Question

Given the following entropy values (in J K⁻¹ mol⁻¹)
at 298 K and 1 atm :H₂(g) : 130.6, Cl₂(g) : 223.0,
HCl(g) : 186.7.The entropy change (in J K⁻¹ mol⁻¹) for the reaction
H₂(g) + Cl₂ (g)➝2HCl (g) is
(a) +540.3 (b) +727.0
(c) –166.9 (d) +19.8

Answer 1

Answer:

c is correct answer....I hope help you.......

Answer 2

The entropy change (in J K⁻¹ mol⁻¹) for the reaction is (d) +19.8

Explanation:

The reaction given in the question is:

H₂ (g) + Cl₂ (g) ➝ 2 HCl (g)

The change in entropy is given by the formula is given as:

ΔS = ∑Sm° (P) - ∑Sm° (R)

The change in entropy the given reaction is:

ΔS° = (2 × Sm° (HCl)) - [Sm° (Cl₂) + Sm° (H₂)]

Where,

Sm° (HCl)) = 186.7

Sm° (Cl₂) = 223.0

Sm° (H₂) = 130.6

On substituting the values, we get,

ΔS° = (2 × 186.7) - [223.0 + 130.6]

ΔS° = 373.4 - 353.6

∴ ΔS° = 19.8 J K⁻¹ mol⁻¹