1. 1. The equation for the combustion of ethanol in air is: C 2 H 5 OH(l) + 3O 2
ID: 77801 • Letter: 1
Question
1. 1. The equation for the combustion of ethanol in air is:
C2H5OH(l) + 3O2(g) %u21922CO2(g) + 3H2O(l)
Calculate the enthalpy change for this reaction using the average bond enthalpy values given in the table.
Bond
Average bond enthalpy/
kJ mol%u20131
Bond
Average bond enthalpy/
kJ mol%u20131
C%u2013H
+412
C%u2013C
+348
C%u2013O
+360
O%u2013H
+463
O=O
+496
C=O
+743
2.Complete the electronic configuration of a sulfur atom: 1s2.....
Bond
Average bond enthalpy/
kJ mol%u20131
Bond
Average bond enthalpy/
kJ mol%u20131
C%u2013H
+412
C%u2013C
+348
C%u2013O
+360
O%u2013H
+463
O=O
+496
C=O
+743
Explanation / Answer
there is 1 C-C bond, 1 C-O bond,5 C-H bonds and 1 O-H bond.... hence, average enthalpy of ethanol in liquid state(as it is ethanol's standard state)=5*412+360+463+348=3231 KJ/mol
now for O2 in gaseous state, total enthalpy=496 KJ/mol
there are 2 C=O in CO2, hence, enthalpy=1486KJ/mol
similarly, for H2O in liquid state, enthalpy=463*2=926 KJ/mol
now enthalpy change= (sum_{}^{}coefficient * H_{i, product}-coefficient*H_{i,reactant})
=-(2*1486+3*926-3231-3*496)=-1031KJ/mol
b)sulphur:1s2 2s2 2p6 3s2 3p4
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