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?

For the equilibrium at 298 K: N_2O_(4(g)) hArr 2NO_(2(g)), G_(N_2O_4)^

1.	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?
Answer» Reaction proceeds in forward direction `K_c=1` `DELTAG`= - 0.55 kJ , `DeltaG^@=0` At equilibrium `[N_2O_4]` = 4.894 M and `[NO_2]`= 2.212 M Solution :`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