Question

For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar 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 ° = − 32.8 kJ / mol . What is Δ G for this reaction at 298 K when the partial pressures are:
PNO = 0.500 bar , PO2 = 0.250 bar , and PNO 2 = 0.800 bar
DeltaG = ?

Answer 1

Answer : The value of
`\Delta G_(rxn)` is, -27.0kJ/mole

Explanation :

The formula used for
`\Delta G_(rxn)` is:

`\Delta G_(rxn)=\Delta G^o+RT\ln K_p` ............(1)

where,

`\Delta G_(rxn)` = Gibbs free energy for the reaction

`\Delta G_^o` = standard Gibbs free energy = -32.8 kJ

/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

`K_p` = equilibrium constant

First we have to calculate the value of
`K_p`.

The given balanced chemical reaction is,

`2NO(g)+O_2(g)\rightarrow 2NO_2(g)`

The expression for equilibrium constant will be :

`K_p=((p_(NO_2))^2)/((p_(NO))^2* (p_(O_2)))`

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

`K_p=((0.800)^2)/((0.500)^2* (0.250))`

`K_p=10.24`

Now we have to calculate the value of
`\Delta G_(rxn)` by using relation (1).

`\Delta G_(rxn)=\Delta G^o+RT\ln K_p`

Now put all the given values in this formula, we get:

`\Delta G_(rxn)=-32.8kJ/mol+(8.314* 10^(-3)kJ/mole.K)* (298K)\ln (10.24)`

`\Delta G_(rxn)=-27.0kJ/mol`

Therefore, the value of
`\Delta G_(rxn)` is, -27.0kJ/mole