Objective is to measure, using a calorimeter, the energy changes accompanying neutralization reactions....
Purpose To investigate the heat of ionization of a weak acid. When acids and bases are combined a neutralization reaction to occurs which produces salt, and water and the evolution of heat. In order to measure the quantity of heat produced, a thermally insulated container known as a calorimeter is used. This experiment utilized a Styrofoam calorimeter to isolate and measure this heat formed by reactions involving two different acids: the strong acid hydrochloric acid , and the weak acid dichloroacetic acid , when mixed with the strong base sodium hydroxide. HCl + NaOH NaCl + H2O Cl2CH2COOH + NaOH Cl2CH2COONa + H2O
However since the Styrofoam cup is a decent insulator and not perfect some of the heat will be absorbed by calorimeter itself. In order to account for this the heat capacity of the calorimeter is determined by measuring the temperature change when a known amount of warm water and cold water is added to the calorimeter. The amount of heat lost to calorimeter is the difference between the heat lost by the warm water and that gained by the cool water. The temperature data from this reaction will be used to calculate the heat produced for each mole of the acid used, also known as the molar heat of neutralization, allowing for a comparison to be made between these two different acids. This information will allow for determination of what has a greater heat of neutralization: the reaction between a strong acid and a strong base, or a weak acid and a strong base. Method Part 1: Heat Capacity of Calorimeter A graduated cylinder was used to obtain 50 mL of distilled water. This water was then transferred to the Styrofoam cup calorimeter, which was mounted to a stir plate using a stand and clamp. A stirring rod was put into the cup and a digital thermometer (Calibrated to +/-0.1°C) was used to find the temperature of the water. 50 mL of warm water in a graduated cylinder, was obtained from a warm water bath. This warm water was added immediately to the Styrofoam cup
calorimeter and a stopwatch was initiated once all the water had been added. The temperature was noted at 15 second intervals until 4 minutes was reached. Part 2: The heats of neutralization A graduated cylinder was used to measure out 50 mL of NaOH, which was transferred to the Styrofoam cup calorimeter. A stirring rod was added to the cup and a digital thermometer was used to record the temperature. A graduated cylinder was used to measure out 50 mL of HCl, which was allowed to stand until at room temperature. The temperature was then recorded using a digital thermometer. The HCl was then quickly added and a stopwatch was initiated when the first drop was added. The temperature was recorded at 15 second intervals until 4 minutes was reached. Part 2 was repeated, except HCl was substituted with dichloroacetic acid. Results 32.3 32.2
f(x) = - 0x + 32.19 R² = 0.55
32.1 32 Temperature of mixture (°C) 31.9 31.8 31.7 31.6
105 135 165 195 225 90 120 150 180 210 240
Time after mixing (s)
Figure 1: Graph of the temperature versus time data for addition of hot water and cold water in the calorimeter. This reaction is used to determine the heat capacity of the Styrofoam cup calorimeter; the heat which is lost to the calorimeter.
30 f(x) = 0.01x + 28.01 R² = 0.13
28 26 Temperature of mixture (°C)
24 22 15 0
105 135 165 195 225 120 150 180 210 240
Time after mixing (s) Figure 2: Graph of the temperature versus time data for the reaction between the strong acid hydrochloric acid (HCl) and strong base sodium hydroxide (NaOH). 30 28 26 24 Temperature of mixture (°C) 22 20
f(x) = 0.01x + 27.16 R² = 0.12
105 135 165 195 225 120 150 180 210 240
Time after mixing (s) Figure 3: Graph of the temperature versus time data for the reaction between the weak acid dichloroacetic acid (DCA) and strong base sodium hydroxide (NaOH). Table 1: The intial temperature, final temperature and change in temperature of the reactants in each of three calorimeter reactions. Initial Final Change in Temperature, Temperature, Temperature, Ti Tf ΔT Addition of hot Hot water 43.3 +/- 0.1°C 32.19 +/- 0.1°C -11.11+/- 0.1°C water to cold water Cold water 22.0 +/- 0.1°C 10.19 +/- 0.1°C Neutralization of Hydrochloric acid 21.5 +/- 0.1°C 28.01 +/- 0.1°C -6.51+/- 0.1°C strong acid-strong (HCl) base Sodium hydroxide 22.8 +/- 0.1°C 5.21 +/- 0.1°C (NaOH) Neutralization of Dichloracetic acid 21.4 +/- 0.1°C 27.16 +/- 0.1°C -5.76 +/- 0.1°C weak acid-strong (Cl2CH2COOH) base Sodium hydroxide 21.8 +/- 0.1°C 5.36 +/- 0.1°C (NaOH)
Heat capacity of calorimeter Initial temperature of hot water is 43.3 °C (316.30 K) and cold water is 22.0 °C (495 K) | heat lost by hot water | = | heat absorbed by cold water | + | heat absorbed by calorimeter | Vh ρ CpH2O ΔT = Vc ρ CpH2O ΔT1 + Cpcal ΔT1 ΔE hot water = ΔE cold water + ΔE calorimeter mcΔT = -(mcΔT + cΔT) | (50.00 mL)(1.00g/mL)(4.18 J/g K)(-11.11)°C | = -[ |(50.00mL)(4.18 J/g K)(1.00g/mL)(10.19) °C | + | Cpcal (10.19) °C |] 2321.99 J = 2129.71 J + Cpcal (10.19) Cpcal = 18.87 J/K Therefore the Cp of the calorimeter is 18.87 J/K.
