Chemical Thermodynamics Question 430
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 $
BETA
