At 25 ∘c25 ∘c, the equilibrium partial pressures for the reaction 3a(g)+2b(g)↽−−⇀c(g)+2d(g) 3a(g)+2b(g)↽−−⇀c(g)+2d(g) were found to be pa=5.84pa=5.84 atm, pb=4.47pb=4.47 atm, pc=4.17pc=4.17 atm, and pd=4.32pd=4.32 atm. what is the standard change in gibbs free energy of this reaction at 25 ∘c25 ∘c ?

9742.37 J/mol

Explanation:

We follow the expression for Gibbs free energy:

First we calculate Q with the following expression for the reaction:

Therefore, Q=0.0196

Now we continue with the first equation, and as we are on equilibrium, we know that the Gibbs free energy is zero, therefore:

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.