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JAMB Chemistry 1997 Question 26

Question 26

Energy Changes

½N2(g) + ½O2(g) → NO(g) ΔH° 89 kJ mol−1. If the entropy change for the reaction above at 25°C is 11.8 J mol−1, calculate the change in free energy, ΔG°, for the reaction at 25°C.

  1. A.88.71 kJ
  2. B.85.48 kJ
  3. C.−204.00 kJ
  4. D.−3427.40 kJ.
Show the answer and explanation

Answer: B (85.48 kJ)

Explanation

ΔG = ΔH − TΔS. Change ΔS to kJ: 11.8 J = 0.0118 kJ. ΔG = 89 − 298 × 0.0118 = 89 − 3.52 = 85.48 kJ.