Thermodynamics Question 98
Question: If at 298 K the bond energies of $ C-H,C-C, $ $ C=C $ and $ H-H $ bonds are respectively 414, 347, 615 and 435 kJ $ mo{{l}^{-1}} $ , the value of enthalpy change for the reaction $ H _{2}C=CH _{2}(g)+H _{2}(g)\to H _{3}C-CH _{3}(g) $ at 298 K will be [AIEEE 2003]
Options:
A) + 250 kJ
B) ? 250 kJ
C) + 125 kJ
D) ? 125 kJ
Show Answer
Answer:
Correct Answer: D
Solution:
$ CH _{2}=C{H _{2(g)}}+{H _{2(g)}}\to H _{3}C-C{H _{3(g)}} $
$ 414\times 4=1656 $
$ 414\times 6=2484 $
$ 615\times 1=615 $
$ 347\times 1=347 $
$ 435\times 1=\underset{\overline{\underline{2706}}}{\mathop{435}} $ 2831 $ \Delta H=2706-2831=-125kJ $
