Respuesta :
Answer: The [tex]\Delta G[/tex] for the reaction is -467.595 kJ/mol
Explanation:
For the given balanced chemical equation:
[tex]2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(g);\Delta G^o=-229kJ/mol[/tex]
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 = [tex]-229kJ\times 2=-458kJ=-485000J[/tex] (Conversion factor: 1kJ = 1000J)
The expression of [tex]K_p[/tex] for the given reaction:
[tex]K_p=\frac{(p_{H_2O})^2}{p_{H_2}^2\times p_{O_2}}[/tex]
We are given:
[tex]p_{H_2O}=1.00atm\\p_{H_2}=4.00atm\\p_{O_2}=3.00atm[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{(1.00)^2}{(4.00)^2\times 3.00}\\\\K_p=0.0208[/tex]
To calculate the Gibbs free energy of the reaction, we use the equation:
[tex]\Delta G=\Delta G^o+RT\ln K_p[/tex]
where,
[tex]\Delta G[/tex] = Gibbs' free energy of the reaction = ?
[tex]\Delta G^o[/tex] = Standard Gibbs' free energy change of the reaction = -458000 J
R = Gas constant = [tex]8.314J/K mol[/tex]
T = Temperature = [tex]25^oC=[25+273]K=298K[/tex]
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = 0.0208
Putting values in above equation, we get:
[tex]\Delta G=-458000J+(8.314J/K.mol\times 298K\times \ln(0.0208))\\\\\Delta G=-467595.14J/mol=-467.595kJ/mol[/tex]
Hence, the [tex]\Delta G[/tex] for the reaction is -467.595 kJ/mol
