A reaction: A(aq) + B(aq) <-> C(aq)

has a standard free-energy change of –4.65 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?

Im not even sure where to start. Can you walk me through this please?

Sure! To solve this problem, we'll need to use the concept of equilibrium constant and the expression for the equilibrium constant (K) in terms of concentrations.

Step 1: Write down the balanced chemical equation:
A(aq) + B(aq) ↔ C(aq)

Step 2: Write down the expression for the equilibrium constant (K):
K = [C] / ([A] * [B])

Step 3: Identify the initial concentrations given in the problem:
[A] = 0.30 M (concentration of A)
[B] = 0.40 M (concentration of B)
[C] = 0 M (concentration of C at the beginning)

Step 4: Recall that at equilibrium, the concentrations of reactants and products are related by the equilibrium constant expression.

Since we know K and the initial concentrations, we can use the equation to determine the equilibrium concentrations.

Step 5: Substitute the values into the equilibrium constant expression:
K = [C] / ([A] * [B])

Step 6: Rearrange the equation to solve for [C]:
[C] = K * [A] * [B]

Step 7: Substitute the given values and calculate [C]:
[C] = (-4.65 kJ/mol) * (0.30 M) * (0.40 M)

Note: The units of kJ/mol cancel out, leaving the concentration units of M.

Step 8: Perform the calculation:
[C] = -4.65 * 0.30 * 0.40 M

Step 9: Calculate the value of [C].

[C] = -0.558 M

Step 10: The equilibrium concentrations are:
[A] = 0.30 M
[B] = 0.40 M
[C] = -0.558 M

Note: The negative value for [C] indicates that the concentration of C is very small at equilibrium.

In summary, the equilibrium concentrations of A, B, and C are 0.30 M, 0.40 M, and -0.558 M, respectively.

Sure! To solve this problem, we can use the concept of the equilibrium constant (K) and the standard free-energy change (ΔG°) to find the concentrations of A, B, and C at equilibrium. Here's how you can do it step by step:

Step 1: Write down the balanced chemical equation:
A(aq) + B(aq) ↔ C(aq)

Step 2: Write down the expression for the equilibrium constant (K):
K = [C(aq)] / ([A(aq)] * [B(aq)])

Step 3: Use the relationship between ΔG° and K:
ΔG° = -RT * ln(K)
where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 + 273.15 K) since 25 °C = 298.15 K, and ln denotes natural logarithm.

Step 4: Convert the given ΔG° from kJ to J:
ΔG° = -4.65 kJ/mol = -4650 J/mol

Step 5: Calculate K:
-4650 J/mol = -8.314 J/(mol·K) * 298.15 K * ln(K)
Solve the equation for ln(K):
ln(K) = -4650 J/mol / (-8.314 J/(mol·K) * 298.15 K)
Simplify:
ln(K) = 5.932

Step 6: Take the antilogarithm of both sides to find K:
K = e^(ln(K)) = e^(5.932)
K ≈ 377.93

Step 7: Use the law of mass action to set up an expression for the concentrations at equilibrium:
K = [C(aq)] / ([A(aq)] * [B(aq)])
We know that:
[A(aq)] = 0.30 M (initial concentration of A)
[B(aq)] = 0.40 M (initial concentration of B)
[C(aq)] = unknown (concentration of C at equilibrium)

Step 8: Substitute the known values into the equilibrium expression and rearrange to solve for [C(aq)]:
377.93 = [C(aq)] / (0.30 M * 0.40 M)
377.93 = [C(aq)] / 0.12 M^2
[C(aq)] = 377.93 * 0.12 M^2
[C(aq)] ≈ 45.35 M

So, at equilibrium, the concentrations of A, B, and C will be approximately 0.30 M, 0.40 M, and 45.35 M, respectively.

I can show you how.

dG = -RT*lnK
dG = -4.65 kJ/mol = 4650 J/mol
substitute in the above and solve for K. Then
............A + B ==> C
I..........0.3..0.4...0
C..........-x...-x....x
E........0.3-x...4-x..x
Substitute the E line into the Keq expression (use K from the first calculation with delta G) and solve for x. Then x, 0.3-x and 0.4-x will give you the equilibrium concns.