Enthalpy of Neutralization

Introduction Enthalpy change, ΔH, is defined as the heat output of a system as it goes through a reaction under constant pressure. It is an important aspect of thermochemistry, which is the study of energy changes during a chemical or physical reaction . When we calculate enthalpy change, we always assume that the pressure is constant. We are able to calculate enthalpy change numerous ways, notably by the increase in heat, Q, given by an exothermic reaction or the heat absorbed by an endothermic reaction. To do this, we use the concept of calorimetry, the measuring of heat of chemical reactions or physical changes. For this concept, we use a device called a calorimeter, which is a device that creates an isolated system that enables the user to accurately measure the change in temperature, ΔT. We use the following formula in order to calculate Q, quantity of heat, given the mass of the substance, m; its specific heat capacity, c; and the change in temperature, ΔT, given in °C:
Q = m · c · Δt [1] Using this equation, we are able to calculate the heat lost or gained by one of the substances in the reaction. Since heat lost = heat gained in a reaction, we are able to therefore say that the heat lost by the first substance, is equal to the heat gained by the second substance, and therefore:
Qsubstance 1 = -Qsubstance 2 [2]
Since the law of conservation of energy states that energy cannot be created or destroyed, the change in energy must equal zero and therefore the left side of this equation must be equal in magnitude but opposite in sign of the right side. By substituting equation [1] into equation [2] we are able to derive: m1 · c1 · Δt1 = - (m2 · c2 · Δt2) [3] and thus we are able to calculate an unknown given the other 5 variables. The determination of the change in temperature can be done by taking the temperatures in intervals and giving a graphical interpretation of these values. In these experiments, the initial temperature of the substance in question is taken before the second substance is mixed into the calorimeter. Therefore, the determination of the final temperature is slightly more complex. In the perfect case, the final temperature would be able to be calculated immediately after the substances were mixed, and we would be able to form a graph that consists of a straight line with constant slope. In the case of a dissolution, a curve is formed, because the temperature change occurs as the substance is dissolving into the solvent. Therefore, when a plateau has been reached, this is the final temperature. We can find enthalpy using many different methods. We are able to find the enthalpy of neutralization, a reaction between an acid and a base, by measuring the heat absorbed or produced per mole of a certain substance. Therefore we can use the following equation, where ΔHN is equal to the enthalpy of neutralization; QN is the heat absorbed or produced; and nsub is the number of moles of the substance under consideration:
ΔHN = QN / nsub [4] Furthermore, we can measure the enthalpy of a solution, or of a dissolution, by the of addition the heat absorbed or produced per mole (n) of substance dissolved dissolved by the two substances involved in the reaction. In all cases, a positive enthalpy value results in an endothermic reaction because the solution absorbs energy, and a negative enthalpy value produces an exothermic reaction because the solution loses energy (Waterloo University).
Procedure:
The experiment was performed following the procedure outlined in Lab # 3: The Enthalpy of Various Reactions. No modifications were made (CHM 1311 Laboratory Manual , 2013).
Observations:
Table 1: Enthalpy of Copper
Measurement
Trial 1
Trial 2
Mass of metal (g)
10.68
10.51
Mass of calorimeter (g)
10.87
10.87
Mass of Lids (g)
7.088
7.088
Volume of Distilled Water (mL)
20.0
20.5
Mass H2O and Calorimeter (g)
30.97
31.48
----
-----
----
Initial Temp. of H2O (°C)
22.4
23.2
0.5 min temp.
22.4
23.2
1:00 min temp.
22.5
23.3
1.5 min temp.
22.5
23.4
2:00 min temp
22.5
23.4
2.5 min temp.
22.5
23.4
Temp. Boiling H2O (°C)
98.7
98.8
Temperature Intervals (min)
(°C)
(°C)
3:00

