Question

The equilibrium constant of a system, K, can be related to the standard free energy change, ΔG∘, using the following equation:

ΔG∘=−RTlnK

where T is a specified temperature in kelvins (usually 298 K) and R is equal to 8.314 J/(K⋅mol).

Under conditions other than standard state, the following equation applies:

ΔG=ΔG∘+RTlnQ

In this equation, Q is the reaction quotient and is defined the same manner as K except that the concentrations or pressures used are not necessarily the equilibrium values.

At 25 ∘C the reaction from Part A has a composition as shown in the table below. Substance Pressure (atm)

C2H2(g) 5.25

H2(g) 4.35

C2H6(g) 4.25×10−2

What is the free energy change, ΔG, in kilojoules for the reaction under these conditions?

Answer 1

C2H2 + 2H2------------> C2H6

ΔG∘rxn   = ΔG⁰f products - ΔG⁰f reactants

              = -32.82 -(209.2+2*0)   = -242.02KJ/mole   = -242020J/mole

Qp   = PC2H6/PC2H2*P^2H2

         = 4.25*10^-2 /5.25*(4.35)^2

         = 0.0004278

ΔG=ΔG∘+RTlnQ

      = -242020+8.314*298ln0.0004278

     = -242020 + 8.314*298*-7.7568

     = -261238J/mole

    = -261.238KJ/mole >>>>answer