At 1090K, `K_(p)` for the reaction
`CO_(2)(g) +C(s) rarr 2CO(g)` is 10 atm,
At the constant temperature ,Equilibrium position will shift, when pressure change occurs.
In the above reaction, if we use catalyst. Equilibrium constant value `:`

Options

A. increases
B. decreases
C. does no change
D. change is rapid

Explanation

To solve the question regarding the equilibrium constant \( K_p \) for the reaction:

\[ \text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g) \]

at a temperature of 1090 K, we need to analyze the effect of a catalyst on the equilibrium constant.

### Step-by-Step Solution:

1. **Understanding the Equilibrium Constant**:
The equilibrium constant \( K_p \) is defined for a given reaction at a specific temperature. It is expressed in terms of the partial pressures of the gaseous products and reactants.

2. **Effect of Temperature**:
The value of \( K_p \) is dependent only on the temperature of the system. If the temperature remains constant, \( K_p \) will not change regardless of the conditions of the reaction.

3. **Role of a Catalyst**:
A catalyst speeds up the rate of both the forward and reverse reactions equally. It does not affect the position of equilibrium or the value of the equilibrium constant.

4. **Conclusion**:
Since the question states that a catalyst is used and the temperature remains constant, the equilibrium constant \( K_p \) will not change. Therefore, the correct answer is that the equilibrium constant value does not change.

### Final Answer:
The equilibrium constant value does not change when a catalyst is used at constant temperature.