Hess Law

IB CHEMISTRY LAB REPORT
PRACTICAL 13 : HESS’SLAW
PRACTICAL ASSESSMENT : DCP,CE

Collecting raw data:

Part 1

Weight of magnesium ribbon,Mg,m/g
(± 0.01)
Initial temperature of hydrocloric acid,HCl,T/C
(± 0.05)
Final tmperature of
HCl + Mg,T/C
(± 0.05)

0.31
26.00
57.50

Part 2

Weight of magnesium oxide,MgO,m/g
(± 0.01)
Initial temperature of hydrocloric acid,HCl,T/C
(± 0.05)
Final tmperature of
HCl + Mg),T/C
(± 0.05)

0.50
27.50
40.50

Processing Raw Data :

Part 1

 Equation of the reaction;

Mg(s) + 2HCl(ℓ) → MgCl2(aq) + H2(g)

 no of moles :

no of moles of Mg = mass/molar mass = 0.31g___ 24.31g/mol = 0.01275 moles no of moles of HCl = molarity x volume = 1.0M x (25/1000)mL = 0.025 moles  heat of reaction will be 0.025 moles as HCL act as the limiting reactant.

 Enthalpy change of the reaction,H2

Volume of HCL solution = 25 mL

Assume the solution has the same density and specific heat capacity as water.

Volume of HCL solution = 25 mL ρ = mass/volume 1.0g/cm³ = mass/ 25 cm³ m = 25g = 0.025kg ± 0.00001

Temperature change, T = 57.50 – 26.00 = 31.5C ± 0.1

Heat evolved,Q = mHCLcwT = 0.025kg x 4200Jkg-1K-1x 31.5C = 3.3075 kJ

 H = -Q/1000n = -3.3075kJ/0.025 mol = -132.3 kJmol-1

 Uncertainty of enthalpy change of reaction of H2

Part 1

The uncertainty for the reaction above is as follows:

Uncertainty of ∆H2 = = 0.01 + 0.1 25.00 31.5 = ± 0.004

Part 2

 Equation of the reaction;

MgO(s) + 2HCl(ℓ) → MgCl2(aq) + H2O(g)

 no of moles :

no of moles of MgO = mass/molar mass = 0.50g___ 40.31g/mol = 0.01240 moles no of moles of HCl = molarity x volume = 1.0M x (25/1000)mL = 0.025 moles  heat of reaction will be 0.0025 moles as HCL act as the limiting reactant.

 Enthalpy change of the reaction,H3

Volume of HCL solution = 25 mL

Assume the solution has the same density and specific heat capacity as water.

Volume of HCL solution = 25 mL ρ = mass/volume 1.0g/cm³ = mass/ 25 cm³ m = 25g = 0.025kg ± 0.00001

Temperature change, T = 40.50 – 27.50 = 13.0C ± 0.1

Heat evolved,Q = mHCLcwT = 0.025kg x 4.2Jkg-1K-1x 13.0C = 1.365 kJ  H = -Q/1000n = -1.365kJ/0.025 mol = -54.6 kJmol-1

 Uncertainty of enthalpy change of reaction,H3 The uncertainty for the reaction above is as follows:

Uncertainty of ∆H3 = = ( 0.01 + 0.1) x546 25.00 13.0 = ± 0.008

 uncertainty H =(uncertainty H2 +uncertainty H3) = ± ( 0.004 + 0.008) = ± 0.012

Calculation of Enthalpy of Reaction

By using Hess’s Law, the enthalpy of reaction of

Mg(s) + H2O(ℓ) → MgO(aq) + H2(g) can be calculated

Mg(s) + 2HCl(ℓ) → MgCl2 (aq)+ H2(g) = -132.3 ± 0.004 kJ mol-1 MgCl2(aq) + H2O (ℓ) → MgO(s) + 2HCl(ℓ) = 54.6 ± 0.008 kJ mol-1 Mg(s) + H2O(ℓ) → MgO(aq) + H2(g) = -77.7 ± 0.012kJ mol-1

Therefore, the ∆rxn for the reaction is = -77.7 ± 0.012 kJ mol-1

 Degree of Accuracy Experimental error = Experimental value-Real value ×100% real value

=  -77.7 – (-361.0)  x 100% -361.0 = 78.48 %

Presenting Processing Data :

Part 1

No of moles Mg 0.01275 moles HCl 0.0025 moles
Heat evolved in the
Reaction,Q
3.3075 kJ
Enthalpy change in reaction, H
- 132.3 kJmol-1
Uncertainty of enthalpy change
± 0.004

Part 2

No of moles Mg 0.01240 moles HCl 0.0025 moles
Heat evolved in the
Reaction,Q
1.365 kJ
Enthalpy change in reaction, H
- 54.6 kJmol-1
Uncertainty of enthalpy change
± 0.008

Conclusion

By using the Hess’s Law, the enthalpy change of a reaction that cannot be determined straight forward in the laboratory can be found.Heat released by both reaction is detected by the rising of the temperature.As the enthalpy change in reaction get from the experiment is negative,this shows that both reaction is exothermic reaction.The experimental error of the experiment will also more than 50%.This is result from the systematic error as well as random error.

Evaluation

Limitations/weaknesses Suggestions/improvements magnesium ribbon used may consist of a little leftover oxidation parts Polishing the ribbon efficiently will ensure a pure ribbon by using sandpaper
Heat loss Providing an insulator
Parallax error on taking temperature reading repeating the experiment would improve the data collected.
Cup is not rinsed before it’s used Rinsed should be rinsed with distilled water