Thermo-Chemistry Calculations
Data Sheet
Q1.
| Avg (°C） | Absolute Uncertainty | Relative uncertainty | |
| T1 | 21.1 | 0.2 | 0.95% |
| T2 | 40.5 | 0.2 | 0.49% |
| T3 | 31.6 | 0.2 | 0.63% |
| T4 | 21.15 | 0.2 | 0.95% |
| T5 | 26.35 | 0.2 | 0.76% |
| T6 | 21.9 | 0.2 | 0.91% |
| T7 | 18.85 | 0.2 | 1.06% |
Calculations
Average temperatures (°C）
T1= (21.2+21.0)/2= 21.1
Absolute Uncertainty= (0.1+0.1) = 0.2
21.1°C± 0.2 °C
Relative Uncertainty= (0.2/21.1)*100%= 0.95%
T2= (40.1+40.0)/2= 40.5°C
Absolute Uncertainty= 0.2
40.5 °C ±0.2 °C
Relative Uncertainty= (0.2/40.5)*100%= 0.49%
T3= (31.6+31.6)/2= 31.6
Absolute Uncertainty= (0.1+0.1) = 0.2
31.6 °C ± 0.2 °C
Relative Uncertainty= (0.2/31.6)*100%= 0.63%
T4= (21.20+ 21.10)/2= 21.15 °C
Absolute Uncertainty= 0.2
21.15 °C±0.2 °C
Relative Uncertainty= (0.2/21.15)*100%= 0.95%
T5= (26.30+26.40)/2= 26.35 °C
Absolute Uncertainty= (0.1+0.1) = 0.2
26.35°C±0.2 °C
Relative Uncertainty= (0.2/26.35)*100%= 0.76%
T6= (22.0+21.8)/2= 21.9 °C
Absolute uncertainty= (0.1+0.1) = 0.2
21.9°C± 0.2°C
Relative Uncertainty= (0.2/21.9)*100%= 0.91%
T7= (18.80+18.90)/2= 18.85 °C
Absolute Uncertainty= (0.1+0.1) = 0.2
18.85°C±0.2 °C
Relative Uncertainty= (0.2/18.85)*100= 1.06%
Q2.
Part C
NH4Cl
Avg mass in grams Absolute Uncertainty Relative Uncertainty
4.9906 0.0004 0.01%
Calculations
Avg mass= (4.9908+4.9904)/2 = 4.9906g
Absolute Uncertainty= (5.0000-4.9906) g= 0.0004g
Relative Uncertainty= (0.0004/5)*100= 0.01%
Report Sheet
Q1.
Moles of NH4Cl
Moles= mass/RMM
RMM= 14+4+35.5= 53.5
4.9906/53.5= 0.0933moles
Avg moles=0.0933moles
The number of moles when 5g of the NH4Cl was used
5/53.5= 0.0935moles
Absolute uncertainty= (0.0935-0.0933) = 0.0002
Relative uncertainty= (0.0002/0.0935)*100%
0.21%
Moles of Water (moles of cold water= moles of hot water)
Mass of water= volume*density
=50.00cm3*1g/cm3= 50.00g
Moles=mass/RMM= 50.00g/18= 2.78moles
Absolute uncertainty according to the instructions is 0.5
Therefore, relative uncertainty= (0.5/2.78)*100%= 17.99%
Q2.
2.78*75.31(31.6-21.1) +2.78*75.31*(31.6-40.5) +Ccal (31.6-21.1) =0
2198.30-1863.32+10.5Ccal=0
10.5Ccal=-334.98
Ccal=-31.90JK
Absolute Uncertainty= 0.2+0.2+0.2+0.2+0.2+0.2= 1.2
Relative uncertainty= (1.2/31.9)*100%= 3.76%
Q3.
Moles of 1M NH4OH
1*50/1000
=0.05moles
NH4OH (aq) +HCl (aq) =NH4Cl (aq) +H2O (l)
Mole ratio= 1:1
Moles of water=0.05moles
Absolute uncertainty= 0.5
Relative uncertainty= (0.5*0.05)*100%= 2.5%
ΔH2+nCp (T5-T4) +Ccal (T5-T4) =0
ΔH2+0.05*72.72(26.35-21.15) + -31.90(26.35-21.15) =0
ΔH2+18.9072-165.88=0
ΔH2=+146.9728
=+146.97J/K
Absolute uncertainty=2.0
Relative uncertainty= (2.0/146.97)*100%= 1.36%
Molar enthalpy=146.97/0.05
=+2939.4Jmol-1K
Absolute uncertainty= 1.2+4(0.2) = 2.0
Relative uncertainty= (2/2939.4)*100% =0.0068%
ΔH3+nCp (T7-T6) +Ccal (T7-T6) =0
ΔH3+0.0933*70.74(18.85-21.9)-31.90(18.85-21.9) =0
ΔH3-13.53+65.395=0
ΔH3=-51.865JK
Absolute uncertainty=2.0
Relative uncertainty= (2.0/51.865)*100%= 3.86%
Molar enthalpy= -51.865/0.0933
=-555.9Jmol-1K
Absolute uncertainty=2.0
Relative uncertainty= (2/555.9)*100%=
=0.36%
Q4.
