This is an incomplete question, here is a complete question.
Calculate the free energy ΔG at 25 °C for the nonstandard conditions at point B where the reaction quotient Q is 2.75 × 10⁻⁵.
In solving Part A, we found that ΔG = -40.82 kJ/mol
Answer : The value of [tex]\Delta G[/tex] is, -66.8 kJ/mol
Explanation :
The expression for free energy is:
[tex]\Delta G=\Delta G^o+RT\ln Q[/tex]
where,
[tex]\Delta G[/tex] = free energy = ?
[tex]\Delta G_^o[/tex] = standard Gibbs free energy = -40.82 kJ/mol
R = gas constant = [tex]8.314\times 10^{-3}kJ/mole.K[/tex]
T = temperature = 298 K
Q = equilibrium constant = 2.75 × 10⁻⁵
Now put all the given values in the above formula, we get:
[tex]\Delta G=-40.82kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (2.75\times 10^{-5})[/tex]
[tex]\Delta G=-66.8kJ/mol[/tex]
Therefore, the value of [tex]\Delta G[/tex] is, -66.8 kJ/mol
