For the reaction, `A(g)+2B(g)hArr2C(g)` one mole of A and 1.5 mol of B are taken in a 2.0 L vessel. At equilibrium, the concentration of C was found to be 0.35 M. The equilibrium constant `(K_(c))` of the reaction would be

Options

A. `0.295 M^(-1)`
B. `0.673 M^(-1)`
C. `1.178 M^(-1)`
D. `2.36 M^(-1)`

Answer

To solve the problem, we need to determine the equilibrium constant \( K_c \) for the reaction:

\[ A(g) + 2B(g) \rightleftharpoons 2C(g) \]

### Step 1: Determine Initial Concentrations
We start with 1 mole of \( A \) and 1.5 moles of \( B \) in a 2.0 L vessel.

- Initial concentration of \( A \):
\[ [A]_0 = \frac{1 \text{ mol}}{2.0 \text{ L}} = 0.5 \, M \]

- Initial concentration of \( B \):
\[ [B]_0 = \frac{1.5 \text{ mol}}{2.0 \text{ L}} = 0.75 \, M \]

- Initial concentration of \( C \):
\[ [C]_0 = 0 \, M \]

### Step 2: Set Up the Change in Concentrations
Let \( x \) be the amount of \( A \) that reacts at equilibrium. According to the stoichiometry of the reaction:

- Change in concentration of \( A \): \( -x \)
- Change in concentration of \( B \): \( -2x \)
- Change in concentration of \( C \): \( +2x \)

### Step 3: Write Equilibrium Concentrations
At equilibrium, the concentrations will be:

- Concentration of \( A \):
\[ [A] = 0.5 - x \]

- Concentration of \( B \):
\[ [B] = 0.75 - 2x \]

- Concentration of \( C \):
\[ [C] = 2x \]

### Step 4: Use Given Information
We know that at equilibrium, the concentration of \( C \) is 0.35 M:

\[ 2x = 0.35 \implies x = 0.175 \]

### Step 5: Calculate Equilibrium Concentrations
Now we can substitute \( x \) back into the expressions for \( [A] \) and \( [B] \):

- Concentration of \( A \):
\[ [A] = 0.5 - 0.175 = 0.325 \, M \]

- Concentration of \( B \):
\[ [B] = 0.75 - 2(0.175) = 0.75 - 0.35 = 0.4 \, M \]

### Step 6: Calculate the Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:

\[ K_c = \frac{[C]^2}{[A][B]^2} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(0.35)^2}{(0.325)(0.4)^2} \]

Calculating \( K_c \):

1. Calculate \( (0.35)^2 = 0.1225 \)
2. Calculate \( (0.4)^2 = 0.16 \)
3. Now calculate \( K_c \):

\[ K_c = \frac{0.1225}{0.325 \times 0.16} = \frac{0.1225}{0.052} \approx 2.35 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) is approximately:

\[ K_c \approx 2.35 \]

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