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 1: Convert Temperature to Kelvin
The given temperature is \( 27^{\circ}C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

So,

\[ T = 27 + 273 = 300 \, K \]

**Hint:** Always convert Celsius to Kelvin when dealing with thermodynamic equations.

### Step 2: Identify ΔH and ΔU
We know that:

- ΔH is the change in enthalpy (heat change at constant pressure).
- ΔU is the change in internal energy (heat change at constant volume).

The relationship between ΔH and ΔU is given by:

\[ \Delta H = \Delta U + \Delta N_g RT \]

Where:
- \( \Delta N_g \) is the change in the number of moles of gas.
- \( R \) is the universal gas constant (approximately \( 8.314 \, J/(mol \cdot K) \)).
- \( T \) is the temperature in Kelvin.

### Step 3: Calculate ΔNg
To find \( \Delta N_g \), we need to calculate the change in the number of moles of gaseous products minus the change in gaseous reactants.

From the reaction:
- Products: \( SO_3(g) \) contributes 1 mole of gas.
- Reactants: \( \frac{3}{2} O_2(g) \) contributes \( \frac{3}{2} \) moles of gas.

Since \( S_8 \) is a solid, it does not contribute to \( \Delta N_g \).

Thus,

\[ \Delta N_g = \text{(moles of products)} - \text{(moles of reactants)} = 1 - \frac{3}{2} = 1 - 1.5 = -0.5 \]

**Hint:** Only consider gaseous species when calculating \( \Delta N_g \).

### Step 4: Substitute into the Equation
Now we can substitute \( \Delta N_g \) into the equation for \( \Delta H - \Delta U \):

\[ \Delta H - \Delta U = \Delta N_g RT \]

Substituting the values we have:

\[ \Delta H - \Delta U = (-0.5)(R)(300) \]

### Step 5: Calculate the Result
Now we can simplify this expression:

\[ \Delta H - \Delta U = -150R \]

### Conclusion
The difference of heat change at constant pressure and constant volume at \( 27^{\circ}C \) is:

\[ \Delta H - \Delta U = -150R \]

### Final Answer
The answer is \( -150R \).

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