# Work cited

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Work Cited s's_Law Appendix a To calculate the temperature change of each, one must take the maximum temperature minus the initial temperature. - NaOH + HCl: 36.00 ˚C- 22.20 ˚C = 13.80 ˚C - NaOH + NH 4 Cl: o Trial 1: 22.90˚C – 21.40 ˚C = 1.50˚C o Trial 2: 22.10 ˚C – 19.70 ˚C = 2.40 ˚C - HCl + NH 3 : 33.40 ˚C – 22.70 ˚C = 10.70 ˚C b To calculate the q surroundings of each one must use the q = Cp x m x Δ T. - NaOH + HCl: 4.18 (J/g x ˚C) x 103 (g) x 13.80˚C = 5941.452 J - NaOH + NH 4 Cl: o Trial 1: 4.18 x 103 x 1.50 = 645.81 J o Trial 2: 4.18 x 103 x 2.40 = 1033.296 J - HCl + NH 3: 4.18 x 103 x 10.70 = 4606.778 J c All q reaction (J) is, is q surroundings but negative. It is negative because all the chemical reactions are exothermic, therefore they are giving off energy, which is negative. d ΔHrxn(kJ) [per mole] d was calculated by multiplying the q reaction by 10, and dividing it by 1000 to get the answer in kJ/mol. - NaOH + HCl: - 5941.452 x 10 / 1000 = -59.41 kJ/mol - NaOH + NH 4 Cl Trial 1:-645.81 x 10 / 1000 = -6.45 kJ/mol - NaOH + NH 4 Cl Trial 2:- 1033.296 x 10 / 1000 = -10.33 kJ/mol - HCl + NH 3 : -4606.778 x 10 / 1000 = -46.07 kJ/mol e To calculate the Average ΔHrxn(kJ) [per mole], one must take the sum of trial 1 and 2 together for NaOH + HCl NH 4 Cl, and divide by two, for reaction 2. For the other two reactions, the average ΔHrxn(kJ) [per mole will be the same as the ΔHrxn(kJ) [per mole. - 6.45 +− 10.33 2 = -8.39 kJ/mol f To find the kJ/mol using Hess’s Law, one must take the Average ΔHrxn(kJ) [per mole], of NaOH + HCl and NaOH + NH 4 Cl, and add them together. -59.41 kJ/mol + 8.39 kJ/mol = -51.02 kJ/mol

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• Fall '07
• KEISER,JOSEPHTHOHECKMAN,KIMBERLYPROIA,MICHAELANTSANDERS,REBECCALO'NEILL,RYANSHAW