A reaction A(aq)+B(aq)<-->C(aq) has a standard free-energy change of –3.35 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?

dG = -RTlnK. Solve for K. Then

.......A + B ==> C
I..0.30.0.0.40...0
C....-x....-x....x
E..0.30-x..0.40-x..x

Substitute the E line into Keq expression and solve for x, then evaluate concn A and B.
Don't forget to use dG in J/mol and not kJ/mol

Ah, equilibrium...the point at which the reaction gets tired and decides to take a break. Let me crunch some numbers and bring out the clown calculator!

Given that we have a standard free-energy change of -3.35 kJ/mol at 25°C, we can use the equation ΔG = -RT ln(Q), where ΔG is the change in free energy, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.

We know that at equilibrium, the reaction quotient Q is equal to the equilibrium constant K. So let's solve for Q:

ΔG = -RT ln(Q)
-3.35 kJ/mol = -(8.314 J/K/mol)(298 K) ln(Q)

Now, converting kJ to J and solving for ln(Q), we get:

-3350 J/mol = -2476 J/mol ln(Q)
ln(Q) = (-3350 J/mol) / (-2476 J/mol)
ln(Q) ≈ 1.352

To find Q, we can take the exponential of both sides:

Q = e^(ln(Q))
Q ≈ e^1.352
Q ≈ 3.866

Now, let's use the equilibrium expression to relate the concentrations:

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

We can plug in the given concentrations to find K:

K = 0 M / (0.30 M * 0.40 M)
K = 0 / (0.12 M^2)
K = 0

Since K = 0 and Q ≠ K, we know that the reaction is not at equilibrium yet. So, sadly, I can't provide you with the concentrations of A, B, and C at equilibrium. This reaction still has some work to do before it can sit back and relax!

To determine the concentrations of A, B, and C at equilibrium, we can use the concept of the reaction quotient (Q) and compare it to the equilibrium constant (K).

The reaction quotient Q is calculated by dividing the concentrations of the products by the concentrations of the reactants, using the stoichiometric coefficients as exponents.

Q = [C] / ([A] x [B])

Given that the initial concentration of C is 0 M, we can substitute the initial concentrations of A and B into the reaction quotient:

Q = 0 / (0.30 M x 0.40 M)
Q = 0

Since the reaction has a negative free-energy change, it is energetically favorable and the equilibrium constant (K) will be greater than 1.

The equilibrium constant expression for this reaction can be written as K = [C] / ([A] x [B]).

To calculate the concentrations at equilibrium, we can rearrange the expression:

[C] = K x [A] x [B]

Given that K = 0.150 (which can be calculated from the standard free-energy change), and the initial concentrations of A and B are 0.30 M and 0.40 M respectively, we can substitute these values into the expression to obtain the concentrations at equilibrium:

[C] = 0.150 x 0.30 M x 0.40 M
[C] = 0.018 M

Therefore, at equilibrium, the concentrations of A, B, and C are:
[A] = 0.30 M
[B] = 0.40 M
[C] = 0.018 M

To determine the concentrations of A, B, and C at equilibrium, we need to use the concept of equilibrium constants and the reaction quotient. The reaction quotient (Q) is calculated in the same way as the equilibrium constant (K), but it is used to analyze the concentrations of reactants and products at any given time during the reaction, not just at equilibrium.

The equilibrium constant (K) is related to the standard free-energy change (∆G°) through the equation:

∆G° = -RT ln(K)

Where:
- ∆G° is the standard free-energy change (-3.35 kJ/mol in this case),
- R is the gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin (25 + 273 = 298 K),
- ln denotes the natural logarithm.

Rearranging the equation, we have:

ln(K) = -∆G° / RT

Now, let's use the given values to solve for ln(K):

ln(K) = -(-3.35 kJ/mol) / (8.314 J/(mol·K) * 298 K)
ln(K) = 3.35 kJ/mol / (8.314 J/(mol·K) * 298 K)
ln(K) ≈ 0.443

Finding the value of K:
K = e^(ln(K)) = e^(0.443) ≈ 1.556

With the value of K, we can write the expression for the equilibrium constant:

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

At equilibrium, the concentrations of A, B, and C are represented as [A], [B], and [C] respectively. We don't know their exact values, but we can express them in terms of 'x' as follows:

[A] = 0.30 - x
[B] = 0.40 - x
[C] = x

Substituting these values into the equilibrium expression, we get:

1.556 = x / ((0.30 - x) * (0.40 - x))

To solve for 'x', we need to rearrange the equation and solve the resulting quadratic equation. However, let's simplify the equation by assuming that the value of 'x' is small compared to 0.30 and 0.40:

1.556 ≈ x / (0.30 * 0.40)
1.556 ≈ x / 0.12

Now, solve for 'x':

x = 0.12 * 1.556 ≈ 0.18672

Since 'x' represents the concentration of C, we have found that the concentration of C at equilibrium is approximately 0.18672 M. Substituting this value back into the expressions for [A] and [B], we get:

[A] = 0.30 - x = 0.30 - 0.18672 ≈ 0.11328 M
[B] = 0.40 - x = 0.40 - 0.18672 ≈ 0.21328 M

Therefore, the concentrations of A, B, and C at equilibrium are approximately 0.11328 M, 0.21328 M, and 0.18672 M, respectively.

.298, .398, .00177