Consider the following process ½A → B; ΔH = +150 kJ mol⁻¹ 3B → 2C + D; ΔH = −125 kJ mol⁻¹ E + A → 2D; ΔH = +350 kJ mol⁻¹ For B + D → E + 2C, ΔH will be
- A−325 kJ/mol
- B325 kJ/mol
- C525 kJ/mol
- D−175 kJ/mol
✓ Correct answer: D
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