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In the equation provided, \( \Delta n_g \) represents the change in the number of moles of gaseous substances during the reaction. This term is used to account for the difference in the number of gas molecules between the products and reactants, which affects the calculation of internal energy change (\( \Delta U \)) from the enthalpy change (\( \Delta H \)).

For the reaction:
\[
2\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)}
\]

1. **Count moles of gaseous reactants**: In the reactants, we have \( 3 \) moles of \( \text{H}_2 \) (g).
2. **Count moles of gaseous products**: In the products, we have \( 1 \) mole of \( \text{C}_2\text{H}_6 \) (g).

Thus:
\[
\Delta n_g = (\text{moles of product gases}) - (\text{moles of reactant gases}) = 1 - 3 = -2
\]

This \( \Delta n_g = -2 \) is then used in the equation:
\[
\Delta H = \Delta U + \Delta n_g \cdot R \cdot T
\]
to adjust the enthalpy change (\( \Delta H \)) to obtain the internal energy change (\( \Delta U \)).

Would you like a further breakdown of the steps involved? Here are some related questions to consider:

1. Why do we subtract reactant gas moles from product gas moles for \( \Delta n_g \)?
2. How does the value of \( \Delta n_g \) affect the calculation of \( \Delta U \)?
3. Why is \( R \) (the gas constant) used in this equation?
4. How does the temperature factor into this calculation of internal energy?
5. What would happen if \( \Delta n_g \) was zero (no change in gas moles)?

**Tip:** When dealing with enthalpy and internal energy, always look at the phases of substances—only gases affect \( \Delta n_g \).

Understand how to calculate Δng in thermodynamic reactions, involving gas moles for the reaction 2C(s) + 3H₂(g) → C₂H₆(g). Learn about the use of ΔH = ΔU + ΔngRT and the First Law of Thermodynamics. Suitable for Grades 11-12 students.

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