Question

Calculate enthalpy change for the change 8S(g) S8 (g),
given that

H₂S₂ (g) → 2H(g) + 2S(g), ∆H = 239.0 k calmol⁻¹

H₂S(g) → 2H(g) + S(g), ∆H = 175.0 k calmol⁻¹
(a) + 512.0 k cal (b) – 512.0 k cal
(c) 508.0 k cal (d) – 508.0 k cal

Answer 1

The enthalpy change for the change 8S(g) ---> S8 (g) is  -512.0 k.cal

Option (B) is correct.

Explanation:

We are given that:

H₂S₂ (g) → 2H(g) + 2S(g), ∆H = 239.0 k calmol⁻¹

H₂S(g) → 2H(g) + S(g), ∆H = 175.0 k calmol⁻¹

We have to calculate enthalpy change for 8S ---> S8

Thus we know that:

ΔH1 - ΔH2 = 1 S - S

Bond Energy = 239-175 = 64

Now  for 8S → S8  we can calculate it as:

(-64) x 8 = -512.0 kcal

Thus the enthalpy change for the change 8S(g) ---> S8 (g) is  -512.0 k.cal

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