Question

For the reaction,
$2A_{(g)}+B_{(g)}\to 2D_{(g)}, \Delta U^{\circ}=-10.5\,KJ, \Delta \,S^{\circ}=-44.1 \,J\,K^{-1}$ $\Delta \,G^{\circ}$ for the reaction is

• A. $0.16 \, kJ\,mol^{-1}$
• B. $5.5 \,kJ\,mol^{-1}$
• C. $4.2\,kJ\,mol^{-1}$
• D. $8.52\,kJ\,mol^{-1}$

Answer 1

Answer: (A)

$\Delta\,H^{\circ} = \Delta\,U^{\circ}+\Delta\,n_{g}RT$
$\delta\, U^{\circ} = 10.5\,KJ, R = 8.314\,J\,K^{-1}\,mol^{-1}, T=298\,K$
$\Delta\,n_{g} = 2-(2+1)=-1$
$\therefore \Delta\,H^{\circ}=-10.5 -(1) \times 8.314\times 10^{-3} \times 298$
$-10.5-2.48=-12.98\,KJ\,mol^{-1}$
Now $\Delta \,G^{\circ} = \Delta\,H^{\circ}-T \Delta\,S^{\circ}=-12.98-298 \times (-44.1 \times 10^{-3})$
$=-12.98+13.14=0.16\,kJ\,mol^{-1}$