3:10
25.7
26.6
3:20
25.7
26.6
3:30
25.7
26.6
3:40
25.7
26.5
3:50
25.7
26.5
4:00
25.7
26.5
4:10
25.6
26.5
4:20
25.6
26.5
4:30
25.6
26.5
4:40
25.6
26.5
4:50
25.6
26.5
5:00
25.6
26.5
5:10
25.6
26.5
5:20
25.6
26.5
5:30
25.6
26.5
5:40
25.5
26.5
5:50
25.5
26.5
6:00
25.5
26.5
6:30
25.5
26.5
7:00
25.5
26.5
Observations
Temperature of the water rose to 25.7 °C from 22.5°C when copper was added (Trial 1) and temperature rose to 26.6°C from 23.4 °C when copper was added (Trial 2)
Temperature goes down gradually until reaching equilibrium at 25.5 °C for Trial 1 and 26.5 °C for Trial 2.
Table 2: Enthalpy of Neutralization (HNO3 Acid)
Measurement
Trial 1
Trial 2
Volume 1.1 M HNO3 (mL)
50.0
50.0
Concentration NaOH (M)
1.0
1.0
Volume NaOH (mL)
50.0
50.0
Mass of Calorimeter + Base (g)
63.38
63.38
-------
-------
-------
Initial Temp. NaOH (°C)
22.2
22.1
0.5 min temp
22.2
22.1
1:00 min temp.
22.2
22.2
1.5 min temp.
22.3
22.2
2:00 min temp.
22.3
22.2
2.5 min temp.
22.3
22.2
Temperature Intervals (min)
(°C)
(°C)
3:00

3:10
29.7
29.3
3:20
29.7
29.3
3:30
29.7
29.2
3:40
29.7
29.2
3:50
29.7
29.2
4:00
29.7
29.2
4:10
29.7
29.2
4:20
29.7
29.2
4:30
29.7
29.2
4:40
29.7
29.2
4:50
29.7
29.2
5:00
29.6
29.1
5:10
29.6
29.1
5:20
29.6
29.1
5:30
29.6
29.1
5:40
29.6
29.1
5:50
29.6
29.1
6:00
29.6
29.1
6:30
29.6
29.1
7:00
29.6
29.1
Observations
Temperature goes up from 22.3°C to 29.7°C between mixing in Trial 1 and from 22.2 °C to 29.3°C in trial 2.
Temperature settled at 29.6°C in trial 1 and 29.1 °C in trial 2.
Table 3: Enthalpy of Unknown Salt B
Measurement
Trial 1
Trial 2
Mass of salt B (g)
1.501
1.497
Mass of Calorimeter (g)
10.87
10.87
Volume H2O (mL)
20.5
20.0
Mass of Calorimeter and H2O (g)
30.96
30.39
Initial temp. H2O (°C)
22.9
23.2
0.5 min temp.
22.9
23.2
1:00 minute temp.
22.9
23.2
1.5 min temp.
22.9
23.2
2:00 min temp.
22.9
23.2
2.5 min temp.
22.9
23.2
Temperature Intevals (min)
(°C)
(°C)
3:00

3:10
17.9
21.1
3:20
17.8
21.2
3:30
17.8
21.2
3:40
17.8
21.3
3:50
17.8
21.3
4:00
17.8
21.3
4:10
17.8
21.3
4:20
17.8
21.4
4:30
17.8
21.4
4:40
17.8
21.4
4:50
17.8
21.4
5:00
17.8
21.4
5:20
17.9
21.5
5:30
18.0
21.6
5:40
18.0
21.6
5:50
18.0
21.6
6:00
18.1
21.6
6:30
18.2
21.7
7:00
18.3
21.8
7:30
18.4
21.8
8:00
18.4
21.8
Observations
Temperature goes down from 22.9°C to 18.0 °C after mixing in trial 1 and from 23.2°C to 21.1°C after mixing in trial 2.
Temperature gradually increases over intervals.
Table 4: Enthalpy of Neutralization (HCl)
Measurement
Trial 1
Trial 2
Volume 1.1 M HCl (mL)
50.0
50.0
Concentration NaOH (M)
1.0
1.0
Volume NaOH (mL)
50.0
50.0
Mass of Calorimeter + Base (g)
63.50
63.28
-------
-------
-------
Initial Temp. NaOH (°C)
22.1
22.4
0.5 min temp
22.1
22.4
1:00 min temp.
22.1
22.4
1.5 min temp.
22.1
22.4
2:00 min temp.
22.1
22.4
2.5 min temp.
22.1
22.4
Temperature Intervals (min)
(°C)
(°C)
3:00

