The following reaction is performed at 298 K. 2NO(g) + O2(g) ⇌ 2NO2(g), Kp = 1.6 × 1012The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) (in J mol–1) at 298 K ?

To solve this problem, we need to use the relationship between the equilibrium constant (Kp) and the standard free energy change (ΔG°) for a reaction. This relationship is given by the equation:

ΔG° = -RT ln(Kp)

where R is the universal gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin.

First, we need to find the standard free energy change for the reaction. We can do this by substituting the given values into the equation:

ΔG° = - (8.314 J/(mol·K)) * (298 K) * ln(1.6 × 10^12) = - (8.314 J/(mol·K)) * (298 K) * 27.63 = -68834.8 J/mol

This is the standard free energy change for the reaction as a whole. However, we want to find the standard free energy of formation of NO2(g).

The standard free energy change for the reaction is the difference between the standard free energies of formation of the products and the reactants. In this case, the reaction is:

2NO(g) + O2(g) ⇌ 2NO2(g)

So, the standard free energy change for the reaction is:

ΔG° = 2ΔGf°(NO2) - 2ΔGf°(NO) - ΔGf°(O2)

We know that the standard free energy of formation of O2(g) is zero, because O2(g) is in its standard state. So, the equation simplifies to:

ΔG° = 2ΔGf°(NO2) - 2ΔGf°(NO)

We can rearrange this equation to solve for the standard free energy of formation of NO2(g):

ΔGf°(NO2) = (ΔG° + 2ΔGf°(NO)) / 2

Substituting the given and calculated values into this equation gives:

ΔGf°(NO2) = (-68834.8 J/mol + 2 * 86600 J/mol) / 2 = 52182.6 J/mol

So, the standard free energy of formation of NO2(g) at 298 K is 52182.6 J/mol.