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The difference between ΔH and ΔE at 300 K for the reaction
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
Options
(a) 300 ˣ 8.31 J/mol
(b) -300 ˣ 8.314 J/mol
(c) 3 ˣ 300 ˣ 8.314 J/mol
(d) -3 ˣ 300 ˣ 8.314 J/mol
Correct Answer:
-3 ˣ 300 ˣ 8.314 J/mol
Explanation:
ΔH = ΔE + Δn(g)RT
or ΔH – ΔE = Δn(g)RT,
where Δn(g) = (3) – (5 + 1) = -3.
Thus, ΔH – ΔE = -3 * 8.31 *300 J/mol.
Related Questions: - In Cu-ammonia complex, the state of hybridization of Cu²⁺ is
- Which one of the following has maximum laevorotatory nature
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Cu-ammonia complex, the state of hybridization of Cu²⁺ is
- Which one of the following has maximum laevorotatory nature
- Carbon percentage is maximum in
- When the first ionisation energies are plotted against atomic number
- In which of the following hydrocarbons, hydrogen is most acidic
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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