3:10
29.2
29.4
3:20
29.2
29.4
3:30
29.2
29.3
3:40
29.2
29.3
3:50
29.2
29.3
4:00
29.2
29.3
4:10
29.2
29.3
4:20
29.1
29.3
4:30
29.1
29.3
4:40
29.1
29.2
4:50
29.1
29.2
5:00
29.1
29.2
5:10
29.1
29.2
5:20
29.1
29.2
5:30
29.1
29.2
5:40
29.1
29.2
5:50
29.1
29.2
6:00
29.1
29.1
6:30
29.0
29.1
7:00
29.0
29.1

Observations
Temperature goes up from 22.1°C to 29.2°C between mixing in Trial 1 and from 22.4 °C to 29.4°C in trial 2.
Temperature settled at 29.0°C in trial 1 and 29.1 °C in trial 2.

Results:
See appendix A for results.
Calculations
Part 1: Enthalpy of Copper
1. Δtwater = t2 – t1
Trial 1: = 25.5 °C – 22.4 °C Trial 2: = 26.5 °C – 23.2 °C = 3.1°C = 3.3 °C

2. q=mcΔt
Trial 1: q=(mass of calorimeter+H2O - mass of calorimeter)(4.184 J/g°C)(3.1°C) =(30.97 – 7.088) )(4.184 J/g°C)(3.1°C) = 309.8J
Trial 2: q=(31.48 – 7.1750) )(4.184 J/g°C)(3.3°C) = 336.8 J

3. Δtmetal = t2 – t1
Trial 1: Δt= (98.7 °C – 25.5 °C) Trial 2: Δt= (98.8°C – 26.5°C) = 73.2 °C = 72.3 °C

4. qgained = -qlost qmetal = mcΔt
Trial 1: 309.8 J = (10.68g)(c)(73.2°C) Trial 2: 336.8 J = (10.51g)(c)(72.3°C ) c = 0.396 J/g °C c = 0.443 J/g °C

5. c · MM = 24.9 J/(K mol)
Trial 1: MM = 24.9 / c Trial 2: MM = 24.9 / 0.443 = 24.9 / 0.396 = 56.2 g/mol = 62.8 g/mol

6. MMcopper = 63.546 g/mol
Percent Yield = Actual yield – theoretical yield x100 Theoretical yield Trial 1: 62.8 – 63.546 x100 Trial 2: 56.2 – 63.546 x100 63.546 63.546 =-1.17 % = -11.6 %
Part 2: Enthalpy of Neutralization for HNO3
7. Δtsolution = t2 – t1
HNO3: Δt = 29.6 °C – 22.2 °C HCl: Δt = 29.0 °C – 22.1 °C = 7.4 °C = 6.9 °C

8. Volumefinal = VNaOH + Vacid
HNO3: = 50.0 mL NaOH + 50.0 mL HNO3 HCl: V = 50.0 mL NaOH + 50.0 mL HCl = 100.0 mL = 100 mL

9. Massfinal = Volumefinal (Assuming 1.0g/1.0 mL)
HNO3: m = 100.0 g HCl: m = 100.0 g

The mass measured in the experiment is higher because we included the weight of the calorimeter and lid in the measurement. Therefore, the measured mass would be off by that amount of the mass of the calorimeter and lid. They would both be equally accurate once the calorimeter and lids were subtracted off the final volume, barring there was no extra weight in the calorimeter (see discussion for errors).

