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

Step-by-step 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 =
`((p_(NO_2))^2)/((p_(NO))^2* 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* 298K* \ln (((0.600)^2)/((0.500)^2* 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