Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction N2(g) + 3 H2(g) = 2 NH3(g) the standard change in Gibbs free energy is AGⓇ = -72.6 kJ/mol. What is AG for this reaction at 298 K when the partial pressures are PN, = 0.200 bar, Ph, = 0.150 bar, and PnHz = 0.800 bar. kJ AG = mol

Answer 1

$\Delta G$ = $\Delta G^{0}$ + RTlnQ

Q =
_{PH Pi, x PN2}

Putting the given values

Q = = 948.15

(0.800) (0.150)3 x (0.200)

$\Delta G^{0}$ = - 72.6 KJ/mol.

So,

$\Delta G$ = - 72×1000 + 8.314 × 298 × ln (948.15)

Or, $\Delta G$ = - 72000 + (8.314 × 298 × 6.854)

Or, $\Delta G$ = - 72000 + 16982 J/mole

Or, $\Delta G$ = - 55018 J/mole.

Or, $\Delta G$ = - 55.018 KJ/mole.

Hence, $\Delta G$ = -55.02 KJ/mole ( 4 significant figure).