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When enthalpy and entropy change for a chemical reaction are – 2.5 x 10³ cals and 7.4 cals deg⁻¹ respectively predict the reaction at 298 K is
Options
(a) irreversible
(b) reversible
(c) spontaneous
(d) non spontaneous
Correct Answer:
spontaneous
Explanation:
Enthalpy change, ∆H = -2.5 ⨯ 10³ cal
Entropy change, ∆G = 7.4 kcal deg⁻¹
Temperature, T = 298 K
As, ∆G = ∆H -T∆S
⇒ ∆G = -2.5 ⨯ 10³ – 298 ⨯ 7.4 = -ve value
For spontaneity of reaction, negative value of ∆G is required, so the reaction is spontaneous.
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- How many mole of MNO₄⁻ ion will react with 1 mole of ferrous oxalate in acidic medium
- The compound that does not produce blue colour in Lassaigne’s test is
- The reduction of benzoyl chloride with H₂/Pd BaSO₄ gives
- Who modified Bohr’s theory by introducing elliptical orbits for electron path
- How will you separate a solution (miscible) of benzene + CHCl₃
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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