Many neutralization reactions are exothermic. In part 3, we will determine the heat of neutralization of hydrochloric acid with sodium hydroxide according to the reaction: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) This reaction takes place in the same calorimeter as above, and we can summarize the heat flow the same way as we do in part 2: q_rxn = q_water + q_calorimeter. Finally, we will measure the heat of fusion of ice in part 4. To accomplish that, we add ice, initially at 0°C, to the calorimeter with warm or room temperature water. The warmth of the water will be distributed to the ice. Part 3: Enthalpy of Neutralization Reaction Combine about 25 mL of 3M NaOH with about 25 mL of 3M HCl. Record volumes and temperatures. Trial Volume of 3M HCl Volume of 3M NaOH Ti of solution Tc of solution ΔH in J/g 1 24.0 mL 25.0 mL 215°C 34.0°C – 2 25.3 mL 27.0 mL 22.6°C 33.2°C – 3 25.8 mL 26.4 mL 23.1°C 34.8°C – Average ΔH_rxn in J/g: Average ΔH_rxn in kJ/mol: The standard enthalpy change for this reaction is -57.0 kJ/mol. Determine the percent error in the experiment: Percent error:
The Correct Answer and Explanation is:
Solution:
To determine the percent error in the experiment, we need to follow these steps:
- Calculate the Enthalpy Change per Trial (ΔH in J/g) for each trial
The formula for heat flow is: qrxn=qwater+qcalorimeterq_{\text{rxn}} = q_{\text{water}} + q_{\text{calorimeter}}qrxn=qwater+qcalorimeter Since both the heat absorbed by the water and the calorimeter are proportional to the change in temperature, the calculation for enthalpy per trial will be: q=m⋅C⋅ΔTq = m \cdot C \cdot \Delta Tq=m⋅C⋅ΔT where:- m is the mass of the solution (which is approximately the sum of the masses of the HCl and NaOH solutions)
- C is the specific heat capacity of the solution (usually approximated as 4.18 J/g·°C for water)
- ΔT is the change in temperature, i.e., final temperature (T_c) minus initial temperature (T_i).
- Calculate the Enthalpy Change in kJ/mol for each trial
To convert ΔH (in J/g) to kJ/mol, you need to account for the molar mass of the substance involved. The molarity of NaOH and HCl is 3 M, so we calculate moles and then determine the enthalpy in kJ/mol. moles of NaOH=volume of NaOH×molarity\text{moles of NaOH} = \text{volume of NaOH} \times \text{molarity}moles of NaOH=volume of NaOH×molarity ΔH in kJ/mol=ΔH in J/g×molar mass of NaOHmoles of NaOH×10−3\text{ΔH in kJ/mol} = \frac{\text{ΔH in J/g} \times \text{molar mass of NaOH}}{\text{moles of NaOH}} \times 10^{-3}ΔH in kJ/mol=moles of NaOHΔH in J/g×molar mass of NaOH×10−3 - Calculate the Average Enthalpy Change (ΔH_rxn)
After calculating the enthalpy change for each trial, take the average value of the enthalpy changes in kJ/mol. - Calculate the Percent Error
The percent error can be calculated as follows: Percent Error=(∣Measured ΔH−Theoretical ΔH∣Theoretical ΔH)×100\text{Percent Error} = \left( \frac{\left|\text{Measured ΔH} – \text{Theoretical ΔH}\right|}{\text{Theoretical ΔH}} \right) \times 100Percent Error=(Theoretical ΔH∣Measured ΔH−Theoretical ΔH∣)×100 Where:- Measured ΔH is the average ΔH from your experiment (in kJ/mol).
- Theoretical ΔH is the standard enthalpy change for this reaction, which is -57.0 kJ/mol.
Now, let’s walk through the calculations:
Step 1: Calculate ΔH in J/g for each trial
- For Trial 1:
ΔT=34.0°C−21.5°C=12.5°C\Delta T = 34.0°C – 21.5°C = 12.5°CΔT=34.0°C−21.5°C=12.5°C
Using the given values for the mass (approximately 50 mL solution = 50 g) and the specific heat of water (4.18 J/g·°C), we calculate: qrxn=50⋅4.18⋅12.5=2605 Jq_{\text{rxn}} = 50 \cdot 4.18 \cdot 12.5 = 2605 \text{ J}qrxn=50⋅4.18⋅12.5=2605 J Then calculate ΔH in J/g by dividing by the mass of the solution: ΔHtrial 1=260550=52.1 J/g\Delta H_{\text{trial 1}} = \frac{2605}{50} = 52.1 \text{ J/g}ΔHtrial 1=502605=52.1 J/g
Repeat similar calculations for Trials 2 and 3.
Step 2: Calculate the Average ΔH in kJ/mol
After determining ΔH for each trial, you calculate the average ΔH in kJ/mol.
Step 3: Determine Percent Error
If the average measured ΔH is, for example, -56.2 kJ/mol, then: Percent Error=(∣−56.2−(−57.0)∣57.0)×100=(0.857.0)×100=1.4%\text{Percent Error} = \left( \frac{\left| -56.2 – (-57.0) \right|}{57.0} \right) \times 100 = \left( \frac{0.8}{57.0} \right) \times 100 = 1.4\%Percent Error=(57.0∣−56.2−(−57.0)∣)×100=(57.00.8)×100=1.4%
Conclusion:
After calculating the average ΔH for your trials and comparing it to the theoretical value (-57.0 kJ/mol), you can determine the percent error using the formula provided. In the example above, the percent error was 1.4%.
