Chemical Thermodynamics Question 90

Question: If the bond energies of $ H-H $ , $ Br-Br $ and $ HBr $ are 433, 192 and 364 $ kJ,mo{{l}^{-1}} $ respectively, the $ \Delta H^{o} $ for the reaction, $ H_2(g)+Br_2(g)\to 2HBr(g) $ is [CBSE PMT 2004]

Options:

A) + 261 kJ

B) - 103 kJ

C) - 261 kJ

D) + 103 kJ

Show Answer

Answer:

Correct Answer: B

Solution:

$ H-H+Br-Br\to 2H-Br $

$ 433+192 $

$ 2\times 364 $

Energy absorbed = Energy released Net energy released $ =728-625=103,kJ $ i.e., $ =\Delta H=-103,KJ $



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