`1` mole each of `H_(2)(g) "and" I_(2)(g)` are introduced in a `1L` evacuated vessel at `523K` and equilibrium`H_(2)(g)+I_(2)(g)hArr2Hi(g)` is established. The concentration of `HI(g)` at equilibrium:

Options

A. Changes on changing pressure.
B. Changes on changing temperature.
C. Changes on changing volume of the vessel
D. Is same even if only `2` mol of `HI(g)` were introduced in the vessel in the beginning.
Is same even when a platinum gauze is introduced to catalyst the reaction.

Explanation

To find the concentration of HI at equilibrium for the reaction:

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

we can follow these steps:

### Step 1: Write the Initial Conditions
Initially, we have:
- 1 mole of \( H_2 \)
- 1 mole of \( I_2 \)
- 0 moles of \( HI \)

Since the volume of the vessel is 1 L, the initial concentrations are:
- \([H_2] = 1 \, \text{mol/L}\)
- \([I_2] = 1 \, \text{mol/L}\)
- \([HI] = 0 \, \text{mol/L}\)

### Step 2: Set Up the Change in Concentrations
Let \( x \) be the amount of \( H_2 \) and \( I_2 \) that reacts to form \( HI \). According to the stoichiometry of the reaction:
- \( H_2 \) and \( I_2 \) will decrease by \( x \)
- \( HI \) will increase by \( 2x \)

At equilibrium, the concentrations will be:
- \([H_2] = 1 - x\)
- \([I_2] = 1 - x\)
- \([HI] = 2x\)

### Step 3: Write the Equilibrium Expression
The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]

### Step 4: Substitute the Equilibrium Concentrations
Substituting the equilibrium concentrations into the expression gives:

\[ K_c = \frac{(2x)^2}{(1 - x)(1 - x)} = \frac{4x^2}{(1 - x)^2} \]

### Step 5: Determine the Equilibrium Constant
At 523 K, the equilibrium constant \( K_c \) for this reaction is typically known to be around 50 (this value can vary based on specific data sources, but we will use this for our calculation).

Thus, we set:

\[ 50 = \frac{4x^2}{(1 - x)^2} \]

### Step 6: Solve for \( x \)
Cross-multiplying gives:

\[ 50(1 - x)^2 = 4x^2 \]

Expanding and rearranging:

\[ 50(1 - 2x + x^2) = 4x^2 \]
\[ 50 - 100x + 50x^2 = 4x^2 \]
\[ 50x^2 - 100x + 50 = 0 \]

Dividing through by 2:

\[ 25x^2 - 50x + 25 = 0 \]

### Step 7: Factor or Use the Quadratic Formula
This can be factored as:

\[ (5x - 5)^2 = 0 \]

Thus, \( x = 1 \).

### Step 8: Calculate the Concentration of HI
Now substituting \( x \) back into the expression for \( [HI] \):

\[ [HI] = 2x = 2(1) = 2 \, \text{mol/L} \]

### Conclusion
The concentration of \( HI \) at equilibrium is:

\[ \boxed{2 \, \text{mol/L}} \]