log K_c therefore K_c=1`
Now, `{:(N_2O_4,hArr , 2NO_2),(5,,2),((5-x),,(2+2x)):}`
`therefore 1=(2+2x)^2/((5-x))` or x=0.106
`[NO_2]`= 2+ 2x = 2+ 2 x 0.106 = 2.212 M
`[N_2O_4]` = (5-x) =5- 0.106 = 4.894 M