At a certain temperature the equilibrium constant `K_C` is 0.25 for the reaction `A(g)+B(g)hArrC(g)+D(g)` If we take 1 mole of each of four gases in a 10 litre container , what woluld be the equilibrium concentration of A(g) ?

Options

A. 0.331 M
B. 0.033M
C. 0.133M
D. 1.33M

Answer

To solve the problem step by step, we will follow the equilibrium concept and the relationship between the equilibrium constant \( K_C \) and the concentrations of the reactants and products.

### Step 1: Write the reaction and identify initial concentrations
The reaction is given as:
\[ A(g) + B(g) \rightleftharpoons C(g) + D(g) \]

We are given that initially, we have 1 mole of each gas in a 10-liter container. Therefore, the initial concentration of each gas is:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} = \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \text{ M} \]
So, the initial concentrations are:
- \([A] = 0.1 \, \text{M}\)
- \([B] = 0.1 \, \text{M}\)
- \([C] = 0.1 \, \text{M}\)
- \([D] = 0.1 \, \text{M}\)

### Step 2: Write the expression for the reaction quotient \( Q_C \)
The reaction quotient \( Q_C \) is given by:
\[ Q_C = \frac{[C][D]}{[A][B]} \]
Substituting the initial concentrations:
\[ Q_C = \frac{(0.1)(0.1)}{(0.1)(0.1)} = 1 \]

### Step 3: Compare \( Q_C \) with \( K_C \)
We are given that \( K_C = 0.25 \). Since \( Q_C (1) > K_C (0.25) \), the reaction will shift to the left (towards the reactants) to reach equilibrium.

### Step 4: Set up the equilibrium expression
Let \( x \) be the change in concentration of \( A \) and \( B \) as the reaction shifts to the left. Therefore, at equilibrium:
\[ [A] = 0.1 - x, \quad [B] = 0.1 - x, \quad [C] = 0.1 + x, \quad [D] = 0.1 + x \]

### Step 5: Write the equilibrium expression using \( K_C \)
Substituting the equilibrium concentrations into the \( K_C \) expression:
\[ K_C = \frac{[C][D]}{[A][B]} = \frac{(0.1 + x)(0.1 + x)}{(0.1 - x)(0.1 - x)} = 0.25 \]

### Step 6: Simplify the equation
This can be rewritten as:
\[ \frac{(0.1 + x)^2}{(0.1 - x)^2} = 0.25 \]
Taking the square root of both sides:
\[ \frac{0.1 + x}{0.1 - x} = 0.5 \]

### Step 7: Cross-multiply and solve for \( x \)
Cross-multiplying gives:
\[ 0.1 + x = 0.5(0.1 - x) \]
Expanding and rearranging:
\[ 0.1 + x = 0.05 - 0.5x \]
\[ x + 0.5x = 0.05 - 0.1 \]
\[ 1.5x = -0.05 \]
\[ x = -\frac{0.05}{1.5} = -0.0333 \text{ M} \]

### Step 8: Calculate the equilibrium concentration of \( A \)
Now, substituting \( x \) back to find the equilibrium concentration of \( A \):
\[ [A]_{eq} = 0.1 - x = 0.1 - (-0.0333) = 0.1 + 0.0333 = 0.1333 \text{ M} \]

### Final Answer
The equilibrium concentration of \( A(g) \) is approximately:
\[ \boxed{0.133 \, \text{M}} \]