H+ + Cl- + Na+ + OH- → HOH + Na+ + Cl- (Helmenstine, 2007)
After eliminating the common ions, the chemical reaction is essentially:
H++ OH- → HOH (Helmenstine, 2007)
This simply means that H+ is the limiting reactant and since there is a 1:1 ratio between H+ and H2O, the amount of moles of water must be used in the calculations. As a result, the ΔH is the number of kJ/mol of water formed. Thus, the number of moles of H20 is equal to the product of the concentration and volume of the limiting reagent (n=CV) (Helmenstine, 2007). As well, the value of ΔH for this reaction, and essentially all neutralization reactions involving strong electrolytes, is -55.90 kJ/mol. Also, keep in mind that the sign for Q must change before calculating ΔH (Qrxn = -Qcalorim). The Qrxn is the actual ions and molecules that react in the reaction, while the Qcalorim contains the water, since no heat is released or absorbed because of the calorimeter (Helmenstine, 2007).
The purposes of this experiment were to determine the value of ΔH for the neutralization of NaOH with various acids to determine the unknown concentration of an HCl solution. For each of the neutralization reactions, a percent error can be calculated by comparing the calculated value of ΔH with the accepted value of ΔH for all strong acid-base neutralization reactions, which is approximately -55.90 kJ/mol. For the neutralization of phenol, the heat of reaction is -30.6kJ/mol. This value is calculated based on the fact that weak electrolyte reactions involve two steps (Helmenstine, 2007). One of the steps is the same as the neutralization of NaOH and strong acids (H++ OH- → HOH). The other step produces a heat of reaction value of 25.3kJ/mol, which means the ionization process is endothermic (Helmenstine, 2007). The sum of the two reactions produces the final result.
A percent error must have a positive number, so the subtraction part of the equation can be reversed (absolute value of the difference) (Curran, 2007). Percent error is determined using the following equation:
| (Calculated Value-Accepted Value) |/ (Accepted Value) *100 (Curran, 2007)
Procedure
The experimental procedure used for this experiment was outlined in the CHEM 120L lab manual, under Experiment #4. All steps were followed with no changes in the procedure.
Results and Observations
Table 1: Neutralization of NaOH with HCl – Trial 1
Table 2: Neutralization of NaOH with HCl – Trial 2
Table 3: Neutralization of NaOH with HNO3 – Trial 3
Table 4: Neutralization of NaOH with HNO3 – Trial 4
Table 5: Neutralization of NaOH with phenol – Trial 5
Table 6: Neutralization of NaOH with phenol – Trial 6
Table 7: Neutralization of NaOH with unknown concentration of HCl – Trial 7
Table 8: Neutralization of NaOH with unknown concentration of HCl – Trial 8
Table 9: Summary of Results
Sample Calculations
Part A – Trial 1
∙ Moles of H2O formed
n=CV
(The number of moles of the acid is the same as the number of moles of H2O)
n = (2.360 M)(0.04L)
= 0.094 mol.
∙ Number of kJ/mole of H2O formed
Total mass =90 g
Specific Heat Capacity (s) = 4.184 J/°Cg
Δ T = 17 °C
Q = (mass)(s)( Δ T)
= (90 g)(4.184 J/°Cg)(17 °C)
= 6401.52 J
= 6.402 kJ
Qrxn = -Qcalorim
Qrxn = -6.402kJ
Δ H = ΔQ/n
= -6.402kJ/0.094 mol
= -67.81 kJ/mol
Part B – Trial 3
∙ Moles of H2O formed
n =CV
= (1.868 M)(0.04L)
= 0.075 mol.
∙ Number of kJ/mole of H2O formed
Total mass =90 g
Specific Heat Capacity (s) = 4.184 J/°Cg
Δ T = 13 °C
ΔQ = (mass)(s)( ΔT)
= (90 g)(4.184 J/°Cg)(13 °C)
= 4895.28 J
= 4.895 kJ
Qrxn = -Qcalorim
Qrxn = -4.895kJ
Δ H = ΔQ/n
= -4.895 kJ/0.075 mol
= -65.52 kJ/mol
Part C – Trial 5
∙ Moles of H2O formed
n=CV
= (0.5M)(0.05L)
= 0.025 mol.
