A reaction Alaa)+Bag)clag has a standard free-energy change of-5.02 kJ/mol at 25

Question

A reaction Alaa)+Bag)clag has a standard free-energy change of-5.02 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? Number Number Number How would your answers above change if the reaction had a standard free-energy change of +5.02 kJlmol? O All concentrations would be highet O There would be no change to the answers. O There would be less A and B but more C. O There would be more A and B but less C. O All concentrations would be lower.

Explanation / Answer

initially we must calculate the equilibrium constant of the K reaction:

K = exp [- (?G / R * T)] = exp [- (- 5020 J / mol / 8.314 J / mol K * 298 K)] = 7.59

Next, starting from the equation of K in equilibrium, we have an expression:

K = [C] / [A] * [B]

Defining each of them, taking A as a limit reagent:

[A] = [Ao] * (1-x)

[B] = [Ao] * (1-x)

[C] = [Ao] * x

replacing:

K = [Ao] * x / [Ao] * (1-x) * [Ao] * (1-x)

Fixing the equation we have left:

x / (1-x) ^ 2 = [Ao] * K = 2.28 = Da

We clear to calculate the X and we have a polynomial:

DaX ^ 2 + (-2Da + 1) X + Da = 0

and solving the equation of the second degree, we have:

X = 0.5217

and our equilibrium concentrations are calculated with the deduced equations:

[A] = 0.1435 M

[B] = 0.1435 M

[C] = 0.1565 M

For the case that ?G = 5.02 Kj / mol we perform the same calculations and we have:

K = 0.1318

Da = 0.03954

X = 0.0367

[A] = 0.289 M

[B] = 0.289 M

[C] = 0.011 M

The concentrations of A and B are greater and that of C decreases.

initially we must calculate the equilibrium constant of the K reaction:

K = exp [- (?G / R * T)] = exp [- (- 5020 J / mol / 8.314 J / mol K * 298 K)] = 7.59

Next, starting from the equation of K in equilibrium, we have an expression:

K = [C] / [A] * [B]

Defining each of them, taking A as a limit reagent:

[A] = [Ao] * (1-x)

[B] = [Ao] * (1-x)

[C] = [Ao] * x

replacing:

K = [Ao] * x / [Ao] * (1-x) * [Ao] * (1-x)

Fixing the equation we have left:

x / (1-x) ^ 2 = [Ao] * K = 2.28 = Da

We clear to calculate the X and we have a polynomial:

DaX ^ 2 + (-2Da + 1) X + Da = 0

and solving the equation of the second degree, we have:

X = 0.5217

and our equilibrium concentrations are calculated with the deduced equations:

[A] = 0.1435 M

[B] = 0.1435 M

[C] = 0.1565 M

For the case that ?G = 5.02 Kj / mol we perform the same calculations and we have:

K = 0.1318

Da = 0.03954

X = 0.0367

[A] = 0.289 M

[B] = 0.289 M

[C] = 0.011 M

The concentrations of A and B are greater and that of C decreases.

initially we must calculate the equilibrium constant of the K reaction:

K = exp [- (?G / R * T)] = exp [- (- 5020 J / mol / 8.314 J / mol K * 298 K)] = 7.59

Next, starting from the equation of K in equilibrium, we have an expression:

K = [C] / [A] * [B]

Defining each of them, taking A as a limit reagent:

[A] = [Ao] * (1-x)

[B] = [Ao] * (1-x)

[C] = [Ao] * x

replacing:

K = [Ao] * x / [Ao] * (1-x) * [Ao] * (1-x)

Fixing the equation we have left:

x / (1-x) ^ 2 = [Ao] * K = 2.28 = Da

We clear to calculate the X and we have a polynomial:

DaX ^ 2 + (-2Da + 1) X + Da = 0

and solving the equation of the second degree, we have:

X = 0.5217

and our equilibrium concentrations are calculated with the deduced equations:

[A] = 0.1435 M

[B] = 0.1435 M

[C] = 0.1565 M

For the case that ?G = 5.02 Kj / mol we perform the same calculations and we have:

K = 0.1318

Da = 0.03954

X = 0.0367

[A] = 0.289 M

[B] = 0.289 M

[C] = 0.011 M

The concentrations of A and B are greater and that of C decreases.

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