The difference between `DeltaH and DeltaU (DeltaH - DeltaU)`, when the combustion of one mole of heptane (l ) is carried out at a temperature T, is equal to: `–xRT`. The value of x is_____.

Options

A. a.4
B. b.6
C. c.1
D. d.NONE OF THE ABOVE

Explanation

To solve the problem, we need to find the difference between the enthalpy change (ΔH) and the internal energy change (ΔU) during the combustion of one mole of heptane (C7H16) at a temperature T. We will use the relationship between ΔH and ΔU, which is given by the equation:

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

where Δn_g is the change in the number of moles of gas during the reaction, R is the gas constant, and T is the temperature.

### Step-by-Step Solution:

1. **Write the balanced combustion reaction of heptane:**
The combustion of heptane can be represented as:
\[ C_7H_{16(l)} + 11 O_{2(g)} \rightarrow 7 CO_{2(g)} + 8 H_2O_{(l)} \]

2. **Identify the number of moles of gaseous reactants and products:**
- **Reactants:** 11 moles of \(O_2\) (gaseous)
- **Products:** 7 moles of \(CO_2\) (gaseous)

The water produced is in liquid form, so it does not contribute to Δn_g.

3. **Calculate Δn_g:**
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 7 - 11 = -4 \]

4. **Substitute Δn_g into the equation for ΔH:**
Now, we can substitute Δn_g into the equation:
\[ \Delta H = \Delta U + (-4)RT \]
Rearranging gives:
\[ \Delta H - \Delta U = -4RT \]

5. **Identify the value of x:**
According to the problem, we have:
\[ \Delta H - \Delta U = -xRT \]
Comparing both equations, we find that:
\[ -xRT = -4RT \]
Thus, \(x = 4\).

### Final Answer:
The value of \(x\) is **4**.

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