For the reaction `1/8 S_(8)(s) + 3/2O_(2)(g) rightarrow SO_(3)(g),` the difference of heat change at constant pressure and constant volume at `27 ^(@)C` will be.

Options

A. `+ 150R`
B. `-150 R`
C. `+450 R`
D. `-450 R`

Answer

To solve the problem, we need to find the difference between the heat change at constant pressure (ΔH) and constant volume (ΔU) for the given reaction:

\[ \frac{1}{8} S_8(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) \]

### Step-by-Step Solution:

1. **Identify the Reaction:**
The reaction is given as:
\[ \frac{1}{8} S_8(s) + \frac{3}{2} O_2(g) \rightarrow SO_3(g) \]

2. **Understand the Relationship Between ΔH and ΔU:**
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) at constant pressure and constant volume is given by:
\[ \Delta H = \Delta U + \Delta n_g R T \]
where \( \Delta n_g \) is the change in the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

3. **Calculate Δng (Change in Moles of Gas):**
- **Products:** The only gaseous product is \( SO_3 \), which contributes 1 mole.
- **Reactants:** The gaseous reactant is \( O_2 \) contributing \( \frac{3}{2} \) moles. The solid \( S_8 \) does not contribute to \( \Delta n_g \).
- Therefore, the change in moles of gas is:
\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} = 1 - \frac{3}{2} = 1 - 1.5 = -0.5 \]

4. **Convert Temperature to Kelvin:**
The temperature given is \( 27^\circ C \). To convert to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

5. **Substitute Values into the Equation:**
Now we can substitute \( \Delta n_g \), \( R \), and \( T \) into the equation:
\[ \Delta H - \Delta U = \Delta n_g R T \]
Substituting the values:
\[ \Delta H - \Delta U = (-0.5) R (300) \]
\[ \Delta H - \Delta U = -150 R \]

6. **Final Answer:**
Therefore, the difference of heat change at constant pressure and constant volume is:
\[ \Delta H - \Delta U = -150 R \]