| ⇦ |
| ⇨ |
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: - Vanadium (III) oxide is a strong
- 5 g of hydrogen peroxide is present in 56 mL of solution. The solution
- The correct order of increasing bond length of C–H, C–O, C–C and C=C is
- Which of the following shows bond in silicone?
- The electronic configuration of transition elements is exhibited by
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Vanadium (III) oxide is a strong
- 5 g of hydrogen peroxide is present in 56 mL of solution. The solution
- The correct order of increasing bond length of C–H, C–O, C–C and C=C is
- Which of the following shows bond in silicone?
- The electronic configuration of transition elements is exhibited by
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