Use the following bond enthalpies to construct enthalpy cycles and calculate ent
ID: 529990 • Letter: U
Question
Use the following bond enthalpies to construct enthalpy cycles and calculate enthalpy changes for the following reactions. You may assume that all species are in the gaseous state.
Bond
H–H
O=O
C–C
C=C
C–H
F–F
H–O
Br–Br
C–Br
H–Br
C=O
KJ mol-1
436
496
348
612
412
158
463
193
276
366
743
(i)
(ii) H2C=CH2(g) + 3O2(g) 2CO2(g) + 2H2O(g)
Bond
H–H
O=O
C–C
C=C
C–H
F–F
H–O
Br–Br
C–Br
H–Br
C=O
KJ mol-1
436
496
348
612
412
158
463
193
276
366
743
Explanation / Answer
Energy is required to break the bonds and it is released when new bonds are formed.
Ethene has one C=C bond and four C-H bonds
oxygen has one O=O bond
CO2 has two C=O bonds
H2O has two O-H bonds
The enthalpy change for the reaction will be
=612 + 4(412) + 3(496) + 4(-743) + 4(-463)
=-1076 KJ
So the reaction enthalpy for the combustion of ethene is -1076 KJ
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