Each question contains STATEMENT-1 (Assertion) and STATEMENT-2( Reason).
Examine the statements carefully and mark the correct answer according to the instruction given below:
STATEMENT-1: For the reaction `H_(2)(g)+I_(2)(g) iff 2HI(g)` if the volume of vessel is reduced to half of its original volume, equilibrium concentration of all gases will be doubled.
STATEMENT-2: According to Le- Chatelier's principle, reaction shifts in a direction that tends to minimized the effect of the stess.

Options

A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1
B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE
D. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Answer

To solve the question, let's analyze both statements step by step.

### Step 1: Analyze Statement 1
**Statement 1**: For the reaction \( H_2(g) + I_2(g) \iff 2HI(g) \), if the volume of the vessel is reduced to half of its original volume, the equilibrium concentration of all gases will be doubled.

- When the volume of the reaction vessel is reduced, the concentrations of all gaseous reactants and products will increase because concentration is defined as moles per unit volume.
- If the initial volume is \( V \), and we reduce it to \( \frac{V}{2} \), the concentration of each gas will be calculated as follows:
- For \( H_2 \):
\[ \text{New concentration} = \frac{N_1}{\frac{V}{2}} = \frac{2N_1}{V} \]
- For \( I_2 \):
\[ \text{New concentration} = \frac{N_2}{\frac{V}{2}} = \frac{2N_2}{V} \]
- For \( HI \):
\[ \text{New concentration} = \frac{N_3}{\frac{V}{2}} = \frac{2N_3}{V} \]
- Thus, the equilibrium concentrations of all gases will indeed be doubled.

### Step 2: Analyze Statement 2
**Statement 2**: According to Le-Chatelier's principle, the reaction shifts in a direction that tends to minimize the effect of the stress.

- Le-Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust itself to counteract that change and restore a new equilibrium.
- In this case, reducing the volume increases the pressure, and the system will respond by shifting the equilibrium position to the side with fewer moles of gas to reduce the pressure.
- However, in this specific reaction, there are equal moles of reactants (1 mole of \( H_2 \) and 1 mole of \( I_2 \)) and products (2 moles of \( HI \)). Therefore, the shift does not favor either side in terms of moles.

### Conclusion
- **Statement 1** is true.
- **Statement 2** is also true, but it does not correctly explain why the concentrations of gases double when the volume is halved. The doubling of concentrations is due to the change in volume, not a shift in equilibrium.

### Final Answer
Both statements are true, but Statement 2 is not the correct explanation for Statement 1.