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

Question

Consider the following changes : $\text{SnO}_2 \rightarrow \text{SnO}$ $\Delta G_1^o$ and $\text{PbO}_2 \rightarrow \text{PbO}$ $\Delta G_2^o$. Select the correct option

Answer 1

The problem relates to the thermodynamics of chemical reactions, specifically focusing on the relationship between changes in internal energy ($∆U^Θ$) and changes in enthalpy ($∆H^Θ$) during the combustion of methane. Let's explore the relationship between these two quantities to determine the correct answer.

In general, the relationship between the change in enthalpy ($∆H$) and the change in internal energy ($∆U$) for a reaction at constant pressure can be given by the formula:

∆H = ∆U + ∆nRT

∆H is the change in enthalpy.
∆U is the change in internal energy.
∆n is the change in the number of moles of gas during the reaction.
R is the ideal gas constant, approximately 8.314 J/(mol·K).
T is the temperature in Kelvin.

For the combustion of methane, the balanced chemical equation is:

CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)

In this reaction:

The initial number of moles of gas: 1 (from methane) + 2 (from oxygen) = 3 moles.
The final number of moles of gas: 1 from $CO_2$.

Thus, the change in moles of gas ($∆n$) is:

∆n = 1 - 3 = -2

Because the combustion of methane involves a decrease in the number of moles of gas ($∆n$ is negative), the term $∆nRT$ will be negative. This indicates:

∆H = ∆U + (-2RT) \Rightarrow ∆H < ∆U

This means that the enthalpy change ($∆H^Θ$) for the combustion of methane will be less than the internal energy change ($∆U^Θ$). Therefore, the correct answer to the question is:

∆H^Θ < ∆U^Θ

Answer 2