Question

\(∆U^Θ\) of combustion of methane is – \(X\ kJ \ mol^{–1}\). The value of \(∆H^Θ\) is

• A. \(= ∆U^Θ\)
• B. \(>∆U^Θ\)
• C. \(<∆U^Θ\)
• D. \(=0\)

Answer 1

The Correct Option is C

The given problem is about finding the relationship between the change in internal energy (\(∆U^Θ\)) and the change in enthalpy (\(∆H^Θ\)) for the combustion of methane. Understanding these thermodynamic concepts is crucial:

1. In thermodynamics, the relationship between the change in enthalpy (\(∆H\)) and the change in internal energy (\(∆U\)) is given by: \(∆H = ∆U + ∆n_gRT\), where \(\Delta n_g\) represents the change in the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
2. For the combustion of methane (\(CH_4\)), the balanced chemical reaction is: \(CH_4 (g) + 2O_2 (g) \rightarrow CO_2 (g) + 2H_2O (l)\).
3. Count the gaseous molecules involved:
   • In reactants: 1 mole of CH₄ + 2 moles of O₂ = 3 moles of gas.
   • In products: 1 mole of CO₂ = 1 mole of gas.
4. The change in the number of moles of gas, \(∆n_g\), is calculated as: \(∆n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - 3 = -2\).
5. Given \(∆n_g\) is negative, this implies a decrease in the number of gaseous moles during the reaction.
6. Substitute \(∆n_g\) in the formula: \(∆H = ∆U + (-2)RT\).
7. Since \(RT\) is positive (as \(R\) and \(T\) are positive constants), the term \(-2RT\) will reduce the value of \(∆H\), making \(∆H\) less than \(∆U\).
8. Therefore, the correct relationship is: \(∆H^Θ < ∆U^Θ\).

Thus, the correct answer is \(<∆U^Θ\).