In a system `P_(s) hArr 2Q_(g) + R_(g)`, at equilibrium is concentration
of 'Q' is doubled than how many times the concentration of R at equilibrium will be :-

Options

A. Double of its original concentration
B. `(1)/(4)` of its orignal concentration
C. `(1)/(2) `of its original concentration
D. 4 times of its orignal concentration

Explanation

To solve the problem, we will analyze the equilibrium system given by the equation:

\[ P_{(s)} \rightleftharpoons 2Q_{(g)} + R_{(g)} \]

### Step-by-Step Solution:

1. **Identify the Equilibrium Expression**:
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[Q]^2 [R]}{1} \]
Since \( P \) is a solid, its concentration is constant and does not appear in the expression.

2. **Let the Initial Concentrations be Defined**:
Let the initial concentration of \( Q \) be \( [Q] \) and the initial concentration of \( R \) be \( [R] \). Thus, the equilibrium expression can be written as:
\[ K_c = [Q]^2 [R] \]

3. **Consider the Change in Concentration of \( Q \)**:
According to the problem, the concentration of \( Q \) is doubled at equilibrium. Therefore, the new concentration of \( Q \) is:
\[ [Q]_{new} = 2[Q] \]

4. **Substitute the New Concentration into the Equilibrium Expression**:
The new equilibrium expression with the doubled concentration of \( Q \) becomes:
\[ K_c = (2[Q])^2 [R_{new}] \]
Simplifying this gives:
\[ K_c = 4[Q]^2 [R_{new}] \]

5. **Set the Two Equilibrium Expressions Equal**:
Since \( K_c \) remains constant, we can set the two expressions for \( K_c \) equal to each other:
\[ [Q]^2 [R] = 4[Q]^2 [R_{new}] \]

6. **Cancel Out \( [Q]^2 \)**:
Dividing both sides by \( [Q]^2 \) (assuming \( [Q] \neq 0 \)):
\[ [R] = 4[R_{new}] \]

7. **Solve for the New Concentration of \( R \)**:
Rearranging gives:
\[ [R_{new}] = \frac{[R]}{4} \]

### Conclusion:
The concentration of \( R \) at the new equilibrium is \( \frac{1}{4} \) times its original concentration.

### Final Answer:
The concentration of \( R \) at equilibrium will be \( \frac{1}{4} \) times the original concentration.

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