When enthalpy and entropy change for a chemical reaction are – 2.5 x 10³ cals

⇦

⇨

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:

1. Chloropicrin is obtained by the reaction of
2. The correct set of four quantum numbers for the valence electron of rubidium atom
3. A ideal gas will have maximum density when
4. Cell reaction is spontaneous when
5. Identify the feasible reaction among the following

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