In the system `A_((s))hArr2B_((g))+3C_((g))`, if the concentration 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` times its original value
D. `(1)/(2sqrt2)` times that the original value

Answer

To solve the problem, we will analyze the equilibrium reaction step by step. The reaction is given as:

\[ A_{(s)} \rightleftharpoons 2B_{(g)} + 3C_{(g)} \]

### Step 1: Write the expression for the equilibrium constant (K)

The equilibrium constant \( K \) for the reaction can be expressed in terms of the concentrations of the gaseous products (B and C). Since A is a solid, it does not appear in the expression:

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

### Step 2: Define the initial concentrations at equilibrium

Let the equilibrium concentrations of B and C be represented as \( [B] = b \) and \( [C] = c \). Thus, we can write:

\[ K = c^3 b^2 \]

### 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:

\[ [C] = 2c \]

### Step 4: Write the new equilibrium expression

With the new concentration of C, we can express the new equilibrium constant \( K' \):

\[ K' = [C]^3[B]^2 = (2c)^3[B]^2 = 8c^3[B]^2 \]

### Step 5: Set the equilibrium constants equal

Since the equilibrium constant \( K \) is dependent only on temperature and not on concentration changes, we can set \( K \) equal to \( K' \):

\[ c^3 b^2 = 8c^3 b'^2 \]

where \( b' \) is the new concentration of B at equilibrium after the change.

### Step 6: Cancel \( c^3 \) from both sides

Assuming \( c \neq 0 \), we can simplify the equation:

\[ b^2 = 8b'^2 \]

### Step 7: Solve for \( b' \)

Now, we can solve for \( b' \):

\[ b' = \frac{b}{2\sqrt{2}} \]

### Conclusion

The equilibrium concentration of B changes to \( \frac{b}{2\sqrt{2}} \), which indicates that the concentration of B decreases as a result of the increase in the concentration of C.

### Final Answer

The equilibrium concentration of B changes to \( \frac{b}{2\sqrt{2}} \).

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