Heats of reaction NaOH(s)  NaOH(aq) H= -43.5 kJ Mol NaOH 0.1 Mol HCl H = cmT =(4.18)(200)(T) T= 5.2 4347 J T= 11.6 9698 J T= 6.0 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 NaOH dissolves (goes from solid to aqueous) NaOH dissolves and there is a reaction (neutralization) between NaOH and HCl The reaction between NaOH and HCl

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

For more lessons, visit www.chalkbored.com 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 For more lessons, visit www.chalkbored.com