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Thermochemistry 103
C2H2(g) (5/2) O2(g) l 2 CO2(g)H2O(l) H 1299.8 kJ
Find the enthalpy change for: 2 C(s) H2(g) l C2H2(g) Answer:
Since we need 2 carbon atoms, we will multiply the first equation by 2:
2 (C(s) O2(g) l CO2(g)) 2(393.5 kJ)
H2(g) (1/2) O2(g) l H2O(l) 285.8 kJ
Since C 2H 2 appears on the product side we will reverse the third reaction and
change the sign of H:
2CO 2(g)H 2O(l) l C 2H 2(g) (5/2) O 2(g) (1299.8 kJ)
Adding the reactions and H’s: 2 C(s) H 2(g) l C 2H 2(g) 227.0 kJ
6-4 Utterly Confused About Calorimetry
Quick Tip
Many people have trouble with the problems concerning this material. The major
cause is that many of the problems are too simple. You need to analyze carefully
what the questions are really asking. Unit conversions are exceedingly important
in working these problems.
Calorimetry problems appear in the thermochemistry chapter of many texts.
These problems often appear intimidating, but this need not be the case. We
will begin with the following example: The burning of a 1.5886-g sample of glucose
(C 6H 12O 6) in a bomb calorimeter resulted in a temperature increase of
3.682C. The heat capacity of the calorimeter was 3.562 kJ/C, and the calorimeter
contained 1000. g of water. Find the molar heat of reaction per mole of glucose
for the following reaction:
C 6H 12O 6(s) 6 O 2(g) l 6 CO 2(g) 6 H 2O(l)
The question is “Find the molar heat of reaction.” This means we need the
enthalpy change (H) in kJ/mol (or J/mol) of glucose. The increase in temperature
means that this is an exothermic process (negative enthalpy change).