For the reaction `H_(2)(g)+I_(2)(g)hArr2HI(g)`
the equilibrium constant `K_(p)` changes with

Options

A. total pressure
B. catalyst
C. the amount of `H_(2)` and `I_(2)` present
D. temperature

Answer

To solve the question regarding the equilibrium constant \( K_p \) for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \), we need to analyze how \( K_p \) is affected by various factors.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The given reaction is \( H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \). This is a gaseous equilibrium reaction.

2. **Defining the Equilibrium Constant**:
The equilibrium constant \( K_p \) for this reaction can be expressed in terms of the partial pressures of the gases involved:
\[ K_p = \frac{(P_{HI})^2}{(P_{H_2})(P_{I_2})} \]
where \( P_{HI} \), \( P_{H_2} \), and \( P_{I_2} \) are the partial pressures of HI, H2, and I2, respectively.

3. **Factors Affecting \( K_p \)**:
- \( K_p \) is influenced by changes in temperature.
- \( K_p \) is not affected by changes in pressure, volume, or concentration of reactants/products.

4. **Conclusion**:
The only factor that changes the equilibrium constant \( K_p \) is temperature. Therefore, if the temperature of the system changes, \( K_p \) will also change.

5. **Final Answer**:
The equilibrium constant \( K_p \) changes only with a change in temperature.