Question

For a reaction with ΔH∘=−24 kcal/mol and ΔS∘=0.07 kcalmol−1K−1, calculate the equilibrium constant at: (1.) 30 ∘C and (2.) 150 ∘C.

Answer 1

(1) We know that ΔG∘ = ΔH∘ - T ΔS∘

Given ΔH∘ = -24 kCal/mol =

        ΔS∘ = 0.07 kCal/mol-K

         T = temperature in kelvin = 30 oC = 30+273 = 303 K

Plug the values we get ΔG∘ = -24 - ( 303x0.07)

                                         = -45.21 Kcal/mol

                                         = -45.21x10³ Cal /mol

We know that ΔG∘ = -RT ln K

Where R = gas constant = 2 Cal/mol-K

T = temperature = 303 K

K = Equilibrium constant = ?

Plug the values we get ln K = -ΔG∘/ (RT)

                                        = -(-45.21x10³) / ( 2x303)

                                      = 74.6

                                    K = e^74.6

                                       = 2.51x10³²

Simillarly do the remaining