| ⇦ | 
| ⇨ |
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: - Oxidation numbers of P in PO₄³⁻, of S in SO₄²⁻ and that of Cr in Cr₂O₇²⁻
- Which of the following is not a thermodynamic function
- During the crystallization of a solid from the aqueous solution, the following statement
- At 0K, (i) ¹²C and (ii) a mixture of ¹²C and ¹⁴C will
- If Avogadro number NA, is charged from 6.022 ˣ 10²³ mol⁻¹ to 6.022 ˣ 10²⁰ mol⁻¹
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Oxidation numbers of P in PO₄³⁻, of S in SO₄²⁻ and that of Cr in Cr₂O₇²⁻
- Which of the following is not a thermodynamic function
- During the crystallization of a solid from the aqueous solution, the following statement
- At 0K, (i) ¹²C and (ii) a mixture of ¹²C and ¹⁴C will
- If Avogadro number NA, is charged from 6.022 ˣ 10²³ mol⁻¹ to 6.022 ˣ 10²⁰ mol⁻¹
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