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.