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Consider the following processes: ΔH (kJ/mol) 1/2 A → B: +150 3B → 2C + D: −125 E + A → 2D: +350 For B + D → E + 2C, ΔH will be:

1. 525 kJ/mol
2. 175 kJ/mol
3. −325 kJ/mol
4. 325 kJ/mol

Correct answer: −325 kJ/mol

Solution

Using Hess's Law, we combine the given reactions to match B + D → E + 2C. The enthalpy change is calculated as ΔH = −125 − 150 − 350 = −325 kJ/mol.

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