`underset(S^(@)(298K)K^(-1)mol^(-1))(H^(+)(aq))+underset(-10.7)(OH^(-)(aq))rarrunderset(+70)(H_(2)O(l))`
Standard entropy change for the above reaction is

Options

A. `60.3 JK^(-1) mol^(-1)`
B. `80.7 JK^(-1) mol^(-1)`
C. `-70 JK^(-1) mol^(-1)`
D. `+10.7 JK^(-1) mol^(-1)`

Explanation

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

\[ \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \]

we will follow these steps:

### Step 1: Identify the standard entropies of the reactants and products.
- The standard entropy of \(\text{H}^+(aq)\) is given as \(S^°(H^+) = 0 \, \text{J K}^{-1} \text{mol}^{-1}\) (since it is a reference state).
- The standard entropy of \(\text{OH}^-(aq)\) is given as \(S^°(OH^-) = -10.7 \, \text{J K}^{-1} \text{mol}^{-1}\).
- The standard entropy of \(\text{H}_2\text{O}(l)\) is given as \(S^°(H_2O) = 70 \, \text{J K}^{-1} \text{mol}^{-1}\).

### Step 2: Write the formula for the standard entropy change.
The standard entropy change for the reaction can be calculated using the formula:

\[ \Delta S° = S^°_{\text{products}} - S^°_{\text{reactants}} \]

### Step 3: Substitute the values into the formula.
Substituting the values we have:

\[ \Delta S° = S^°(H_2O) - \left(S^°(H^+) + S^°(OH^-)\right) \]

\[ \Delta S° = 70 \, \text{J K}^{-1} \text{mol}^{-1} - \left(0 + (-10.7 \, \text{J K}^{-1} \text{mol}^{-1})\right) \]

### Step 4: Simplify the equation.
Now, simplifying the equation:

\[ \Delta S° = 70 \, \text{J K}^{-1} \text{mol}^{-1} + 10.7 \, \text{J K}^{-1} \text{mol}^{-1} \]

\[ \Delta S° = 70 + 10.7 = 80.7 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 5: State the final answer.
Thus, the standard entropy change for the reaction is:

\[ \Delta S° = 80.7 \, \text{J K}^{-1} \text{mol}^{-1} \]

---