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 step by step, we will analyze the equilibrium reaction and calculate the equilibrium constant \( K_c \).

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ A(g) + 2B(g) \rightleftharpoons 2C(g) \]

### Step 2: Determine initial moles and concentrations
Given:
- Initial moles of \( A = 1 \) mole
- Initial moles of \( B = 1.5 \) moles
- Volume of the vessel = 2.0 L

To find the initial concentrations:
- Concentration of \( A \) at \( t = 0 \):
\[ [A] = \frac{1 \text{ mole}}{2.0 \text{ L}} = 0.5 \, M \]
- Concentration of \( B \) at \( t = 0 \):
\[ [B] = \frac{1.5 \text{ moles}}{2.0 \text{ L}} = 0.75 \, M \]

### Step 3: Set up the change in concentration at equilibrium
Let \( x \) be the amount of \( A \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- For \( A \): \( [A] = 0.5 - x \)
- For \( B \): \( [B] = 0.75 - 2x \)
- For \( C \): \( [C] = 0 + 2x \)

### Step 4: Use the given equilibrium concentration of \( C \)
We are given that the concentration of \( C \) at equilibrium is \( 0.35 \, M \):
\[ [C] = 2x = 0.35 \]
From this, we can solve for \( x \):
\[ x = \frac{0.35}{2} = 0.175 \]

### Step 5: Calculate the equilibrium concentrations of \( A \) and \( B \)
Now, substituting \( x \) back to find the concentrations of \( A \) and \( B \):
- Concentration of \( A \):
\[ [A] = 0.5 - x = 0.5 - 0.175 = 0.325 \, M \]
- Concentration of \( B \):
\[ [B] = 0.75 - 2x = 0.75 - 2(0.175) = 0.75 - 0.35 = 0.4 \, M \]

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2}{[A][B]^2} \]

### Step 7: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we found:
\[ K_c = \frac{(0.35)^2}{(0.325)(0.4)^2} \]

### Step 8: Calculate \( K_c \)
Calculating the numerator and denominator:
- Numerator:
\[ (0.35)^2 = 0.1225 \]
- Denominator:
\[ (0.4)^2 = 0.16 \]
\[ K_c = \frac{0.1225}{0.325 \times 0.16} = \frac{0.1225}{0.052} \approx 2.356 \]

### Final Result
Thus, the equilibrium constant \( K_c \) is approximately:
\[ K_c \approx 2.356 \, M^{-1} \]