Hess’s Law Lab Report PDF

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31st January 2017

Caitlin Bettenay

Hess’s Law: ABSTRACT: INTRODUCTION: The following experiment has been designed to determine the enthalpy for MgO(s) using Hess’s Law and simple calorimetry. Hess’s Law states that: regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes which is clearly depicted in Figure 1 below. This law is a manifestation that enthalpy is a state function. Chemical reactions involve the breaking of chemical bonds in a given set of reactants, and the formation of other different chemical bonds in a given set of products. A thorough examination of the many roles that energy changes have in chemical processes can lead to insights into other chemical phenomena.

Figure 1: Diagram displaying the process of how Hess’s Law works

According to Hess's Law, if two or more reactions can be added to give a net reaction, Δ H° for the reaction is the sum of the Δ H°'s for the reactions which are added. In this lab, finding the enthalpy of formation for MgO (s) was done by finding the enthalpies for the net of the reaction that gives MgO as a product. Here are the reactions: Mg (s) + 2HCl (aq) ---> MgCl2 (aq) + H2 (g) (1) MgCl2 (aq) + H2O (l) ---> MgO (s) + 2HCl (aq) (2) H2 (g) + 1/2 O2 (g) ---> H2O (l) (3) Mg (s) + 1/2 O2 (g) ---> MgO (s) (4) The purpose of this experiment is to determine the heat for the reactions

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1 and 2 experimentally, and then to calculate the Δ H° for the reactions. Then, to use the known value of formation of water (-285.9 kJ/mol) to calculate Δ H° for the reaction 4 which is the enthalpy of formation of MgO.

DATA/RESULTS: Table 1: Table displaying the temperature change in reactions between Magnesium (Mg) and Hydrochloric Acid (HCl), and Magnesium Oxide (MgO) and HCl. Mg

Mass (g) Tinitial (°C) Tfinal (°C) Δ T (°C)

MgO

Trial 1

Trial 2

Trial 1

Trial 2

0.533 21.4 59.42 38.1

0.480 21.4 57.26 35.8

0.806 21.5 34.76 13.2

0.845 21.6 35.78 14.2

CALCULATONS: q sol =C ×m ×∆ T q rxn=−q sol m n= molar mass q ∆ H= rxn n

(1)

Mg (s)+2 HCl (aq )−−→ MgCl2 (aq)+ H 2(g )

Trial 1: q sol=4.18×(0.533 + 50)×38.1=8047 J =8.05 kJ q rxn=−8.05 kJ 0.533 n= =0.022 mol 24.305 −8.05 ΔH= =−365.9 kJ /mol 0.022 Trial 2 : q sol=4.18×(0.480 + 50)×35.8=7554 J =7.55 kJ q rxn=−7.55 kJ 0.480 n= =0.019 mol 24.305

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Caitlin Bettenay

ΔH=

−7.55 =−397.4 kJ /mol 0.019

ΔH=

(−365.9 +(−397.4 )) =−381.7 kJ / mol 2

31st January 2017

(2)MgO(s)+2 HCl (aq)−−→ Mg (s)+1/2 O 2(g) Trial 1: q sol=4.18×(0.806 +50)×13.2=2803 J =2.8 kJ q rxn=−2.8 kJ 0.806 n= =0.02mol 40.3 −2.8 ΔH= =−140 kJ /mol 0.02 Trial 2 : q sol=4.18×(0.845 + 50)×14.2=3018 J =3.0 kJ q rxn=−3.0 kJ 0.845 n= =0.021mol 40.3 −3.0 =−142.9 kJ /mol ΔH= 0.021 ΔH=

(−140 +(−142.9)) =−141.5 kJ /mol 2

Mg( s) +2 HCl ( aq) −−→ MgCl 2 (aq )+ H 2 ( g ) Δ H =−381.7 kJ /mol MgCl 2( aq ) + H 2 O (l ) −−→ MgO ( s )+2 HCl ( aq ) Δ H =141.5 kJ /mol

H 2(g)+1/2 O2(g)−−→ H 2O (l) Δ H=−285.9 kJ /mol Mg(s)+1/2 O 2(g)−−→ MgO(s ) Δ H =−381.7+141.5 +(−285.9 )=−526.1 kJ /mol ∴The enthalpy of formation for MgO(s)is−526.1 kJ /mol .

DISCUSSION: The purpose of this experiment was to find the enthalpy of formation for MgO (s) using Hess's Law. Hess's Law is implemented when the enthalpy of the substance cannot be determined directly, but can be found by adding the enthalpies of the reactions that create net reactions, and as a product, provide a substance with an unknown enthalpy. In this experiment, the enthalpies of reactions 1, 2 and 3 were added to get the enthalpy of the formation for MgO (s). The second reaction was done in

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the opposite direction, so the negative value of the enthalpy of the second reaction can be used in calculations. This experiment contained a deficiency in that it was conducted through using open coffee cup calorimetry. As there was no lid, some of the heat was lost from the system, thus the overall temperature change was calculated as a lesser value than it actually was. The heat flow of the solutions were positive in these cases which led to a greater heat flow of the reactions which in turn, led to a greater enthalpy of the reactions. The standard enthalpy of the formation for MgO (s) was calculated as -601.8 kJ / mol. The result of the experiment was -526.1 kJ / mol, so as previously described, the possible source of error, the missing lid, caused the value of the experiment to be higher than expected. ADDITIONAL QUESTIONS: 1. The exothermic reaction is the reaction where the temperature observed rises. 2. In the exothermic reaction, Δ H is negative, as the heat flow of the reaction is negative, since the heat flow of the solution is positive. 3. Reaction 1 is exothermic. Reaction 2 is endothermic. 4. The quantity, q, was positive and it's measured using kilojoules (kJ). The enthalpy, Δ H was negative and it's measured in kilojoules per mole (kJ / mol). a) The mass of the alloy would first have to be obtained. Then it would be dissolved in HCl so the Mg can react with it. Since the other metal won't react with HCl, it could be dried and measured. When the mass of the metal that didn’t react with the HCl is subtracted from the mass of the alloy, it results in the mass of the Mg. Using proportion (ratios), the percent of Mg in the alloy can be calculated: mass of Mg × 100 % percent Mg= mass of the alloy b)

Mass of Mg=

percent Mg×mass of the alloy 100 %

m (Mg)=30 %∗5 g /100 % m (Mg)=1.5 g Δ H =−381.7 kJ /mol 1.5 g g /mol=0.062mol n= 24.305 qrxn=Δ H ×n=−381.7 × 0.062=−23.67 kJ qsol=23.67 kJ

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Caitlin Bettenay

31st January 2017

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