For the reaction
`C_(3)H_(8)(g)+5O_(2)(g)rarr3CO_(2)(g)+4H_(2)O(l)`
at constant temperature, `DeltaH-DeltaU` is

Options

A. `+3RT`
B. `-RT`
C. `+RT`
D. `-3RT`

Explanation

To solve the problem, we need to calculate the difference between the change in enthalpy (ΔH) and the change in internal energy (ΔU) for the given reaction:

**Reaction:**
\[ C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(l) \]

### Step-by-Step Solution:

1. **Understand the Relationship between ΔH and ΔU:**
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) is given by the equation:
\[ ΔH = ΔU + ΔN_{g}RT \]
where \( ΔN_{g} \) is the change in the number of moles of gas.

2. **Calculate ΔN_{g}:**
To find \( ΔN_{g} \), we need to determine the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

- **Reactants:**
- \( C_{3}H_{8}(g) \): 1 mole
- \( O_{2}(g) \): 5 moles
- **Total moles of gaseous reactants = 1 + 5 = 6 moles**

- **Products:**
- \( CO_{2}(g) \): 3 moles
- \( H_{2}O(l) \): 4 moles (liquid, not counted for \( ΔN_{g} \))
- **Total moles of gaseous products = 3 moles**

Now, we can calculate \( ΔN_{g} \):
\[ ΔN_{g} = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 3 - 6 = -3 \]

3. **Substitute ΔN_{g} into the ΔH and ΔU Relationship:**
Now that we have \( ΔN_{g} \), we can substitute it into the equation:
\[ ΔH - ΔU = ΔN_{g}RT \]
Substituting \( ΔN_{g} = -3 \):
\[ ΔH - ΔU = -3RT \]

4. **Final Answer:**
Therefore, the value of \( ΔH - ΔU \) is:
\[ ΔH - ΔU = -3RT \]

### Conclusion:
The answer to the question is:
\[ ΔH - ΔU = -3RT \]