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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.
Related Questions: - The number of moles of acidified KMnO₄ required to oxidise one mole
- Which of the following is not a common component of Photochemical Smong
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The number of moles of acidified KMnO₄ required to oxidise one mole
- Which of the following is not a common component of Photochemical Smong
- Which of the following pairs has both members from the same group of periodic
- If 900 J/g of heat is exchanged at boiling point of water, then what is increase
- The volume of 2.8g of carbon monoxide at 27⁰C and 0.821 atm pressure is
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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