One mole of `A_((g))` is heated to `300^(@)` C in a closed one litre vessel till the following equilibrium is reached. `A_((g)) hArr B_((g)` . The equilibrium constant of the reaction at `300^(@)` C is 4. What is the conc. of B `("in. mole. lit"^(-1))`at equilibrium ?

Options

A. 0.2
B. 0.6
C. 0.8
D. 0.1

Answer

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

### Step 1: Set up the reaction and initial conditions
We have the reaction:
\[ A_{(g)} \rightleftharpoons B_{(g)} \]

Initially, we have 1 mole of \( A \) in a 1-liter vessel, which means the initial concentration of \( A \) is:
\[ [A]_{initial} = 1 \, \text{mol/L} \]
And the initial concentration of \( B \) is:
\[ [B]_{initial} = 0 \, \text{mol/L} \]

### Step 2: Define the change at equilibrium
Let \( X \) be the number of moles of \( A \) that convert to \( B \) at equilibrium. Therefore, at equilibrium:
- The concentration of \( A \) will be:
\[ [A]_{equilibrium} = 1 - X \]
- The concentration of \( B \) will be:
\[ [B]_{equilibrium} = X \]

### Step 3: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[B]_{equilibrium}}{[A]_{equilibrium}} \]
Substituting the expressions from Step 2:
\[ K_c = \frac{X}{1 - X} \]

### Step 4: Substitute the known value of \( K_c \)
We are given that \( K_c = 4 \). Therefore, we can set up the equation:
\[ \frac{X}{1 - X} = 4 \]

### Step 5: Solve for \( X \)
Cross-multiplying gives:
\[ X = 4(1 - X) \]
Expanding this:
\[ X = 4 - 4X \]
Bringing all \( X \) terms to one side:
\[ X + 4X = 4 \]
\[ 5X = 4 \]
Dividing both sides by 5:
\[ X = \frac{4}{5} \]

### Step 6: Find the concentration of \( B \) at equilibrium
Since \( [B]_{equilibrium} = X \), we can substitute the value of \( X \):
\[ [B]_{equilibrium} = \frac{4}{5} = 0.8 \, \text{mol/L} \]

### Final Answer
The concentration of \( B \) at equilibrium is:
\[ \boxed{0.8 \, \text{mol/L}} \]

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