`1` mole of a gas `A` is taken in a vessel of volume `1L`. It dissociates according to the reaction
`A(g)hArrB(g)+C(g) "at" 27^(@)C`. Forward and backward reaction rate constants for the reaction are `1.5xx10^(-2) "and" 3xx10^(-2)` respectively. Find the concentrations of `A,B "and" C` at equilibrium.

Answer

To solve the problem step by step, we will follow the process of determining the equilibrium concentrations of gases A, B, and C based on the given reaction and the rate constants.

### Step 1: Write the Reaction and Identify Variables
The reaction is given as:
\[ A(g) \rightleftharpoons B(g) + C(g) \]

We start with 1 mole of gas A in a 1 L vessel, so the initial concentration of A is:
\[ [A]_0 = 1 \, \text{mol/L} \]
\[ [B]_0 = 0 \, \text{mol/L} \]
\[ [C]_0 = 0 \, \text{mol/L} \]

### Step 2: Define Change in Concentration
Let \( x \) be the amount of A that dissociates at equilibrium. Therefore, at equilibrium, the concentrations will be:
- \( [A] = 1 - x \)
- \( [B] = x \)
- \( [C] = x \)

### Step 3: Write the Expression for Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[B][C]}{[A]} \]

### Step 4: Calculate \( K_c \) from \( K_f \) and \( K_b \)
Given:
- \( K_f = 1.5 \times 10^{-2} \)
- \( K_b = 3 \times 10^{-2} \)

We can find \( K_c \) using the relationship:
\[ K_c = \frac{K_f}{K_b} = \frac{1.5 \times 10^{-2}}{3 \times 10^{-2}} = \frac{1}{2} \]

### Step 5: Substitute Equilibrium Concentrations into \( K_c \) Expression
Substituting the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{x \cdot x}{1 - x} = \frac{x^2}{1 - x} \]

Setting this equal to the calculated \( K_c \):
\[ \frac{x^2}{1 - x} = \frac{1}{2} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:
\[ 2x^2 = 1 - x \]
Rearranging the equation:
\[ 2x^2 + x - 1 = 0 \]

### Step 7: Use the Quadratic Formula
To solve the quadratic equation \( 2x^2 + x - 1 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 2, b = 1, c = -1 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ x = \frac{-1 \pm \sqrt{1 + 8}}{4} \]
\[ x = \frac{-1 \pm 3}{4} \]

Calculating the two possible values:
1. \( x = \frac{2}{4} = 0.5 \)
2. \( x = \frac{-4}{4} = -1 \) (not physically meaningful)

Thus, \( x = 0.5 \).

### Step 8: Calculate Equilibrium Concentrations
Now we can find the equilibrium concentrations:
- \( [A] = 1 - x = 1 - 0.5 = 0.5 \, \text{mol/L} \)
- \( [B] = x = 0.5 \, \text{mol/L} \)
- \( [C] = x = 0.5 \, \text{mol/L} \)

### Final Answer
The concentrations at equilibrium are:
- \( [A] = 0.5 \, \text{mol/L} \)
- \( [B] = 0.5 \, \text{mol/L} \)
- \( [C] = 0.5 \, \text{mol/L} \)