At 25 ∘ C , the equilibrium partial pressures for the reaction 3 A ( g ) + 4 B ( g ) − ⇀ ↽ − 2 C ( g ) + 3 D ( g ) were found to be P A = 4.47 atm, P B = 5.17 atm, P C = 4.08 atm, and P D = 4.83 atm. What is the standard change in Gibbs free energy of this reaction at 25 ∘ C ?

The standard change in Gibbs free energy of this reaction at 25 ∘ C = 8.77 KJ/mol

Explanation:

                         3 A ( g ) + 4 B ( g ) ⇄ 2 C ( g ) + 3 D ( g )  -------(1)

Given

partial pressures (P A) = 4.47 atm, P B = 5.17 atm, P C = 4.08 atm, and P D = 4.83 atm.

                        From the above equation we can write  

                 ⇒  

                 ⇒  

                 ⇒

From equation (1)

              standard change in Gibbs free energy (ΔG°) = -RTlnKp

                                                                        = - 0.0821 x 298 x ln(0.029)

                                                                        = + 86.6 lit. atm / mole

∵ 1 lit.atm = 101.3 joule/mole

So, + 86.6 lit. atm  or 86.6 x 101.3 = 8772.58 joule/mole = 8.77 KJ/mol

the answer is true. hope it .