Relationship Between \(\delta H\) And \(\delta U\)
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)
Select the correct option:
Solution
−198.5kJ/mol
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.
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- relationship between \(\delta h\) and \(\delta u\)
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
−198.5kJ/mol
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.
This medium difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of relationship between \(\delta h\) and \(\delta u\). It appeared in the 2025 exam.
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