Question

Which among the following reactions exhibits $\Delta H=\Delta U$?

• A. $H_{2(g)}+Br_{2(g)}\rightarrow2HBr_{(g)}$
• B. $2CO_{(g)}+O_{2(g)}\rightarrow2CO_{2(g)}$
• C. $PCl_{5(g)}\rightarrow PCl_{3(g)}+Cl_{2(g)}$
• D. $C_{(s)}+2H_{2}O_{(g)}\rightarrow2H_{2(g)}+CO_{2(g)}$

Hint:

Logic Tip: Whenever you are asked to relate $\Delta H$ and $\Delta U$ for a reaction, solely focus on counting the coefficients of the species marked with a $(g)$ state symbol. If the sum of gas coefficients on both sides matches, the enthalpy change equals the internal energy change.

Answer 1

The Correct Option is A

Concept:
The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) at a constant temperature is given by the equation: $$\Delta H = \Delta U + \Delta n_g RT$$ where $\Delta n_g$ is the change in the number of moles of gaseous substances (moles of gaseous products - moles of gaseous reactants). For $\Delta H$ to be exactly equal to $\Delta U$, the term $\Delta n_g RT$ must be zero, which means $\Delta n_g = 0$.
Step 1: Evaluate $\Delta n_g$ for each given reaction option.
We calculate $\Delta n_g = (\text{moles of gas in products}) - (\text{moles of gas in reactants})$. Keep in mind that solid and liquid states are ignored. Option (A): $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$ $$\Delta n_g = 2 - (1 + 1) = 2 - 2 = 0$$ Option (B): $2CO_{(g)} + O_{2(g)} \rightarrow 2CO_{2(g)}$ $$\Delta n_g = 2 - (2 + 1) = 2 - 3 = -1$$ Option (C): $PCl_{5(g)} \rightarrow PCl_{3(g)} + Cl_{2(g)}$ $$\Delta n_g = (1 + 1) - 1 = 2 - 1 = 1$$ Option (D): $C_{(s)} + 2H_{2}O_{(g)} \rightarrow 2H_{2(g)} + CO_{2(g)}$ $$\Delta n_g = (2 + 1) - 2 = 3 - 2 = 1$$ (Note: $C_{(s)}$ is a solid and not counted).
Step 2: Identify the correct reaction.
Since only reaction (A) has $\Delta n_g = 0$, substituting this into the equation yields: $$\Delta H = \Delta U + (0)RT \implies \Delta H = \Delta U$$ Therefore, reaction (A) is the correct answer.