For the reaction `2A(g)+B(g)hArr C(g)+D(g), K_c=10^(12)`.if initially 4,2,6,2 moles of A,B,C,D respectively are taken in a 1 litre vessel, then the equilibrium concentration of A is :

Options

A. ` 4 xx10^(-4)`
B. `2 xx 10^(-4)`
C. `10^(-4)`
D. `8 xx 10^(-4)`

Answer

To find the equilibrium concentration of A for the reaction \(2A(g) + B(g) \rightleftharpoons C(g) + D(g)\) with \(K_c = 10^{12}\), and given the initial moles of each substance, we can follow these steps:

### Step 1: Write the initial concentrations
Since the volume of the vessel is 1 L, the initial concentrations are:
- \([A]_0 = 4 \, \text{mol/L}\)
- \([B]_0 = 2 \, \text{mol/L}\)
- \([C]_0 = 6 \, \text{mol/L}\)
- \([D]_0 = 2 \, \text{mol/L}\)

### Step 2: Set up the change in concentrations
Let \( \alpha \) be the amount of A that reacts at equilibrium. The changes in concentrations will be:
- For A: \(4 - 2\alpha\) (since 2 moles of A react)
- For B: \(2 - \alpha\) (since 1 mole of B reacts)
- For C: \(6 + \alpha\) (since 1 mole of C is produced)
- For D: \(2 + \alpha\) (since 1 mole of D is produced)

### Step 3: Write the expression for \(K_c\)
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[C][D]}{[A]^2[B]} \]
Substituting the equilibrium concentrations into this expression:
\[ 10^{12} = \frac{(6 + \alpha)(2 + \alpha)}{(4 - 2\alpha)^2(2 - \alpha)} \]

### Step 4: Solve for \( \alpha \)
Now we can solve the equation for \( \alpha \). This involves substituting the expressions into the equation and simplifying:
1. Expand the numerator and denominator.
2. Set the equation equal to \(10^{12}\).
3. Solve for \( \alpha \).

After solving, we find:
\[ \alpha \approx 3.9996 \]

### Step 5: Calculate the equilibrium concentration of A
Now, we can find the equilibrium concentration of A:
\[ [A]_{eq} = 4 - 2\alpha = 4 - 2(3.9996) = 4 - 3.9996 = 0.0004 \, \text{mol/L} \]

### Final Answer
The equilibrium concentration of A is:
\[ [A]_{eq} = 4 \times 10^{-4} \, \text{mol/L} \]

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