Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) the standard change in Gibbs free energy is Δ G ° = − 69.0 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are P NO = 0.500 atm , P O 2 = 0.200 atm , and P NO 2 = 0.600 atm ?

Answer 1

Answer: The $\Delta G$ for the given reaction at 298 K is -64.11 kJ/mol

Explanation:

For the given chemical equation:

$2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$

The equation used to Gibbs free energy of the reaction follows:

$\Delta G=\Delta G^o+RT\ln Q_{p}$

where,

$\Delta G$ = free energy of the reaction

$\Delta G^o$ = Standard Gibbs free energy = -69.0 kJ/mol = -69000 J/mol   (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/K mol

T = Temperature = 298 K

$Q_{p}$ = Ratio of partial pressure of products and reactants = $\frac{(p_{NO_2})^2}{(p_{NO})^2\times p_{O_2}}$

$p_{NO_2}=0.600atm\\p_{NO}=0.500atm\\p_{O_2}=0.200atm$

Putting values in above equation, we get:

$\Delta G=-69000J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(0.600)^2}{(0.500)^2\times 0.200}))\\\\\Delta G=-64109.07J/mol=-64.11kJ/mol$

Hence, the $\Delta G$ for the given reaction at 298 K is -64.11 kJ/mol