| ⇦ | 
| ⇨ |
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: - In the reaction : 2N₂O₅ → 4NO₂ + O₂, initial pressure is 500 atm and rate
- Rate determining step in nitration of bezene is
- The heat of neutralization is the highest in the following case
- Which of the following is not a cheracteristic of equilibrium
- Which one of the following ions has electronic configuration[Ar] 3d⁶?
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the reaction : 2N₂O₅ → 4NO₂ + O₂, initial pressure is 500 atm and rate
- Rate determining step in nitration of bezene is
- The heat of neutralization is the highest in the following case
- Which of the following is not a cheracteristic of equilibrium
- Which one of the following ions has electronic configuration[Ar] 3d⁶?
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