The standard entroples of `N_(2)(g), H_(2) (g) and NH_(3) (g)` are `191.5, 130.5, 192.6 Jk^(-1) mol ^(-1)`. The value of `DeltaS^(@)` of formation 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 change in entropy (ΔS°) of the formation of ammonia (NH₃), we can follow these steps:

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

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

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

### Step 4: Calculate the total standard entropy of the reactants and products.
For the products:
- There is 1 mole of NH₃:
\[ S^\circ(\text{products}) = S^\circ(\text{NH}_3) = 192.6 \, \text{J K}^{-1} \text{mol}^{-1} \]

For the reactants:
- There is \( \frac{1}{2} \) mole of N₂ and \( \frac{3}{2} \) moles of H₂:
\[ S^\circ(\text{reactants}) = \left(\frac{1}{2} \times S^\circ(\text{N}_2)\right) + \left(\frac{3}{2} \times S^\circ(\text{H}_2)\right) \]
\[ S^\circ(\text{reactants}) = \left(\frac{1}{2} \times 191.5\right) + \left(\frac{3}{2} \times 130.5\right) \]
\[ S^\circ(\text{reactants}) = 95.75 + 195.75 = 291.5 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 5: Substitute the values into the ΔS° formula.
Now we can calculate ΔS°:
\[ \Delta S^\circ = S^\circ(\text{NH}_3) - S^\circ(\text{reactants}) \]
\[ \Delta S^\circ = 192.6 - 291.5 \]
\[ \Delta S^\circ = -98.9 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Answer:
The standard change in entropy (ΔS°) of the formation of ammonia is:
\[ \Delta S^\circ = -98.9 \, \text{J K}^{-1} \text{mol}^{-1} \]

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