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

`K_(eq)=3.66`

Step-by-step 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`