Areaction a(aq)+b(aq)↽−−⇀c(aq) has a standard free‑energy change of −4.51 kj/mol at 25 °c.
what are the concentrations of a, b, and c at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 m, 0.40 m, and 0 m, respectively?

how would your answers change if the reaction had a standard free-energy change of +4.51 kj/mol?

a. all concentrations would be lower.
b. there would be more a and b but less c.
c. there would be less a and b but more c.

how would your answers change if the reaction had a standard free-energy change of +4.51 kj/mol?

a. all concentrations would be lower.
b. there would be more a and b but less c.
c. there would be less a and b but more c.
d. there would be no change to the answers.
e. all concentrations would be higher.

(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)

Initial:            0.30M   0.40M      0M         (2)                

Equilibrium:  0.3 - x     0.4 - x       x           (3)

The equation of Gibbs free energy of the reaction (1) is the following:

   (4)

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

(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:                          

Notice that the solution x₁=0.69 would have given negative values of the A and B concentrations.  

(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:

Again, the solution x₁=6.9 would have given negative values of the A and B concentrations.

Hence, the correct answer is option B: there would be more A and B but less C.

I hope it helps you!

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

one negative thing it that it can kill water animals. another negative thing is it can cause the destruction of ecosystem. one way water pollution can be reduced is to conserve energy.