Question

At 25°, the free energy of formation of gaseous water is -229 kJ/mol. Calculate ΔG for the following reaction if the hydrogen is supplied at 4.00 atm and the oxygen is supplied at 3.00 atm, while the water produced is at 1.00 atm pressure.

Answer 1

Answer: The
`\Delta G` for the reaction is -467.595 kJ/mol

Step-by-step explanation:

For the given balanced chemical equation:

`2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(g);\Delta G^o=-229kJ/mol`

Standard free energy for 1 mole of formation of gaseous water is -229 kJ.

So, the standard free energy for 2 moles of formation of gaseous water will be =
`-229kJ* 2=-458kJ=-485000J` (Conversion factor: 1kJ = 1000J)

The expression of
`K_p` for the given reaction:

`K_p=((p_(H_2O))^2)/(p_(H_2)^2* p_(O_2))`

We are given:

`p_(H_2O)=1.00atm\\p_(H_2)=4.00atm\\p_(O_2)=3.00atm`

Putting values in above equation, we get:

`K_p=((1.00)^2)/((4.00)^2* 3.00)\\\\K_p=0.0208`

To calculate the Gibbs free energy of the reaction, we use the equation:

`\Delta G=\Delta G^o+RT\ln K_p`

where,

`\Delta G` = Gibbs' free energy of the reaction = ?

`\Delta G^o` = Standard Gibbs' free energy change of the reaction = -458000 J

R = Gas constant =
`8.314J/K mol`

T = Temperature =
`25^oC=[25+273]K=298K`

`K_p` = equilibrium constant in terms of partial pressure = 0.0208

Putting values in above equation, we get:

`\Delta G=-458000J+(8.314J/K.mol* 298K* \ln(0.0208))\\\\\Delta G=-467595.14J/mol=-467.595kJ/mol`

Hence, the
`\Delta G` for the reaction is -467.595 kJ/mol