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.

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 is, -66.8 kJ/mol

Explanation :

The expression for free energy is:

where,

= free energy = ?

=  standard Gibbs free energy  = -40.82 kJ/mol

R = gas constant =

T = temperature = 298 K

Q = equilibrium constant = 2.75 × 10⁻⁵

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

Therefore, the value of is, -66.8 kJ/mol

neither accurate nor precise.