If `S^(@)` for `H_(2),Cl_(2)` and HCl are 0.13, 0.22 and 0.19 KJ `K^(-1)mol^(-1)` respectively. The total change in standard entropy for the reaction
`H_(2)+Cl_(2)rarr 2HCl` is :-

Options

A. `30 JK^(-1) mol^(-1)`
B. `40 JK^(-1) mol^(-1)`
C. `60 JK^(-1) mol^(-1)`
D. `20 JK^(-1) mol^(-1)`

Answer

To find the total change in standard entropy for the reaction:

\[ \text{H}_2 + \text{Cl}_2 \rightarrow 2 \text{HCl} \]

we will follow these steps:

### Step 1: Identify the standard entropies
We have the following standard entropies given:
- \( S^\circ \) for \( \text{H}_2 = 0.13 \, \text{kJ K}^{-1} \text{mol}^{-1} \)
- \( S^\circ \) for \( \text{Cl}_2 = 0.22 \, \text{kJ K}^{-1} \text{mol}^{-1} \)
- \( S^\circ \) for \( \text{HCl} = 0.19 \, \text{kJ K}^{-1} \text{mol}^{-1} \)

### Step 2: Calculate the total standard entropy of the products
The product of the reaction is \( 2 \text{HCl} \). Therefore, the total standard entropy for the products is:

\[ S^\circ_{\text{products}} = 2 \times S^\circ_{\text{HCl}} = 2 \times 0.19 \, \text{kJ K}^{-1} \text{mol}^{-1} = 0.38 \, \text{kJ K}^{-1} \text{mol}^{-1} \]

### Step 3: Calculate the total standard entropy of the reactants
The reactants are \( \text{H}_2 \) and \( \text{Cl}_2 \). Therefore, the total standard entropy for the reactants is:

\[ S^\circ_{\text{reactants}} = S^\circ_{\text{H}_2} + S^\circ_{\text{Cl}_2} = 0.13 \, \text{kJ K}^{-1} \text{mol}^{-1} + 0.22 \, \text{kJ K}^{-1} \text{mol}^{-1} = 0.35 \, \text{kJ K}^{-1} \text{mol}^{-1} \]

### Step 4: Calculate the change in standard entropy for the reaction
The change in standard entropy (\( \Delta S^\circ \)) for the reaction can be calculated using the formula:

\[ \Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]

Substituting the values we calculated:

\[ \Delta S^\circ = 0.38 \, \text{kJ K}^{-1} \text{mol}^{-1} - 0.35 \, \text{kJ K}^{-1} \text{mol}^{-1} = 0.03 \, \text{kJ K}^{-1} \text{mol}^{-1} \]

### Step 5: Convert the result to Joules
Since the options are in Joules, we convert the result from kJ to Joules:

\[ \Delta S^\circ = 0.03 \, \text{kJ K}^{-1} \text{mol}^{-1} \times 1000 \, \text{J/kJ} = 30 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Answer
The total change in standard entropy for the reaction is:

\[ \Delta S^\circ = 30 \, \text{J K}^{-1} \text{mol}^{-1} \]

---