-
Calculate the temperature change, ΔT, of the HCl.
Solution: ∆THCl = ∆THClfinal -∆THClinital
∆TH20 = 27.4°C -22.1°C
∆TH20 = 5.3°C
-
Calculate the Quantity of Heat absorbed by the water in the can, Assume that the specific heat of HCl is 4.18J/g°C the same as water and use the formula of Q=mc∆T
Solution: QHCl = mHCl. cHCl. ∆THCl
QHCl = 100g×4.18J/g°C×5.3°C
QHCl= 418J/°C × 5.3°C
QHCl= 2215.4 J
QH20= 2.2154 kJ
- Calculate the moles of the Mg burned.
Solution: nMg =
nMg =
nMg = 0.005154 ± 0.0798% mol
nMg = 5.154x10-3 ± 0.0798% mol
-
Calculate the molar heat of formation of Mg in kJ/mol. Use Molar heat = Q (in kJ) / mol of wax. Remember to convert to kJ
Solution: ∆HMg =
∆HMg =
∆HMg = -429.841kJ/mol± 8.61%
Table 4: Final Uncertainty Calculations for Trial 1
Therefore the ∆HMg is -4.3 x 102 ± 40.0 kJ/mol
Trial 2:
Calculations for the heat of formation for MgO(S) +2HCl (aq) →MgCl2 (aq) +H2O (g)
Table 5: Uncertainty Calculations
-
Calculate the mass of hydrochloric acid (HCl) heated. Assume that density of HCl is the same as water:1.00g/cm3
Solution: mHCl = DHCl. VHCl
mHCl = 1.00g/cm3. 100cm3
mHCl = 100g
-
Calculate the temperature change, ΔT, of the HCl.
Solution: ∆THCl = ∆THClfinal -∆THClinital
∆TH20 = 33.5°C -21.9°C
∆TH20 = 11.6°C
-
Calculate the Quantity of Heat absorbed by the water in the can, Assume that the specific heat of HCl is 4.18J/g°C the same as water and use the formula of Q=mc ∆T
Solution: QHCl = mHCl. cHCl. ∆THCl
QHCl = 100g×4.18J/g°C×11.6°C
QHCl= 418J/°C × 11.6°C
QHCl= 4848.8 J
QHCl= 4.8488 kJ
- Calculate the moles of the MgO burned.
Solution: nMgO =
nMgO =
nMgO = 0.03737mol ± 0.0066%
nMgO = 3.73x10-2 mol±0.0066%
-
Calculate the molar heat of formation of MgO in kJ/mol. Use Molar heat = Q (in kJ) / mol of wax. Remember to convert to kJ
Solution: ∆HMgO =
∆HMgO=
∆HMgO = -129.75kJ/mol±4.446%
Table 6: Final Uncertainty Calculations for Trial 1
Therefore the ∆HMgO is -1.3 x 102 ± 6.0 kJ/mol
Calculate the ∆H of the equation H2 (g) + ½ O2 (g) → H2O (l)
Using Hess’s Law: ∆Hf (reactants) - ∆Hf (products)
∆Hf (reactants) - ∆Hf (products) = -286 kJ/mol
Table 7: results of ⍙Hf for all equations
To evaluate the ∆Hf of MgO we use th e three equations and Hess’s Law
Mg(s) +2HCl (aq) →MgCl2 (aq) +H2 (g) +∆Hf
MgCl2 (aq) +H2O (g) + ∆Hf → MgO(s) +2HCl (aq)
H2 (g) + ½ O2 (g) → H2O (l) + ∆Hf
Mg(s) + ½ O2 (g) → MgO(s) + ⍙Hf
∆Hf MgO = + ∆Hf (Mg) + ∆Hf (MgO) - ∆Hf (O2)
∆Hf MgO = (-429.841± 37.0515 kJ/mol) + (-286kJ/mol) – (-129 ± 5.7686 kJ/mol)
∆Hf MgO = -715.841kJ/mol +129kJ/mol
∆Hf MgO = -586.841± 42.8201 kJ/mol
Therefore the ∆Hf MgO= -5.9 x 102 ± 43 kJ/mol because we can only claim two significant figures from the difference in temperature from trial1.
Determine the percent error within the experiment
Theoretical value of ∆Hf MgO = -602kJ/mol
=
=
=
= 1.99%
Conclusion and Evaluation
The purpose of this lab was to calculate the heat of formation for magnesium oxide (MgO). There were two reactions that were set up and carried out so that their heat of formation could be calculated. With the first two respective heats of formation and third theoretical value was combined to determine the experimental heat of reaction for magnesium oxide. There were three values that were used to determine the experimental heat. The first was with the reaction of Mg(S) +2HCl (aq) →MgCl2 (aq) +H2 (g) which resulted with a ∆Hf of -4.3 x 102 ± 40 kJ/mol. The second value was from the reaction of MgO(S) +2HCl (aq) →MgCl2 (aq) +H2O (g) which had a value of -1.3 x 102 ±6.0 kJ/mol. The third value or the heat of formation for the equation H2 (g) + ½ O2 (g) → H2O (l) was found to be -286kJ/mol. (see table 7 for all results). By using Hess’s law: ∆Hf MgO = + ∆Hf (Mg) + ∆Hf (MgO) - ∆Hf (O2), the experimental value of MgO was determined to be -5.9 x 102 ± 43 kJ/mol. Keeping in mind that only two significant figures allowed, the final experimental value of MgO was determined to be -5.9 x 102 ± 43 kJ/mol.
The theoretical value of MgO is given as -602kJ/mol. This value was determined from . By determining the percent error from the theoretical value by the experimental there was 1.99% error. By this value we can determine that the sources of error within this lab were quite minimal. However there some things that did attribute to some errors within the experiment; quantitatively there were some uncertainties that could have altered our results. Such ambiguities include the uncertainty of the temperature probe (±0.2°C) and the amount of HCl (±1mL) which was measured in a 100mL graduated cylinder. Other improvements that could be made in order to get more precise results would be to change the type of system that was used to conduct the experiment; because the experiment was carried out in an open system there is a possibility of potential heat loss which could have distorted our results. Another reason to conduct this experiment in a closed system would be because the temperature in the room where the experiment was conducted may have fluctuated creating inconsistency within our calculations. There was also the assumption that the density of the hydrochloric acid was the same as water. If the proper or accurate density was used, there could have been more accurate results.
These minimal errors can be evaded in further investigations by changing some of the equipment used to gather data. To ensure that a proper temperature is recorded, a temperature probe with uncertainty in the hundredth decimal place would give accurate results. Another way to minimize error would be do a different set-up to make certain that heat loss is prevented. To do that, instead of using a Styrofoam cup, a calorimeter can be used. If a calorimeter is unavailable, than adding a lid to the cup would diminish the error significantly. Another way to further improve the experiment would be to determine the actual density and specific heat of HCl, by doing this our results would be more accurate in terms of the solution used.
by
iqraabbasi (student)
