Heats of reaction
Mol NaOH
0.1
0.1
0.1
Mol HCl
0
0.1
0.1
H = cmT
T= 5.2 T= 11.6 T= 6.0
5016 J
=(4.18)(200)(T) 4347 J 9698 J
NaOH(s)  NaOH(aq)
H= -43.5 kJ
NaOH(s) + HCl(aq)  NaCl(aq) + H2O H= -97.0 kJ
NaOH(aq) + HCl(aq) NaCl(aq) + H2O H= -50.2 kJ
2. NaOH dissolves (goes from solid to aqueous)
3. NaOH dissolves and there is a reaction
(neutralization) between NaOH and HCl
4. The reaction between NaOH and HCl
5. Definition of Hess’s law: for any reaction that
can be written in steps, H is equal to the
sum of the Hs for the individual steps
6.
NaOH(s) + HCl(aq)
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
7.
NaOH(s)
 NaOH(aq)
H= -43.5 kJ
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O H= -50.2 kJ
NaOH(s) + HCl(aq)  NaCl(aq) + H2O H= -93.7 kJ
8.
E.g. the NaOH(s) takes some time to dissolve
allowing heat to escape and perhaps giving
artificially low values for changes in temp.
E.g. the calorimeter is not perfectly insulated,
thus larger jumps in temperature would not
show up as high as they should
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