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If the enthalpy change for the transition of liquid water to steam is 30 kJ mol⁻¹ at 27⁰C, the entropy change for the process would be
Options
(a) 10 J mol⁻¹ K⁻¹
(b) 1.0 J mol⁻¹ K⁻¹
(c) 0.1 J mol⁻¹ K⁻¹
(d) 100 J mol⁻¹ K⁻¹
Correct Answer:
100 J mol⁻¹ K⁻¹
Explanation:
We know that ΔG = ΔH – TΔS,
0 = ΔH -TΔS [ThereforeΔG=0],
ΔS = ΔH/T = 30 x 10³ / 300 = 100 J mol⁻¹ K⁻¹.
Related Questions: - Which one of the following is a non-benzenoid aromatic compound
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which one of the following is a non-benzenoid aromatic compound
- The boiling point of glycerol is more than propanol because of
- Which one of the following is present as an active ingredient in bleaching
- Which of the following does not acts as Lewis acid
- Pure nascent hydrogen is based obtained by
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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