At a certain temperature the equilibrium constant `K_(c)` is 0.25 for the reaction
`A_(2)(g)+B_(2)(g)hArrC_(2)(g)+D_(2)(g)`
If we take 1 mole of each of the four gases in a 10 litre container ,what would be equilibrium concentration of `A_(2)` (g)?

Options

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

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ A_2(g) + B_2(g) \rightleftharpoons C_2(g) + D_2(g) \]

### Step 2: Determine initial concentrations
We are given that 1 mole of each gas is placed in a 10-liter container. Therefore, the initial concentrations of each gas can be calculated as:
\[ \text{Initial concentration} = \frac{\text{Number of moles}}{\text{Volume (L)}} = \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \text{ M} \]
Thus, the initial concentrations are:
- \([A_2] = 0.1 \, \text{M}\)
- \([B_2] = 0.1 \, \text{M}\)
- \([C_2] = 0.1 \, \text{M}\)
- \([D_2] = 0.1 \, \text{M}\)

### Step 3: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[C_2][D_2]}{[A_2][B_2]} \]
Given that \( K_c = 0.25 \).

### Step 4: Set up the change in concentrations at equilibrium
Let \( x \) be the amount of \( A_2 \) and \( B_2 \) that reacts at equilibrium. Therefore, the changes in concentration will be:
- \([A_2] = 0.1 - x\)
- \([B_2] = 0.1 - x\)
- \([C_2] = 0.1 + x\)
- \([D_2] = 0.1 + x\)

### Step 5: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the equilibrium concentrations into the \( K_c \) expression gives:
\[ 0.25 = \frac{(0.1 + x)(0.1 + x)}{(0.1 - x)(0.1 - x)} \]

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

### Step 7: Cross-multiply and solve for \( x \)
Cross-multiplying gives:
\[ 0.5(0.1 - x) = 0.1 + x \]
Expanding this gives:
\[ 0.05 - 0.5x = 0.1 + x \]
Rearranging terms:
\[ 0.05 - 0.1 = x + 0.5x \]
This simplifies to:
\[ -0.05 = 1.5x \]
Thus:
\[ x = -\frac{0.05}{1.5} = -\frac{1}{30} \approx -0.0333 \]

### Step 8: Calculate equilibrium concentration of \( A_2 \)
Now, substituting \( x \) back to find the equilibrium concentration of \( A_2 \):
\[ [A_2]_{eq} = 0.1 - x = 0.1 + 0.0333 = 0.1333 \, \text{M} \]

### Final Answer
The equilibrium concentration of \( A_2(g) \) is approximately:
\[ [A_2]_{eq} = 0.133 \, \text{M} \]