Neutralization of strong acid (HCl) and strong base (NaOH) The instantaneous temperature from the reaction between hydrochloric acid and sodium hydroxide is 28.01 °C (301.01 K) which is also considered the final temperature. - Initial temperature of hydrochloric acid is 21.50 °C (294.5 K) - Initial temperature of sodium hydroxide is 22.80 °C (295.8 K)
Therefore the total heat evolved from the acid-base neutralization is 2547.79 J. The number of moles of acid in base neutralized n = c*v number of moles of HCl = (0.999 M)*(0.05L) = 0.0499 moles number of moles NaOH = (0.907M)*(0.05L) = 0.0454 moles (limiting) Neutralization of 1 mole of HCl: 2547.79 J * (1 mol/0.0454 mol) = 56118.72 J/mol = 56.12 kJ/mol Since, NaOH is limiting the number of moles HCl is equal to number of moles of NaOH. Therefore, the molar enthalpy of neutralization between NaOH and HCl is – 56.12 kJ/mol. *since the reaction is exothermic the value becomes negative (Release of heat energy)
Neutralization of weak acid (DCA) and a strong base (NaOH) Number of moles of DCA = (1.007 M)(0.05 L) = 0.0504 moles Number of moles of NaOH = (0.907M)*(0.05L) = 0.0454 moles (limiting) H3O+(aq) + OH- 2H2O(l) - [reaction A] CHCl2COOH + OH- + H2O(l) CHCl2COO- + 2H2O(l) - [reaction B] - to determine amount of heat released it is necessary to know the number of moles of hydronium neutralized HA(aq) + H2O(l) H3O+(aq) + A-(aq) A-(aq) + H2O(l) HA(aq) + OHH2O(l) H3O+(aq) + OHSimultaneous equations:
Five major species: HA, A-, H3O+, OH-, Na+ 1. Ka = [H3O+][A-] / [HA] 2. Kw = [H3O+][OH-] = 1*10-14 [OH] = kw / [H3O+] 3. [HA]+[A] = CA = 1.007 M [HA] = CA – [A-] + 4. [Na ] = 0.907 M 5. [Na+] + [H3O+] = [OH-] + [A-] [A-] = [Na+] + [H3O+] - [OH-] = 0.907 + [H3O+] - kw / [H3O+] Assume [H3O+] >>> [OH-] = 0.907 + [H3O+] - kw / [H3O+] Ka = [H3O+](0.907 + [H3O+]) / 1.007 - (0.907 + [H3O+]) Let [H3O+] = x
From literature Ka2 = 5.62 * 10-2 5.62 * 10-2 = x2 + 0.907x / 0.1 – x 5.62 * 10-2 (0.1 – x) = x2 + 0.907x 5.62 * 10-3 – 0.0562x = x2 + 0.907x 0 = x2 + 0.9632x – 0.00562 Solve quadratic equation a=1 b = 0.9632 c = -0.00562 x = 5.8 * 10-3 Therefore the [H3O+] is 0.0058 Number of moles of [H3O+] = (0.0058 M) * (0.05L) = 2.9 * 10 -4 mol Therefore ther number of moles hydronium is 2.9 * 10 -4
Therefore the total heat evolved from neutralization is 2429.4 J. Heat released by reaction B: CHCl2COOH + OH- + H2O(l) CHCl2COO- + 2H2O(l)
2429.4 J = (56118.72 J/mol )* (2.9 * 10-4 mol ) + | heat released by reaction B | 2429.4 J = 16.27 + | heat released by reaction B | | heat released by reaction B | = 2413.13 J Number of moles of CHCl2COOH moles of H3O+ + moles of DCA = CNaOH * VNaOH 2.9*10-4 + moles of DCA = (0.907M)*(0.05L)
= 0.0451 Therefore 4.51*10-2 moles of DCA produces 2429.4 J of heat. Molar enthalpy of neutralization of CHCl2COOH 2429.4 J * (1/0.0451 mol) = 53866.96 J/mol = 53.87 kJ/mol Therefore the molar enthalpy of neutralization of dichloroacetic acid is -53.87 kJ/mol. The ΔH3 for ionization of CHCl2COOH H3O+(aq) + OH- 2H2O(l) - ΔH1 = – 56.12 kJ/mol CHCl2COOH + OH- + H2O(l) CHCl2COO- + 2H2O(l) - ΔH2 = -53.87 kJ/mol CHCl2COOH + H2O H3O+ + A- - ΔH3 ΔH3 = ΔH2 – ΔH1 = -53.87 kJ/mol – (- 56.12 kJ/mol ) = 2.25 kJ/mol Therefore the molar enthalpy of ionization of DCA is 2.25 kJ/mol. ΔGo of CHCl2COOH ΔGo = -RT ln Ka = - (8.314 * 10-3 ) * (22 + 273 K) * [ln(5.62*10-2)] = - 7.06 kJ/mol Therefore the ΔGo is 7.06 kJ/mol. ΔSo of CHCl2COOH ΔGo = ΔHo - TΔSo 7.06 kJ/mol = 2.25 kJ/mol – 295 K *ΔSo ΔSo = -0.016 kJ/mol K = - 16.31 J/mol K Therfore the ΔSo is -16.31 J/mol K.