At 25 ∘c , the equilibrium partial pressures for the reaction were found to be pa=5.16 bar, pb=5.04 bar, pc=4.11 bar, and pd=4.85 bar . a(g)+2b(g)↽−−⇀4c(g)+d(g) what is the standard change in gibbs free energy of this reaction at 25 ∘c ? δ∘rxn= kjmol

5.85kJ/Kmol.

Explanation:

The balanced equilibrium reaction is

The expression for equilibrium reaction will be,

Now put all the given values in this expression, we get the concentration of methane.

Relation of standard change in Gibbs free energy and equilibrium constant is given by:

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature =

= equilibrium constant = 10.6

Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.

answer is a

explanation:

a is the correct answer