| ⇦ |
| ⇨ |
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 Ca²⁺ and F⁻ are located in CaF₂ crystal,respectively at face centred cubic lattic points
- Hydro-metallurgical process of extraction of metals is based on
- The sulpha drugs are used for
- 100 mL each of 0.5 N NaOH, N/5 HCl and N/10 H₂SO₄ are mixed together
- Triple point of water is
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The Ca²⁺ and F⁻ are located in CaF₂ crystal,respectively at face centred cubic lattic points
- Hydro-metallurgical process of extraction of metals is based on
- The sulpha drugs are used for
- 100 mL each of 0.5 N NaOH, N/5 HCl and N/10 H₂SO₄ are mixed together
- Triple point of water is
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply