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

Options

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

Explanation

To solve the problem, we need to find the value of \(\Delta H - \Delta U\) for the given reaction:

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

### Step 1: Understand the relationship between \(\Delta H\) and \(\Delta U\)

The relationship between the change in enthalpy (\(\Delta H\)) and the change in internal energy (\(\Delta U\)) at constant temperature is given by the equation:

\[ \Delta H = \Delta U + \Delta n_{g}RT \]

where \(\Delta n_{g}\) is the change in the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

### Step 2: Calculate \(\Delta n_{g}\)

Next, we need to calculate \(\Delta n_{g}\), which is the difference between the moles of gaseous products and the moles of gaseous reactants.

- Moles of gaseous products:
- From the reaction, we have \(3CO_{2}(g)\) contributing 3 moles of gas.

- Moles of gaseous reactants:
- From the reaction, we have \(C_{3}H_{8}(g)\) contributing 1 mole and \(5O_{2}(g)\) contributing 5 moles, totaling 6 moles of gas.

Now we can calculate \(\Delta n_{g}\):

\[ \Delta n_{g} = \text{(moles of products)} - \text{(moles of reactants)} = 3 - 6 = -3 \]

### Step 3: Substitute \(\Delta n_{g}\) into the equation

Now we can substitute \(\Delta n_{g}\) into the equation for \(\Delta H\):

\[ \Delta H = \Delta U + (-3)RT \]

This simplifies to:

\[ \Delta H = \Delta U - 3RT \]

### Step 4: Find \(\Delta H - \Delta U\)

To find \(\Delta H - \Delta U\), we rearrange the equation:

\[ \Delta H - \Delta U = -3RT \]

### Conclusion

Thus, the value of \(\Delta H - \Delta U\) is:

\[ \Delta H - \Delta U = -3RT \]

### Final Answer

The correct answer is \(-3RT\).

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