What is the effect of halving the pressure by doubling the volume on the following system at `500^(@)C`?
`H_(2)(g)+I_(2)(g)hArr2HI(g)`

Options

A. Shift to reactant side
B. Shift to product side
C. Liquefaction of HI
D. No effect

Explanation

To solve the problem, we need to analyze the effect of halving the pressure by doubling the volume on the equilibrium system given by the reaction:

\[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \]

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The reaction involves hydrogen gas (\(H_2\)), iodine gas (\(I_2\)), and hydrogen iodide gas (\(HI\)).
- At equilibrium, we need to consider the number of moles of gas on each side of the reaction.

2. **Count the Moles of Gas**:
- On the left side (reactants), we have:
- 1 mole of \(H_2\) + 1 mole of \(I_2\) = 2 moles of gas.
- On the right side (products), we have:
- 2 moles of \(HI\).
- Therefore, the total number of moles of gas on both sides is equal:
- Left: 2 moles (1 from \(H_2\) + 1 from \(I_2\))
- Right: 2 moles (2 from \(2HI\))

3. **Calculate \(\Delta n_g\)**:
- \(\Delta n_g\) (change in moles of gas) is calculated as:
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 2 = 0 \]

4. **Apply Le Chatelier's Principle**:
- According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in pressure, the equilibrium will shift in the direction that counteracts that change.
- In this case, since \(\Delta n_g = 0\), changing the pressure (either increasing or decreasing) will not affect the position of equilibrium.

5. **Effect of Halving the Pressure by Doubling the Volume**:
- Halving the pressure by doubling the volume results in no net change in the equilibrium position because the number of moles of gas on both sides is the same.
- Therefore, the equilibrium will remain unchanged.

6. **Conclusion**:
- The effect of halving the pressure by doubling the volume on the given system at \(500^\circ C\) is that there will be no effect on the equilibrium position.

### Final Answer:
- There will be no effect on the equilibrium position of the reaction.