Chemistry, 28.12.2019 01:31 hrijaymadathil

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.

Answers: 2

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Answers

Answer from: drpharmd1769

(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:

$\Delta G^{\circ} = - RTLn(K_{eq})$ (4)

where ΔG°: is the Gibbs free energy change at standard conditions, R: is the gas constant, T: is the temperature and $K_{eq}$ : 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:

$K_{eq} = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{-\frac{-4.51\cdot 10^{3} J/mol}{8.314 J/K.mol \cdot 298 K}} = 6.17$ (5)

Now, we can calculate the concentrations of A, B, and C at equilibrium using the equilibrium constant calculated (5):

$K_{eq} = \frac{[C]}{[A][B]} = \frac{x}{(0.3 - x)(0.4 - x)}$ (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:

$[A] = 0.3 - x_{2} = 0.3 - 0.17 = 0.13 M$

$[B] = 0.4 - x_{2} = 0.4 - 0.17 = 0.23 M$

[C] = x = 0.17 M

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:

$K_{eq} = e^{-\frac{\Delta G^{\circ}}{RT}} = e^{-\frac{4.51\cdot 10^{3} J/mol}{8.314 J/K.mol \cdot 298 K}} = 0.16$

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:

$[A] = 0.3 - x_{2} = 0.3 - 0.017 = 0.28 M$

$[B] = 0.4 - x_{2} = 0.4 - 0.017 = 0.38 M$

[C] = x = 0.017 M

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!

Answer from: Quest

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Answer from: Quest

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