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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 energy of second Bhor orbit of the hydrogen atom is -328 kJ mol⁻¹, hence the enregy
- Which of the following compounds possesses the C-H bond with the lowest bond dissociation
- Vapour density of a gas is 22. What is its molecular weight
- In which one of the following is not a buffer solution
- Which of the following elements will have the lowest first ionisation energy?
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The energy of second Bhor orbit of the hydrogen atom is -328 kJ mol⁻¹, hence the enregy
- Which of the following compounds possesses the C-H bond with the lowest bond dissociation
- Vapour density of a gas is 22. What is its molecular weight
- In which one of the following is not a buffer solution
- Which of the following elements will have the lowest first ionisation energy?
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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