∙ Number of kJ/mole of H2O formed
Total mass =61.8 g
Specific Heat Capacity (s) = 4.184 J/°Cg
Δ T = 1 °C
ΔQ = (mass)(s)( Δ T)
= (61.8 g)(4.184 J/°Cg)(1 °C)
= 258.57 J
= 0.259 kJ
Qrxn = -Qcalorim
Qrxn = -0.259kJ
Δ H = ΔQ/n
= -0.259kJ/0.025 mol
= -10.34 kJ/mol
∙ Percent error of Trial #1
= (Calculated Value-Accepted Value)/(Accepted Value) *100
= (-67.81kJ/mol – (-55.90kJ/mol)/(-55.90kJ/mol) * 100
= 21.3%
∙ Calculation Question (Trial 1)
Qcalorim = (s)(ΔT)
= (11.7 J/°C)(17°C)
= 198.9 J
= 0.1989 kJ
Qrxn = -Qcalorim
Qrxn = -0.1989 kJ
Δ H = ΔQ/n
= -0.1989 kJ/0.094mol
= -2.11 kJ/mol
Discussion
ΔH = -55.85 kJ/mol (From Part A)
Trial 7 Trial 8
ΔQ = (mass)(s)(ΔT) ΔQ = (mass)(s)(ΔT)
= (100g)(4.184 J/°Cg)(7.4°C) = (100g)(4.184 J/°Cg)(8°C)
= 3096.16 J = 3347.2 J
= 3.096 kJ = 3.347 kJ
Qrxn = -Qcalorim Qrxn = -Qcalorim
Qrxn = -3.096kJ Qrxn = -3.347kJ
n = ΔQ/ΔH n = ΔQ/ΔH
= (- 3.096 kJ)/(-55.85kJ/mol) = (-3.347 kJ/-55.85 kJ/mol)
= 0.0554 mol = 0.0599 mol
C = n/V C = n/V
= (0.0457 mol)/(0.04L) = (0.0494 mol)/(0.04L)
= 1.39M = 1.50 M
Therefore, the unknown concentration of the HCl solution was 1.39M and 1.50M for trials 7 and 8, respectively. Only the concentration found in trial 8 lies in the accepted range of concentrations (1.5M – 3.0M).
For trials 1-4, which involved reactions between a strong base (NaOH) and strong acids, the number of kilojoules/mol of H20 formed were relatively close to one another. These two neutralization reactions were NaOH with HCl and NaOH with HNO3. Of the four trials for the two reactions, trial 1, which was the reaction between NaOH and HCl, generated the most heat per mole of H2O formed (-67.81kJ/mol). However, the values of ΔH for trials 1-4 were all relatively close and only minor errors accounted for the discrepancies. As stated in the introduction, a strong electrolyte completely, or almost completely, ionizes in solution, while a weak electrolyte only partially ionizes in solution. When a weak acid neutralizes with a strong base, part of the heat released is absorbed back into the system. The reason is that the weak acid needs a little bit of energy for its ionization. Therefore, a weak acid-strong base neutralization would release less heat/energy than a strong acid-base reaction. This is proven in the results section, as only -10.34kJ/mol and -20.69 kJ/mol were produced for trials 5 and 6, respectively. Of all the acids, the strong acids were HCl and HNO3 (nitric acid), while phenol was the only weak acid in the experiment.
The percent errors for this trials with the strong acid-base reactions were relatively low. For trial 1, the percent error was 21.3%. For trial 2, the percent error was only 0.09%. For trial 3, the percent error was 17.2% and for trial 4, the percent error was 9.8%. The percent errors for trials 5 and 6 were 66.2% and 32.4%, respectively. It is interesting to note that the second trial for each separate experiment produced the best results for each part. There were certain factors that contributed to the lack of accuracy in the calculated values. One source of error could have come directly from the calorimeters. If there was a hole present in the Styrofoam cup that was undetected, this would mean the cup did not represent a true calorimeter. Since it is no longer a closed and isolated environment, energy that is released would not be kept inside the calorimeter but it would be released into the outside environment. This would lead into discrepancies in the temperature change, resulting in inaccurate ΔH values. A better calorimeter, other than a Styrofoam cup, would help prevent any of these problems. Another source of error could have resulted from the neutralization reactions. If any of the acids did not completely react with NaOH, this would prevent the temperature from reaching its true maximum value, which once again leads to inaccurate ΔH values. The final error could have resulted from insufficient amounts of either the acid or base in the neutralization reactions. If there was not enough acid or base present in any of the reactions, this would mean that the value of the mass used in the calculations would be inaccurate. Any measuring errors would definitely lead to different temperatures, which was evident in the results section with the discrepancies between individual trials.
Conclusion
The first purpose of this experiment was to neutralize NaOH with various acids and determine the value of ΔH for each trial. These values were compared with the actual value of ΔH for all neutralization reactions involving either a strong acid- base reaction or a strong base & weak acid reaction. The percent errors for this experiment ranged from as low as 0.09% to as high as 66.2%. The high percent errors came from the weak acid- strong base neutralization reactions, while the four trials for the strong acid- base reactions produced very good results. The errors that accounted for these discrepancies included human errors and errors associated with the apparatus. The other purpose was to determine the unknown concentration of a HCl solution, which was calculated to be 1.39M and 1.50M for trials 7 and 8, respectively. Only the value for trial 8 did in fact lie in the range of accepetable concentrations, which was from 1.5M to 3M. Further studies in the area of neutralization reactions could include determining ΔH using a weak base and the use of a bomb calorimeter rather than a Styrofoam cup calorimeter.
References
Brain, M. (2007) “How Car Engines Work” 25 November 2007
< >
Curran, G. (2007) “Percent Error” 25 November 2007.
< >
Helmenstine, A. (2007) “Measurement of Heat Flow & Enthalphy Change” 25 November 2007.
< >
Jones, L. (2007) “Heat Notes” 25 November 2007
< >
Tracey, L. (2003) “Is It A Strong or Weak or Nonelectroylte?” 25 November 2007.
< >
University of Florida (2007) “Calorimetry” 25 November 2007
< http://itl.chem.ufl.edu/2045/lectures/lec_9.html >
Gagan Samra Heat of Neutralization
*This is simply used for reading and may not be copied or edited in any way or else it is plagiarism.