Neutralization Reactions Lab Report
This report was to to study Neutralization Reactions
q = ms∆t – 1M HCl
q = (acid mass(g))(base mass(g))(tf-ti)(s)
q = -(50.2+26.1)(8.6)(4.184)
q = -2.746 kJ
x mol of acid & base – 1M HCl
mol HCl = (1M HCl x (50mL))/(1000mL) = 0.05 mol HCl
mol NaOH = (2M NaOH x (25mL)/(1000mL) = 0.05 mol NaOH
mol H20 – 1M HCl
mol H2O – .05mol NaOH + .05mol HCl à .05mol NaCl + .05mol H20
mol H20 =.05 mol
∆H – 1M HCl
∆H = q/mol H20 = -2.746/.05 = -54.920 kJ/mol H20
Standard Enthalpy of Formation – 1M HCl
Hf – [H20 -285.8 NaCl = -385.92 NaOH= -469.6 HCl = -92.3]
∆H = [(H20+NaCl)-(NaOH + HCl)] = -110.27kJ/mol H20
1. Compare results with the group’s results.
The enthalpy results obtained in our experiment was the highest of the class, in the 1M HCl reaction and 2M HCl reaction. There is a 36.3% deviation between our value for enthalpy and the lowest experimental value for the 1M HCl neutralization. There is a 30.1% deviation between our value for enthalpy and the other experimental value for the 2M HCl neutralization.
2. How does concentration of acid correlate with heat released?
The acid concentration has a direct correlation with the heat released, as seen in our two experiments, the 2M HCl neutralization released approximately twice the heat as the 1M HCl neutralization. After determining the experiments’ experimental H/mol value, they had discrepancy of only 1.55%, so this seems very valid.
3. Compare the molar heat of neutralization for the different acids tested and propose explanations for deviations in expectations.
The experimental results show nearly 50% error when compared to the standard enthalphy calculation. This is a very high number. Since the experiment was not capable of being conducted in a 100% closed system, there were obvious errors in calculating the change in temperature. Heat released into the air, into the Styrofoam cup, and etc. That was probably the main source of error.