Given the following entropy values ( in `JK^(-1) "mol"^(-1)`) at 298 K and 1 atm: `H_(2)(g):130.6, Cl_(2)(g):223.0, HCl(g): 186.7`. The entropy change (in `JK^(-1) "mol"^(-1)`) for the reaction
`H_(2)(g) + Cl_(2)(g) to 2HCl(g)` , is
(a)`+540.3`
(b)`+727.0`
(c)`-166.9`
(d)`+19.8`

Options

A. `+540.3`
B. `+727.0`
C. `-166.9`
D. `+19.8`

Explanation

To find the entropy change (ΔS) for the reaction:

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

we will use the formula for the change in entropy:

\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]

### Step 1: Identify the entropy values
We have the following entropy values at 298 K and 1 atm:
- \( S_{H_2(g)} = 130.6 \, \text{JK}^{-1} \text{mol}^{-1} \)
- \( S_{Cl_2(g)} = 223.0 \, \text{JK}^{-1} \text{mol}^{-1} \)
- \( S_{HCl(g)} = 186.7 \, \text{JK}^{-1} \text{mol}^{-1} \)

### Step 2: Calculate the total entropy of the products
Since we have 2 moles of \( HCl(g) \) as products, we calculate the total entropy of the products:

\[ S_{\text{products}} = 2 \times S_{HCl(g)} = 2 \times 186.7 = 373.4 \, \text{JK}^{-1} \text{mol}^{-1} \]

### Step 3: Calculate the total entropy of the reactants
Now, we calculate the total entropy of the reactants:

\[ S_{\text{reactants}} = S_{H_2(g)} + S_{Cl_2(g)} = 130.6 + 223.0 = 353.6 \, \text{JK}^{-1} \text{mol}^{-1} \]

### Step 4: Calculate the change in entropy
Now we can find the change in entropy for the reaction:

\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} = 373.4 - 353.6 = 19.8 \, \text{JK}^{-1} \text{mol}^{-1} \]

### Conclusion
The entropy change for the reaction is:

\[ \Delta S = +19.8 \, \text{JK}^{-1} \text{mol}^{-1} \]

Thus, the correct answer is (d) \( +19.8 \).

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