wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For the reaction: C2H6C2H4+H2 : The reaction enthalpy ΔrH= ______ kJ mol1. (Round off to the Nearest Integer).

[Given: Bond enthalpies in kJ mol1; C – C: 347; C = C : 611; C – H : 414; H – H : 436]

Open in App
Solution

Answer: 131 kJ/mol
ΔH=??
C2H6C2H4+H2
Comparing the reactant and product, reactant has two extra CH bond and a CC bond. In product side we have one C=C bond and one HH bond.

ΔH=[(ECC+6ECH)(EC=C+4ECH+EHH)]
ΔH=[347+6×414](611+4×414+433]=28312700
ΔH=131 kJ/mol

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
First Law of Thermodynamics
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon