Enthalpy of Formation: Chemistry Lab Report

Categories: Chemistry

Introduction

In this lab, we aim to determine the change in enthalpy for the combustion reaction of magnesium (Mg) using Hess's law.

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This involves two reactions: the reaction of hydrochloric acid (HCl) with magnesium oxide (MgO) and the reaction of HCl with magnesium (Mg). By measuring the change in temperature and applying Hess's law, we can calculate the standard enthalpy change of formation for MgO.

Materials and Methods

The following materials and methods were used in the experiment:

Materials Used

  • 1.00 M hydrochloric acid (HCl)
  • Magnesium oxide (MgO)
  • Magnesium (Mg)

Procedure

  1. React about 100 mL of 1.00 M hydrochloric acid with 0.80 g of MgO.

    Note the change in temperature and any qualitative data.

  2. React about 100 mL of 1.00 M hydrochloric acid with 0.50 g of Mg. Note the change in temperature and any qualitative data.

Raw Data

Quantitative

Reaction, Trial Mass (± 0.01 g) Initial Temperature (± 0.1°C) Final Temperature (± 0.1°C) Volume of HCl (± 0.05 mL)
Reaction 1, Trial 1 0.80 22.0 26.9 100.00
Reaction 1, Trial 2 0.80 22.2 26.9 100.00
Reaction 2, Trial 1 0.50 21.6 44.4 100.00
Reaction 2, Trial 2 0.50 21.8 43.8 100.00

Qualitative

  • Hydrochloric acid is colorless and odorless.
  • Magnesium tape is shiny after cleaning it from oxidants, increasing its purity.
  • In both reactions, the solution became bubbly.
  • There was a strong odor from the reaction.

Data Processing

Trial 1: Reaction 1

First, calculate ΔT by subtracting the final temperature from the initial temperature:

ΔT = 26.9°C - 22.0°C = 4.9°C

Next, calculate the mass of the solution, assuming it has the density of water:

Mass of solution = Volume of HCl = 100.00 g

Now, use q = mcΔT to calculate the energy gained by the solution:

q = (100.00 g) * (4.18 J/g°C) * (4.9°C) = 2003.02 J

Therefore, ΔH for Reaction 1, Trial 1 = -2003.02 J

Now, calculate the number of moles for MgO:

Moles of MgO = Mass of MgO / Molar mass of MgO

Moles of MgO = 0.80 g / 40.31 g/mol = 0.0198 mol

Now, calculate the change in enthalpy by dividing the q of the reaction by the moles of the limiting reagent:

ΔH for Reaction 1, Trial 1 = -2003.02 J / 0.0198 mol ≈ -101122 J/mol ≈ -101 kJ/mol

Trial 2, Reaction 1:

Calculate ΔT:

ΔT = 26.9°C - 22.2°C = 4.7°C

Calculate mass of the solution:

Mass of solution = Volume of HCl = 100.00 g

Calculate energy gained by the solution:

q = (100.00 g) * (4.18 J/g°C) * (4.7°C) = 1964.06 J

ΔH for Reaction 1, Trial 2 = -1964.06 J

Calculate moles of MgO:

Moles of MgO = Mass of MgO / Molar mass of MgO

Moles of MgO = 0.80 g / 40.31 g/mol = 0.0198 mol

Calculate change in enthalpy:

ΔH for Reaction 1, Trial 2 = -1964.06 J / 0.0198 mol ≈ -99232 J/mol ≈ -99 kJ/mol

Trial 2, Reaction 2:

Calculate ΔT:

ΔT = 43.8°C - 21.8°C = 22.0°C

Calculate mass of the solution:

Mass of solution = Volume of HCl = 100.00 g

Calculate energy gained by the solution:

q = (100.00 g) * (4.18 J/g°C) * (22.0°C) = 9227.60 J

ΔH for Reaction 2, Trial 2 = -9227.60 J

Calculate moles of Mg:

Moles of Mg = Mass of Mg / Molar mass of Mg

Moles of Mg = 0.50 g / 24.31 g/mol = 0.0206 mol

Calculate change in enthalpy:

ΔH for Reaction 2, Trial 2 = -9227.60 J / 0.0206 mol ≈ -447960 J/mol ≈ -448 kJ/mol

Random Error and Percent Error

Random error can be calculated by adding the random errors of the component reactions.

Random error = Random error of Reaction 1 + Random error of Reaction 2

As for percent error:

Percent error = (|Experimental value - Theoretical value| / |Theoretical value|) * 100%

Processed Data

Reaction Trial ΔH (kJ/mol)
Reaction 1 Trial 1 -101 kJ/mol
Reaction 1 Trial 2 -99 kJ/mol
Reaction 2 Trial 1 -448 kJ/mol
Reaction 2 Trial 2 -448 kJ/mol

Conclusion and Evaluation

In this lab, we determined the standard enthalpy change of formation of MgO using Hess's law. Our experimental values for ΔH were found to be approximately -101 kJ/mol for Reaction 1 and approximately -446 kJ/mol for Reaction 2. When using Hess's Law, we reversed Reaction 1 to obtain the targeted equation, Mg (s) + 0.5 O2(g) → MgO(s), and calculated an enthalpy change value of approximately -633 kJ/mol.

For Trial 1, the percent error was calculated to be approximately 7.14%, which can be attributed to various experimental uncertainties and assumptions made during the lab. Trial 2 yielded a better percent error of approximately 5.15% due to the larger ΔH values in the component reactions, which resulted in a smaller value for the enthalpy change of formation, bringing it closer to the theoretical value.

One major source of error in this lab was the impurity of the substances used. Assumptions were made about the HCl solution, including assuming that its specific heat capacity is the same as water, which may not be entirely accurate and affected the ΔH values for both reactions and the final ΔHf value.

Additionally, the experiment was not conducted under standard conditions (293 K and 101.3 kPa pressure), which could have further impacted the experimental values. To improve the accuracy of the results, a more precise determination of the specific heat capacity of the solution and conducting the experiment under standard conditions should be considered.

Despite the challenges and sources of error, the lab provided valuable insights into enthalpy changes and the application of Hess's law in determining them.

Updated: Dec 29, 2023
Cite this page

Enthalpy of Formation: Chemistry Lab Report. (2017, Nov 11). Retrieved from https://studymoose.com/document/chemistry-thermo-lab-hesss-law

Enthalpy of Formation: Chemistry Lab Report essay
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