The standard entropies of `N_(2)(g),H_(2)(g)` and `NH_(3)(g)` are `191.5, 130.5,192.6JK^(-1)mol^(-1)` . The value of `DeltaS^(0)` during the formation of 1 mole of ammonia is

Options

A. `-98.9 JK^(-1) mol^(-1)`
B. Zero
C. `+129.4 JK^(-1) mol^(-1)`
D. `-29.4 JK^(-1) mol^(-1)`

Answer

To find the standard entropy change (ΔS°) during the formation of 1 mole of ammonia (NH₃), we will follow these steps:

### Step 1: Write the balanced chemical equation for the formation of ammonia.
The formation reaction of ammonia from nitrogen and hydrogen is:
\[ \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g) \]

### Step 2: Identify the standard entropies of the reactants and products.
From the question, we have the following standard entropies:
- \( S^\circ (\text{N}_2) = 191.5 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ (\text{H}_2) = 130.5 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ (\text{NH}_3) = 192.6 \, \text{J K}^{-1} \text{mol}^{-1} \)

### Step 3: Use the formula for the change in standard entropy.
The change in standard entropy for the reaction can be calculated using the formula:
\[ \Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]

For our reaction:
\[ \Delta S^\circ = S^\circ (\text{NH}_3) - \left( \frac{1}{2} S^\circ (\text{N}_2) + \frac{3}{2} S^\circ (\text{H}_2) \right) \]

### Step 4: Substitute the values into the equation.
Substituting the known values:
\[ \Delta S^\circ = 192.6 - \left( \frac{1}{2} \times 191.5 + \frac{3}{2} \times 130.5 \right) \]

Calculating the reactants:
- For \( \text{N}_2 \):
\[ \frac{1}{2} \times 191.5 = 95.75 \, \text{J K}^{-1} \text{mol}^{-1} \]

- For \( \text{H}_2 \):
\[ \frac{3}{2} \times 130.5 = 195.75 \, \text{J K}^{-1} \text{mol}^{-1} \]

Adding these:
\[ 95.75 + 195.75 = 291.5 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 5: Calculate ΔS°.
Now substitute back into the ΔS° equation:
\[ \Delta S^\circ = 192.6 - 291.5 \]
\[ \Delta S^\circ = -98.9 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Answer:
The value of ΔS° during the formation of 1 mole of ammonia is:
\[ \Delta S^\circ = -98.9 \, \text{J K}^{-1} \text{mol}^{-1} \]

---