?U?of combustion of methane is – X kJ mol–1. The value of ?H? is (i) = ?U? (ii) > ?U? (iii) < ?U? (iv) = 0 ?

To determine the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) for the combustion of methane, we need to consider the definitions and the context of these thermodynamic quantities. The combustion of methane can be represented by the following reaction:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Understanding ΔH and ΔU

In thermodynamics, ΔH represents the change in enthalpy, while ΔU signifies the change in internal energy. The relationship between these two quantities can be expressed by the equation:

ΔH = ΔU + ΔnRT

Here, Δn is the change in the number of moles of gas during the reaction, R is the universal gas constant, and T is the temperature in Kelvin. The term ΔnRT accounts for the work done by the system due to volume changes during the reaction.

Analyzing the Combustion of Methane

In the combustion of methane, we start with 1 mole of methane and 2 moles of oxygen, giving us a total of 3 moles of reactants. The products, carbon dioxide and water vapor, yield a total of 2 moles of gas. Therefore, the change in the number of moles of gas (Δn) is:

• Δn = moles of products - moles of reactants
• Δn = 2 - 3 = -1

Since Δn is negative, this indicates that the reaction results in a decrease in the number of gas moles, which means that the system does work on the surroundings. Consequently, the term ΔnRT will be negative.

Relating ΔH and ΔU

Now, substituting Δn into the equation for ΔH, we get:

ΔH = ΔU - RT

This shows that ΔH is less than ΔU because we are subtracting a positive quantity (RT) from ΔU. Therefore, we can conclude that:

Final Conclusion

Based on our analysis, the correct relationship is:

ΔH < ΔU

So, if we were to choose from the options provided, the answer would be (iii) < ΔU.