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For the reaction, S (rhombic)  + O 2 → SO 2 ΔH = - 298   kJ   mol - 1 at 25 ° C and 1   atm . Therefore, Δ E for the reaction should be

Options

  1. A- 298   kJ   mol - 1
  2. B- 298 + 8 . 314 × 298   kJ   mol - 1
  3. C- 298 - 8 . 314 × 298   kJ   mol - 1
  4. D- 298 - 2 × 8 . 314 × 298   kJ   mol - 1

Correct answer

A. - 298   kJ   mol - 1

Step-by-step solution

Given : Δ H = - 298   kJ   mol - 1 T = 25 + 273 = 298   K , P = 1   atm , ΔE = ? ΔH = ΔE + Δn g RT or ΔE = ΔH ∵ Δn g = 0 ΔE = - 298 Δ E = - 298   kJ   mol - 1

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