Relationship Between $\delta H$ And $\delta U$

Question

For the reaction $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$ at 298 K, if $\Delta \cup = -196.0$ kJ/mol, what is the value of $\Delta H$? (R = 8.314 J/mol·K)

RankGuru · Accepted Answer

The relationship between $\Delta H$ and $\Delta U$ for reactions involving gases is: $\Delta H = \Delta \cup + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gaseous species. For the reaction $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$: moles of gaseous products = 2, moles of gaseous reactants = $2 + 1 = 3$, so $\Delta n_g = 2 - 3 = -1$. Now: $\Delta H = -196{,}000 + (-1)(8.314)(298) = -196{,}000 + (-2477.6) = -198{,}477.6$ J $\approx -198.5$ kJ/mol. Option -193.5 kJ/mol arises from using $+\Delta n_g RT$ incorrectly with positive sign. Option -196.0 kJ/mol incorrectly assumes $\Delta H = \Delta U$, ignoring the $p\Delta V$ work correction. Option -200.0 kJ/mol uses an approximate RT of 2500 J instead of the correct 2477.6 J. This is a standard NCERT problem type demonstrating the distinction between constant-volume and constant-pressure thermochemistry. Plausibility check: $\Delta H < \Delta U$ (more negative) when $\Delta n_g < 0$, because the system does less work on surroundings at constant pressure — the sign and magnitude are both consistent.

Relationship Between \(\delta H\) And \(\delta U\)

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