Calculate `K_c` for the reaction: `A(g) +B(g) iff 2C(g)` , if 1 mole of A, 1.4 moles of B and 0.50 mole of C are placed in a one-litre vessel and allowed to reach equilibrium. The equilibrium concentration of C is 0.75 mole per litre.

Answer

To calculate the equilibrium constant \( K_c \) for the reaction \( A(g) + B(g) \iff 2C(g) \), we will follow these steps:

### Step 1: Write the initial concentrations
Initially, we have:
- \( [A] = 1 \, \text{mol/L} \)
- \( [B] = 1.4 \, \text{mol/L} \)
- \( [C] = 0.5 \, \text{mol/L} \)

### Step 2: Set up the change in concentrations
Let \( x \) be the amount of \( A \) and \( B \) that react to form \( C \). The changes in concentration can be expressed as:
- \( [A] \) decreases by \( x \): \( [A] = 1 - x \)
- \( [B] \) decreases by \( x \): \( [B] = 1.4 - x \)
- \( [C] \) increases by \( 2x \): \( [C] = 0.5 + 2x \)

### Step 3: Use the equilibrium concentration of \( C \)
We know that at equilibrium, the concentration of \( C \) is \( 0.75 \, \text{mol/L} \):
\[ 0.5 + 2x = 0.75 \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = 0.75 - 0.5 = 0.25 \\ x = \frac{0.25}{2} = 0.125 \]

### Step 5: Calculate the equilibrium concentrations
Now we can find the equilibrium concentrations of \( A \) and \( B \):
- \( [A] = 1 - x = 1 - 0.125 = 0.875 \, \text{mol/L} \)
- \( [B] = 1.4 - x = 1.4 - 0.125 = 1.275 \, \text{mol/L} \)
- \( [C] = 0.75 \, \text{mol/L} \) (given)

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

### Step 7: Substitute the equilibrium concentrations into the \( K_c \) expression
\[ K_c = \frac{(0.75)^2}{(0.875)(1.275)} \]

### Step 8: Calculate \( K_c \)
Calculating the values:
\[ K_c = \frac{0.5625}{(0.875)(1.275)} = \frac{0.5625}{1.115625} \approx 0.504 \]

### Final Result
Thus, the equilibrium constant \( K_c \) for the reaction is approximately \( 0.50 \).

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