Answer to Question #112989 in General Chemistry for c19

Question

At 25 ∘C , the equilibrium partial pressures for the reaction

3A(g)+4B(g)↽−−⇀2C(g)+2D(g)

were found to be

Answer 1

This question is broken, but i known the normal task.

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