allery∆fH of HBr(g) is -36.4 kJ/mol. What is ∆rH for H2(g) + Br2(l) → 2HBr(g)?
A-72.8 kJ/mol
B+72.8 kJ/mol
C-18.2 kJ/mol
D-36.4 kJ/mol
Answer & Solution
Correct answer: A. -72.8 kJ/mol
Two moles of HBr form, so ∆rH = 2 x (-36.4) = -72.8 kJ/mol.
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