Question

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 ?

Answer 1

Answer:

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  

                         $K_{p} =\frac{P_{C} ^{2} XP_{D} ^{3}}{P_{A} ^{3}XP_{B} ^{4}}$

                 ⇒  $K_{p} = \frac{(4.08)^{2}(4.83)^{3} }{(4.47)^{3}(5.17)^{4} }$

                 ⇒  $K_{p} = \frac{16.64X112.6}{89.31X714.4}$

                 ⇒ $K_{p}= 0.029$

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