We’re being asked to determine the **equilibrium constant (K)** at **298.15 K** for the given reaction:

3 NO_{(g)} → N_{2}O_{(g) }+ NO_{2}_{(g)}

We can use the following equation to solve for ** ΔG˚_{rxn}**:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{prod}}{\mathbf{-}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{react}}}$

Given the values of ΔG_{f}° given below in kJ/mol, calculate the value of ΔG° in kJ for the reaction:

3 NO(g) => N_{2}O(g) + NO_{2}(g)

ΔG_{f}° (NO) = 84

ΔG_{f}° (NO_{2}) = 48

ΔG_{f}° (N_{2}O) = 107

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