Question

Equilibrium concentrations of A, B, and C in A(aq) + B(aq) ⇌ C(aq) with ΔG = –3.35 kJ/mol at 25°C, initial concentrations of A, B, and C are 0.30 M, 0.40 M, and 0 M, respectively.

Answer 1

To find the equilibrium concentrations of species A, B, and C in the reaction \( \text{A(aq)} + \text{B(aq)} \rightleftharpoons \text{C(aq)} \) under the given conditions, we can begin by utilizing the Gibbs free energy change (ΔG) and the initial concentrations.

### Given:
- ΔG = -3.35 kJ/mol
- Temperature = 25°C (298 K)
- Initial concentrations:
- [A] = 0.30 M
- [B] = 0.40 M
- [C] = 0 M

### Step 1: Calculate the equilibrium constant (K)

The relationship between ΔG and the equilibrium constant K is provided by the equation:

\[
\Delta G = -RT \ln K
\]

Where:
- \( R \) = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- \( T \) = 298 K

Rearranging to solve for \( K \):

\[
K = e^{-\Delta G / RT}
\]

Substituting the values:

\[
K = e^{-\left(-3.35 \, \text{kJ/mol}\right) / \left(0.008314 \, \text{kJ/(mol·K)} \cdot 298 \, \text{K}\right)}
\]

Calculating the values in the exponent:

\[
-\Delta G / RT = \frac{3.35}{0.008314 \times 298} \approx \frac{3.35}{2.478} \approx 1.352
\]

Now calculate \( K \):

\[
K = e^{1.352} \approx 3.87
\]

### Step 2: Set up the equilibrium expression

The expression for the equilibrium constant K for the reaction is given by:

\[
K = \frac{[\text{C}]}{[\text{A}][\text{B}]}
\]

At equilibrium, let \( x \) be the concentration of C formed, then the changes in concentrations from initial concentrations will be:

\[
[\text{A}] = 0.30 - x
\]
\[
[\text{B}] = 0.40 - x
\]
\[
[\text{C}] = x
\]

At equilibrium, substituting these into the equilibrium expression gives:

\[
3.87 = \frac{x}{(0.30 - x)(0.40 - x)}
\]

### Step 3: Solve for x

This results in the equation:

\[
3.87 = \frac{x}{(0.30 - x)(0.40 - x)}
\]

Multiplying both sides by \((0.30 - x)(0.40 - x)\):

\[
3.87(0.30 - x)(0.40 - x) = x
\]

Expanding:

\[
3.87(0.12 - 0.70x + x^2) = x
\]

Distributing:

\[
0.4644 - 2.709x + 3.87x^2 = x
\]

Rearranging gives:

\[
3.87x^2 - 3.709x + 0.4644 = 0
\]

### Step 4: Apply the quadratic formula

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Where:
- \( a = 3.87 \)
- \( b = -3.709 \)
- \( c = 0.4644 \)

Calculating the discriminant:

\[
D = b^2 - 4ac = (-3.709)^2 - 4(3.87)(0.4644) \approx 13.76 - 7.198 = 6.562
\]

Calculating \( x \):

\[
x = \frac{3.709 \pm \sqrt{6.562}}{2(3.87)}
\]

Calculating \( \sqrt{6.562} \approx 2.56 \):

So:

\[
x = \frac{3.709 \pm 2.56}{7.74}
\]

Calculating the two possible values for \( x \):

1. \( x = \frac{3.709 + 2.56}{7.74} \approx \frac{6.269}{7.74} \approx 0.81 \) (not possible since concentration cannot exceed initial)
2. \( x = \frac{3.709 - 2.56}{7.74} \approx \frac{1.149}{7.74} \approx 0.148 \)

Thus, we find \( x \approx 0.148 \).

### Step 5: Calculate equilibrium concentrations

Now we can find the equilibrium concentrations:

\[
[\text{A}] = 0.30 - 0.148 \approx 0.152 \, \text{M}
\]
\[
[\text{B}] = 0.40 - 0.148 \approx 0.252 \, \text{M}
\]
\[
[\text{C}] = 0.148 \, \text{M}
\]

### Final Equilibrium Concentrations:

- \([A] \approx 0.152 \, \text{M}\)
- \([B] \approx 0.252 \, \text{M}\)
- \([C] \approx 0.148 \, \text{M}\)

These are the equilibrium concentrations for A, B, and C under the stated conditions.