Heats of reaction

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

=(4.18)(200)(T) 4347 J 9698 J 5016 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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