The Molar Enthalpy Of Combustion Of Magnesium By Using Hess's Law

Prediction

Purpose

The purpose of the investigation is to use calorimetry in order to determine the molar enthalpy of combustion of magnesium by using Hess’s Law.

Question

What is the molar enthalpy of combustion ∆Hc, of magnesium?

Materials Investigation 4.5.1 Pg. 351

Eye protection Steel wool Magnesium oxide powder 10 to 15cm strip of magnesium ribbon 100mL graduated cylinder 1.00 mol/L hydrochloric acid 3 Styrofoam cups (calorimeter) Eye dropper Mass scale GLX temperature probe Fume hood

Procedure

A length of a 0.5g magnesium strip was polished using steel wool 1.00 mol/L hydrochloric acid was obtained from the fume hood 1.00 mol/L hydrochloric acid was added to the graduated cylinder to measure at 100

For both magnesium and magnesium oxide, the final temperatures were greater than the initial temperatures. An endothermic reaction results in a negative ∆ H value.

c) Enthalpy Change per mole of Magnesium

q = mc∆T q = (0.100 kg)(4.18 KJ/kg℃)(30.9℃-21.6℃) q = 3.89kJ

nMg = 0.21g /24.3g/mol nMg = 8.64 x 10-3 mol

∆H = q /n ∆H = 3.89 kJ / 8.64 x 10-3

This occurs since the heat is transferred to the surroundings, not all of the energy released from the reaction would have been transferred to the solution. The calculated enthalpies of reaction would be smaller than the accepted value. MgO on the surface of the magnesium ribbon would result in lower ∆H and ∆T values. The calculated enthalpies of reaction would be inaccurate since the excess MgO would render the Mg impure and yield false results. The enthalpy of combustion for Mg would not be accurate since the MgO coating would also contribute to the mass of the magnesium ribbon. There will be less magnesium to react and release energy so the enthalpy of combustion will be lower.

(i) Another source of error is the uncertainties and imprecisions of the lab equipment. Because of the simple calorimetry apparatus used during the process, heat may have been lost to the surroundings which results in inaccuracies in temperature and mass measurements. In addition, another source of error results from the temperature probe’s use as a stirring rod.

(j) % difference = [(experimental value – accepter value)/(accepted value)] X 100

% difference = [(-605.8 kJ – (-601.9.6 kJ))/(-601.6 kJ)] X