Two moles of each reactant A and B are taken in a reaction flask. They react in the following manner,
`A(g)+B(g)hArr C(g)+D(g)`
At equilibrium, it was found that the concentration of C is triple to that of B the equilibrium constant for the reaction is

Options

A. 4.5
B. 6
C. 9
D. `1//6`

Answer

To find the equilibrium constant \( K_c \) for the reaction \( A(g) + B(g) \rightleftharpoons C(g) + D(g) \), we will follow these steps:

### Step 1: Write the balanced equation and initial conditions
The balanced chemical equation is:
\[ A(g) + B(g) \rightleftharpoons C(g) + D(g) \]

Initially, we have:
- Moles of \( A = 2 \)
- Moles of \( B = 2 \)
- Moles of \( C = 0 \)
- Moles of \( D = 0 \)

### Step 2: Set up the change in moles at equilibrium
Let \( x \) be the number of moles of \( A \) and \( B \) that react to form \( C \) and \( D \). At equilibrium, the moles will be:
- Moles of \( A = 2 - x \)
- Moles of \( B = 2 - x \)
- Moles of \( C = x \)
- Moles of \( D = x \)

### Step 3: Use the information given about the concentrations
According to the problem, the concentration of \( C \) is triple that of \( B \) at equilibrium. This can be expressed as:
\[ [C] = 3[B] \]

### Step 4: Express concentrations in terms of \( x \)
The concentrations can be expressed as:
- Concentration of \( C = \frac{x}{V} \)
- Concentration of \( B = \frac{2 - x}{V} \)

Substituting these into the relationship gives:
\[ \frac{x}{V} = 3 \left( \frac{2 - x}{V} \right) \]

### Step 5: Simplify the equation
Cancel \( V \) from both sides:
\[ x = 3(2 - x) \]

Expanding this gives:
\[ x = 6 - 3x \]

### Step 6: Solve for \( x \)
Rearranging the equation:
\[ x + 3x = 6 \]
\[ 4x = 6 \]
\[ x = 1.5 \]

### Step 7: Calculate moles at equilibrium
Now we can find the moles of each substance at equilibrium:
- Moles of \( A = 2 - x = 2 - 1.5 = 0.5 \)
- Moles of \( B = 2 - x = 2 - 1.5 = 0.5 \)
- Moles of \( C = x = 1.5 \)
- Moles of \( D = x = 1.5 \)

### Step 8: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C][D]}{[A][B]} \]

Substituting the equilibrium concentrations:
\[ K_c = \frac{\left(\frac{1.5}{V}\right)\left(\frac{1.5}{V}\right)}{\left(\frac{0.5}{V}\right)\left(\frac{0.5}{V}\right)} \]

### Step 9: Simplify the expression
This simplifies to:
\[ K_c = \frac{(1.5)^2}{(0.5)^2} \]
\[ K_c = \frac{2.25}{0.25} \]
\[ K_c = 9 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is \( 9 \).

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