Answer:
(a) [A] = 0.13 M, [B]= 0.23 M and [C] = 0.17 M.
(b) Option B.
Explanation:
The reaction given:
A(aq) + B(aq) ⇄ C(aq) (1)
<u>Initial:</u> 0.30M 0.40M 0M (2)
<u>Equilibrium:</u> 0.3 - x 0.4 - x x (3)
The equation of Gibbs free energy of the reaction (1) is the following:
(4)
<em>where ΔG°: is the Gibbs free energy change at standard conditions, R: is the gas constant, T: is the temperature and : is the equilibrium constant </em>
(a) To calculate the concentrations of A, B, and C at equilibrium, we need first determinate the equilibrium constant using equation (4), with ΔG°=-4.51x10³J/mol, T=25 + 273 = 298 K, R=8.314 J/K.mol:
(5)
Now, we can calculate the concentrations of A, B, and C at equilibrium using the equilibrium constant calculated (5):
(6)
Solving equation (6) for x, we have two solutions x₁=0.69 and x₂=0.17, and by introducing the solution x₂ into equation (3) we can get the concentrations of A, B, and C at equilibrium:
<u>Notice that the solution x₁=0.69 would have given negative values of the A and B concentrations.</u>
(b) If the reaction had a standard free-energy change of +4.51x10³J/mol, the equilibrium constant would be:
By solving the equation (6) for x, with the equilibrium constant calculated, we can get again two solutions x₁ = 6.9 and x₂= 0.017, and by introducing the solution x₂ into equation (3) we can get the concentrations of A, B, and C at equilibrium:
<u>Again, the solution x₁=6.9 would have given negative values of the A and B concentrations.</u>
Hence, the correct answer is option B: there would be more A and B but less C.
I hope it helps you!