Question

The reaction N${}_2$O${}_4$ (g) $\rightleftharpoons$ 2NO${}_2$ (g) is
carried out at 298 K and 20 bar. Five moles of each of N${}_2$O${}_4$and
NO${}_2$, are taken initially. Given: $\Delta$${}_f$G${}^{\circ }$
(N${}_2$O${}_4$) = 100 kJ mol${}^{-1}$ and $\Delta$${}_f$G${}^{\circ }$
(NO${}_2$) = 50 kJ mol${}^{-1}$.Choose the appropriate option: The values of
$\Delta$G and K${}^{\circ }$${}_p$ at 298 K are:

• A. G = 0, = 10 bar; The reaction is in equilibrium
• B. 5.7058 kJ, = 1 bar; The reaction goes in the reverse direction
• C. -5.7058 kJ, = 1 bar; The reaction goes in the forward direction
• D. G = 0, = 1 bar; The reaction is in equilibrium

Answer 1

The correct option is B

5.7058 kJ, = 1 bar; The reaction goes in the reverse direction

Some thermodynamic data is given.

$\Delta$G = $\Delta$G${}^{\circ }$ + RTlnQ${}^{\circ }$${}_p$- (1)

R = gas constant. T = temperature; Q${}^{\circ }$${}_p$ = reaction quotient

Q${}^{\circ }$${}_p$ = $\frac{(P_{N{O}_2}{)}^2}{\left(P_{N_2{O}_4}\right)}$ = $\frac{{10}^2}{10}$ =
10 bar

$\Delta$G${}^{\circ }$ (reaction) = 2 $\Delta$G${}^{\circ }$(NO${}_2$) -
$\Delta$G${}^{\circ }$(N${}_2$O${}_4$) = 2 (50 kJ mol${}^{-1}$) - 100 kJ
mol${}^{-1}$ = 0

$\Delta$G = $\Delta$G${}^{\circ }$ + RTlnQ${}^{\circ }$${}_p$ = 0 +
(8.314 J K${}^{-1}$ mol${}^{-1}$) (298 K) In (10) = (8.314 x 298 x 2.303)
Jmol${}^{-1}$ = 5.7058 kJmol${}^{-1}$

$\Delta$G${}^{\circ }$= - RT InK${}^{\circ }$${}_p$ = 0 (from calculation)

This implies that K${}^{\circ }$${}_p$ = 1 bar; Earlier Q${}^{\circ }$${}_p$ =
10 bar.

As Q${}^{\circ }$${}_p$$>$K${}^{\circ }$${}_p$ the reaction will proceed in
the reverse direction. Degree of dissociation will decrease