NaOHs H20 HCL AQ NaCLAQH20 NaOHAQHCLAQ NaCLH20 Reaction 4 Mass Of Water1019707

Naohs h20 hcl aq naclaqh20 naohaqhclaq naclh20

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NaOH(s)+ H20 + HCL (AQ) -> NaCL(AQ)+H20 NaOH(AQ)+HCL(AQ)-> NaCL+H20 Reaction 4 Mass Of Water=101.9707 ΔT= Tf-Ti= 4.6°C Specific Heat = 4.18 qwater (Joules): 1960.69 qwater (kJ): 1960.69/1000=1.961 Moles of NaOH= 1.9707/40=.0493 qwater (kJ/mol): 1.961/.0493=39.777 Q reaction (kJ/mol)= -39.777 Reaction 5 Mass of water = 101.9999 grams (remember to include the ~ 2 grams of solid NaOH) ΔT = Tf – Ti = 34.1-22.5=11.6°C Specific heat = 4.18 qwater (Joules) =4945.727 qwater (kJ) = 4945.727/1000= 4.9457 Moles of NaOH = 1.999/40= .049975 Moles of CH3COOH = .05 qwater (kJ/mol) = 9.0457/.049975= 98.963 qreaction (kJ/mol) = -98.963 = ΔH5 Reaction 6
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Mass of water = 100 grams ΔT = Tf – Ti = 3.4 °C Specific heat = 4.18 qwater (Joules) =1421.2 qwater (kJ) =1.421 Moles of NaOH = .05 Moles of CH3COOH = .05 qwater (kJ/mol) = 53.6 qreaction (kJ/mol) = -53.6 = ΔH6 Evaluating Hess’s Law Measured value for ΔH= -53.6 Value of ΔH for Reaction 3 based on Hess’s law = ΔH2 – ΔH1= -59.188 % Error= (-53.6+59.203)/-59.203 * 100 = -9.46% Reaction: NaOH(s)-> NaOH NaOH(s)+ H20 + CH3COOH (AQ) ->CH3COONA(AQ)+H20 NaOH(AQ)+CH3COOH(AQ)->CH3COONA(AQ) +H20 (L) Conclusion Within this lab, I was not able to prepare solutions of primary standards, and also I was not able to measure heat of a reaction during constant pressure conditions. However, I was able to calculate the enthalpy change using two other reactions by Hess’s Law by using another student’s data tables. I think the error that I had in my experiment was that I did not cover the styrofoam cup with the second cup to minimize the heat loss. Therefore, this ruined my original data, and thus caused the experiment to fail. If I had kept the heat less to a minimum, I think this lab would have ran perfectly.
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