alleryWith H-H 435, Cl-Cl 242, H-Cl 431 kJ/mol, ∆rH for H2(g) + Cl2(g) → 2HCl(g) is:
A+185 kJ
B-185 kJ
C-431 kJ
D-246 kJ
Answer & Solution
Correct answer: B. -185 kJ
∆rH = (435 + 242) - 2 x 431 = 677 - 862 = -185 kJ.
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