In a system `A(s) hArr 2B(g) +3C (g)`
if the conc. Of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to

Options

A. two times the original value
B. one half of its original value
C. `2sqrt2` time the original value
D. `1//2sqrt2` time the original value

Explanation

To solve the problem, we need to analyze the equilibrium reaction given:

\[ A(s) \rightleftharpoons 2B(g) + 3C(g) \]

### Step 1: Understand the Initial Equilibrium
Let’s denote the initial equilibrium concentrations of B and C.
- Let the concentration of C at equilibrium be \( [C] = 3x \).
- Consequently, the concentration of B at equilibrium will be \( [B] = 2x \).

### Step 2: Write the Expression for the Equilibrium Constant
The equilibrium constant \( K_c \) for the reaction can be expressed as:

\[ K_c = \frac{[B]^2 [C]^3}{1} \]

Substituting the equilibrium concentrations into the equation, we have:

\[ K_c = \frac{(2x)^2 (3x)^3}{1} = \frac{4x^2 \cdot 27x^3}{1} = 108x^5 \]

### Step 3: Increase the Concentration of C
According to the problem, the concentration of C is increased by a factor of 2. Therefore, the new concentration of C becomes:

\[ [C] = 3x \times 2 = 6x \]

### Step 4: Set Up the New Equilibrium Expression
Now, we need to find the new equilibrium concentration of B when the concentration of C is increased. The new equilibrium constant expression remains the same:

\[ K_c = \frac{[B]^2 [C]^3}{1} \]

Substituting the new concentration of C into the expression:

\[ K_c = \frac{[B]^2 (6x)^3}{1} = \frac{[B]^2 \cdot 216x^3}{1} \]

### Step 5: Equate the Two Expressions for \( K_c \)
Since \( K_c \) remains constant, we can set the two expressions equal to each other:

\[ 108x^5 = [B]^2 \cdot 216x^3 \]

### Step 6: Solve for the New Concentration of B
Rearranging the equation to solve for \( [B]^2 \):

\[ [B]^2 = \frac{108x^5}{216x^3} = \frac{108}{216} x^2 = \frac{1}{2} x^2 \]

Taking the square root gives us:

\[ [B] = \frac{1}{\sqrt{2}} x \]

### Step 7: Relate to the Original Concentration of B
Since the original concentration of B was \( 2x \), we can express the new concentration of B in terms of the original concentration:

\[ [B] = \frac{1}{\sqrt{2}} x = \frac{1}{\sqrt{2}} \cdot \frac{2}{2} \cdot 2x = \frac{2x}{\sqrt{2}} = \frac{2}{\sqrt{2}} x = \sqrt{2} x \]

### Conclusion
Thus, the equilibrium concentration of B changes to \( \frac{1}{2} \) times its original value.

### Final Answer
The equilibrium concentration of B will change to \( \frac{1}{2} \) times its original value.

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