Question

Calculate the value of the equilibrium constant (K) at 25 °C for the reaction
2 NOBr(g) = N2(g) + O2(g) + Br2()
given that the standard free energy of formation (AG°f) of NOBr(g) is 82.4 kJ/mol​

Answer 1

Answer:

         $K_{eq}=3.66$

Explanation:

The equilibrium constant, Ke, at certain temperature, T, and the standard free energy of formation (ΔG°f) are related by  the equation:

        $\Delta G^0_f=-RT\ln K_{eq}$

From which you can obtain:

        $K_{eq}=e^{-\Delta G^0_f/RT}$

Substituting ΔG°f = 82.4kJ/mol, T = 25 + 273.15 K = 298.15 K and R = 0.008314 kJ·mol⁻¹ K⁻¹:

        $K_{eq}=e^{-82.4kJ.mol^{-1}/(0.008314kJ\cdot mol^{-1}\cdot 298.15K)}$

         $K_{eq}=3.6588$

Rounding to three significant figures:

         $K_{eq}=3.66$