# Lab Report

Part 1: The Dissolving of Solid Sodium Hydroxide in Water

Hydroxide
Volume of Water Temperature Solid Sodium
205 mL (initial) -- --
200 mL (constant temp) 23.2 C (constant temp) --
195 mL (highest temp) 26.5 C (highest temp) 2.526 g
1. NaOH(s) + H2O(I) ---> NaOH(aq)
2. Moles NaOH = 2.526 g x (1mol/40g) = 2.526 g x 0.025 mol = 0.6315
3. q = m c (T2 - T1) ---> 205mL H20 x 1g/1mL = 205g H20 ---> c = 4.184 J/gC ---> T2 = 26.4 ---> T1 = 23.2 ---> Heat Gained by H2O = 205g x 4.18 J/gC x (26.4 - 23.2) ---> 205g x 4.18 J/gC x 3.2 = 2,742.08 J
4. 2,742.08 J/0.6315 mol NaOH = 4.34 x 104 J/mole x 1 kJ/1000 J = 43.4 kJ/mol NaOH

Part II: The Reaction of Sodium Hydroxide Solution with Hydrochloric Acid

0.5 m HC1 0.5 m NaOH -
100 mL (initial) 100 mL (initial) 24.1 C (initial)
- - 27.2 C (final)
1. HC1 + NaOH ---> NaC1 + H2O
2. q = m x c x Δt ---> 200g of solution x 4.18 x (27.2 - 24.1) = 200g x 4.18 x 3.1 = -2,592.6 J/1000 = -2.5926 kJ ---> 1.3 kJ/mol
3. q = m c (T2 - T1) ---> 200mL H20 x 1g/1mL = 200g H20 ---> c = 4.184 J/gC ---> T2 = 27.2 ---> T1 = 24.1 ---> Heat Gained by H2O = 200g x 4.18 J/gC x (27.2 - 24.1) ---> 200g x 4.18 J/gC x 3.1 = 2,592.6 J
4. 2,592.6 J/0.6315 mol NaOH = 4.1 x 104 J/mole x 1 kJ/1000 J = 41 kJ/mol NaOH

Conclusion:

1. a. NaOH(s) + H2O(I) ---> NaOH(aq) 43.4 kJ b. HC1 + NaOH ---> NaC1 + H2O 41 kJ c. 21.7 + 41 = 62.7 kJ
2. (experimental - actual) / actual x 100% ---> (-44.2 - 43.4) / 43.4 x 100% = -201.84
3. (experimental - actual) / actual x 100% ---> (-56 - 41) / 41 x 100% = -236.59
4. Yes, because the accepted rates were negative. Thus, the prediction was vastly off.
5. The changes occurred due to the changes in the chemicals.
6. A gap between the lid and thermometer would cause a temperature change. It would show the final heat as cooler than it actually is. Therefore, the gap would be wider, causing the enthalpy change to increase.