10. q= mcΔt
HNO3: q=(100g)(4.184J/g°C)(7.4°C) HCl: q= (100g)(4.184 J/g°C)((6.9°C) = 3.10 x 103 J = 2.89 x 103 J

11. Concentration NaOH: 1 mol/L
Volume NaOH = 0.05 L nOH = c · V = 0.05 mol OH

12. HCL + NaOH NaCl + H2O
Therefore the ratio is 1:1, and 0.05 moles of H2O is formed (see #11)

13. ΔHN = QN / molx
HNO3: ΔHN = 3096.16 J / 0.05 mol H2O HCl: ΔHN =2886.87 J / 0.05 mol H2O = -61.9 kJ/mol = -57.7 kJ/mol

14. The values for the acids are relatively close to one another. This indicates that the heat capacity for the two acids are considerably close to one another and that they react very similiarly when placed in a solution of NaOH.

15. -57.3 kJ/mol for both HCL and HNO3
Therefore,
HNO3: HCl:
Actual – Theoretical x100 = -57.7 - - 57.3 Theoretical -57.3 = -61.9kJ/mol - -33.38 kJ/molx100 =0.70% - 57.3 kJ/mol
= 8.0 %

Part 3: Enthalpy of Dissolution of a Salt (Salt B)
16. Δtsolution = t2 – t1
Trial 1: Δt = (18.4 °C – 22.9 °C) Trial 2: Δt = (21.8°C – 23.2 °C) = -4.5 °C = -1.4°C

17. qsalt= mcΔt
Trial 1: q = (1.501)(3.877J/g°C)(-4.5) Trial 2: q = (1.497g)( 3.877J/g°C)(-1.4 °C) =-26 J = -8.1 J

18. nsalt = Trial 1: 1.501 g / 101.11 g/mol = 0.0148 mol = Trial 2: 1.497 g / 101.11 g/mol = 0.0148 mol

qwater= mcΔt
Trial 1: q = (30.900g – 1.501g)(4.184J/g°C)(4.5°C) q =550 J
Trial 2: q = (30.393g – 1.497g) (4.184J/g°C)(1.4 °C) q =170 J

ΔHS = ( - cwatermwaterΔtwater) + (- csaltmsaltΔtsalt) nsalt nsalt
Trial 1: ΔHS =[ -550J / 0.0148 mol] + [26 J / 0.0148 mol] = -35.4 kJ/mol Trial 2: ΔHS = [-170 J / 0.0148 mol] + [8.1 J / 0.0148 mol] = -10.9 kJ/mol

19. Enthalpy Value for dissolution of KNO3 (Salt B) = 34.89 kJ/mol (Parker, 1965).

Actual – Theoretical x100 Trial 2 = -10.9 - - 34.89 Theoretical -34.89 = -35.4kJ/mol - -34.89 kJ/molx100 =130% - 34.89 kJ/mol
= 1.5%

