Calculate the standard entropy change for the following reaction `: `
`H_(2)(g) + Cl_(2)(g) rarr 2HCl(g)` at 298 K
Given `S_(H_(2))^(@) = 131J K^(@) mol^(_1), S_(Cl_(2))^(@) = 223 JK^(-1) mol^(-1)` and `S_(HCl)^(@) =187 JK^(-1) mol^(-1)`

Answer

To calculate the standard entropy change (ΔS°) for the reaction:

\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \]

at 298 K, we will use the following formula:

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

### Step 1: Identify the standard entropy values
We are given the standard entropy values for each substance involved in the reaction:
- \( S^\circ_{H_2} = 131 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ_{Cl_2} = 223 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ_{HCl} = 187 \, \text{J K}^{-1} \text{mol}^{-1} \)

### Step 2: Calculate the total standard entropy for the products
The products of the reaction are 2 moles of HCl. Therefore, the total standard entropy for the products is:

\[ S^\circ_{\text{products}} = 2 \times S^\circ_{HCl} = 2 \times 187 \, \text{J K}^{-1} \text{mol}^{-1} = 374 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 3: Calculate the total standard entropy for the reactants
The reactants are 1 mole of \( H_2 \) and 1 mole of \( Cl_2 \). Therefore, the total standard entropy for the reactants is:

\[ S^\circ_{\text{reactants}} = S^\circ_{H_2} + S^\circ_{Cl_2} = 131 \, \text{J K}^{-1} \text{mol}^{-1} + 223 \, \text{J K}^{-1} \text{mol}^{-1} = 354 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 4: Calculate the standard entropy change (ΔS°)
Now we can substitute the values into the formula for ΔS°:

\[ \Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} = 374 \, \text{J K}^{-1} \text{mol}^{-1} - 354 \, \text{J K}^{-1} \text{mol}^{-1} = 20 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Answer
The standard entropy change for the reaction at 298 K is:

\[ \Delta S^\circ = 20 \, \text{J K}^{-1} \text{mol}^{-1} \]