Discussion: In lab 3, Enthalpy of Various Reactions, all four experiments went extremely smooth. We were able to complete all 4 in the time permitted, and thus all of our results are admissible. For part 1, Enthalpy of Copper, we obtained a very good value, taking into consideration our percent error of -1.17%. However in our second trial our percent error was much higher, at -11.6%. The higher percent error in the second trial could be attributed to many areas of the lab, as there was a high risk of error in most of the steps. Human error would be a likely source of error in this experiment. The lid of the calorimeter might not have re-secured in sufficient enough time after the metal was placed into the water, and thus a lot of heat would have been lost. This would result in errors in both our graphs and our results, as the initial temperature would be much higher, and the change in temperature much lower. Thus the calculations involving Δt would all be skewed. Also, it would have been easy to misread the graduated cylinder that contained the water, which would change the mass of the water and change the value we got for q. However, the most likely cause of error would be the quality of the calorimeter. After a certain amount of trials, the styrofoam may have lost its ability to contain 100% of the heat. Also, it was noticed at the end of our experiments that there was water caught between the lids of the calorimeter, thus adding mass to the calorimeter and skewing the results. All in all, our graphs seem to resemble one another, as they both seem to follow the same pattern. This indicates that the experiment did go well, and that we can trust our results. In part 2, Enthalpy of Neutralization, gave us two extremely reliable results. Our percent error was 8.0% and 0.70% in trials 1 and 2, respectively. The theoretical results were from a website called tutorvista, which is a very reliable source. They gave a value for the enthalpy of neutralization per mole of water for HCl and HNO3 of -57.3 kJ/mol. This value is very close to our results, trial 1 being -61.9 kJ/mol and trial 2 being -57.7 kJ/mol. The sources of error in this section of the lab resembles those of part 1. Again, human error involving both the measurement of the acid, and the base, could have drastically affected our results. Also, if the lid was not put on in a decent amount of time, or even not put on securely, heat would have escaped and this would have affected our Δt value severely. Our graphs show the same pattern, and thus indicate our results are very accurate. Part 3, Enthalpy of a Salt, was performed with an unknown salt. The molar mass of this salt was given in Table 2: Unknown Salts for Part 3 of the Experiment providing Molar Masses, Masses Required and Specific Heat Capacities of the Resulting Solutions Formed in the lab manual, as 101.11 g/mol. The specific heat capacity was also given as 3.877 J/g°C. With given molar mass of the unknown salts, the value was compared with values of specific salts in The Journal of the American Chemical Society and I can conclude that unknown salt B, is potassium nitrate, KNO3 (Randall & Rossini, 334-335). The graph for this shows that the salt, when immersed in water, goes down in value, proving it to be an endothermic reaction. With the given literature value of 34.89 kJ/mol KNO3, the percent errors were able to be determined as 1.46% and 130%. For trial 1, the experiment was performed with extreme accuracy; however, the evidence of trial 2 shows that there was huge error. Again, the sources of error, could be due to impurities in the salt in trial 2.
Conclusion:

The molar mass of Copper obtained was 53.9 g/mol (average) a giving a percent error of -6.65%. The enthalpy of neutralization obtained was -57.3 kJ/mol for both HCL and HNO3, giving a percent error of 8.0% and 0.70% respectively with the calculated values of -61.9 kJ/mol (HNO3) and -57.7 kJ/mol (HCl). Finally, I concluded that unknown salt B was KNO3, as its molar mass and other properties resemble that of the unknown salt, and the with the literature value for the dissolution of KNO3 being 34.89 kJ/mol , the percent error for the trials 1 and 2 were 1.5% and 130%, respectively.
Works Cited
Parker, V . B ., Thermal Properties of Uni-Univalent Electrolytes, Natl . Stand . Ref . Data Series — Natl . Bur . Stand .(U .S .), No .2, 1965
Randall, Merle, Rossini, Frederick D., The Journal of the American Chemical Society, Vol. 51(2), February 1929, Table VII, pp.334-335. tutorvista. (2010). Enthalpy of Neutralization. Retrieved Oct 7, 2013, from tutorvista.com: http://www.tutorvista.com/bow/enthalpy-of-neutralisation
Waterloo University. (n.d.). Thermochemistry. Retrieved Oct 7, 2013, from www.science.uwaterloo.ca: http://www.science.uwaterloo.ca/~cchieh/cact/c120/thermosum.html
What in the World Isn 't Chemistry?(2013). General Chemistry. CHM 1311. Laboratory Manual: University of Ottawa.

Cited: Parker, V . B ., Thermal Properties of Uni-Univalent Electrolytes, Natl . Stand . Ref . Data Series — Natl . Bur . Stand .(U .S .), No .2, 1965 Randall, Merle, Rossini, Frederick D., The Journal of the American Chemical Society, Vol. 51(2), February 1929, Table VII, pp.334-335. tutorvista. (2010). Enthalpy of Neutralization. Retrieved Oct 7, 2013, from tutorvista.com: http://www.tutorvista.com/bow/enthalpy-of-neutralisation Waterloo University. (n.d.). Thermochemistry. Retrieved Oct 7, 2013, from www.science.uwaterloo.ca: http://www.science.uwaterloo.ca/~cchieh/cact/c120/thermosum.html What in the World Isn 't Chemistry?(2013). General Chemistry. CHM 1311. Laboratory Manual: University of Ottawa.