| value | units | (±) absolute uncertainty | (±) relative uncertainty % | |
| Mole H2O (50mL) | 2.78 | moles | 0.5 | 17.99 |
| B mole NH4Cl(aq) | 0.05 | moles | 0.5 | 2.5 |
| C mole NH4Cl(S) | 0.0933 | mole | 0.0002 | 0.21 |
| Ccal-Avg | -31.90 | J./K | 1.2 | 3.76 |
| ΔH2-avg | +146.97
|
J/K | 2.0 | 1.36 |
| ΔH2-avg /mole NH4Cl | +2939. 4 | Jmol-1K
|
2.0 | 0.0068 |
| ΔH3-avg | 51.865 | J/K | 2.0 | 3.86 |
| ΔH3-avg/mole NH4Cl | -555.9 | Jmol-1K
|
2.0 | 0.36 |
Q5.
ΔH1: Ʃ (nΔHₒf (products) -Ʃ (nΔHₒf (reactants) = ΔH1
NH4OH (aq) +HCl (aq) = NH4Cl (s) +H2O (l)
NH4OH (aq) +HCl (aq) = NH4Cl (aq) +H2O (l)
NH4Cl (s) = NH4Cl (aq)
Using Hess’s Law, rearrange reaction (2) and (3) and demonstrate how they can be used to calculate ΔH1
The expected equation should be:
NH4OH (aq) +HCl (aq) = NH4Cl (s) +H2O (l)
Rearranging equations to come out with the above equation
NH4OH (aq) +HCl (aq) = NH4Cl (aq) +H2O (l)Reaction (1)
NH4Cl (aq)= NH4Cl (s) Reaction (2)
From the above arrangement of reactions (1) and (2), we can obtain desired final equation after cancellation of NH4Cl (aq)from the product side of equation (1) and the same product from the left side of equation (2). For one to obtain reaction (2) as written above, the initial reaction, NH4Cl (s) = NH4Cl (aq), must be reversed to obtain NH4Cl (aq)= NH4Cl (s). Therefore, the enthalpy of formation of NH4Cl (s)changes from -314.55 Jmol-1K to + 314.55Jmol-1K. After the equation is reversed, it is possible to eliminate the NH4Cl (aq)particles to obtain the final equation. The final desired equation is NH4OH (aq) +HCl (aq) = NH4Cl (s) +H2O (l).
nΔHₒf (products) = (314.55-285.83) Jmol-1K
=+28.72 Jmol-1K
nΔHₒf (reactants) = (-366.25-167.16) Jmol-1K
=-533.41 Jmol-1K
ΔH1= 28.72- (-533.41) = +562.13 Jmol-1K
Absolute uncertainty = (0.005*4) = 0.02
Relative uncertainty = (0.02/562.13)*100% = 0.000036%
Can we carry out reaction (1) experimentally? Explain your reasoning.
According to the results presented (+562.13 Jmol-1K), we can carry out the experiment, NH4OH (aq) +HCl (aq) = NH4Cl (aq) +H2O (l) experimentally since it requires less activation energy for the reaction to start. Therefore, so long as the reactants are heated to provide the necessary energy, the reaction occurs. Besides, one can use a suitable catalyst to change the path of reaction, reducing the activation energy, initiating the reaction.
ΔH1 = ΔH2+ ΔH3
ΔH2 =+2939.4 Jmol-1K
ΔH3 = -555.9 Jmol-1K
ΔH1= (2939. 4- 555.9) Jmol-1K
ΔH1 = +2383.5 Jmol-1K
Absolute uncertainty= 2.0+2.0= 4.0
Relative uncertainty= (4/2383.5)*100%= 0.168%
| value | units | (±) absolute uncertainty | (±) relative uncertainty % | |
| ΔH1 | +562.13 | Jmol-1K | 0.02 | 0.000036% |
| ΔH2-avg /mole NH4Cl + ΔH3-avg /mole NH4Cl | +2383.5 | Jmol-1K | 4.0 | 0.168% |
Was Hess’s Law obeyed (within experimental uncertainty) based on your own data and calculations? Give a brief explanation of your reasoning.
The Hess’s Law states that, regardless of the several steps of a reaction, the total change of enthalpy for the chemical reaction is the addition of all changes. Consequently, at constant pressure, the heat of reaction, enthalpy change in a reaction, is independent of the root between the last and the primary states. Comparing the calculated and experimental results for ΔH1, and the uncertainties between the two values, Hess Law was not obeyed since the two enthalpies are different. Calculated enthalpy of formation of NH4Cl is + 562.13 Jmol-1K. Contrarily, the experimental enthalpy of formation of NH4Cl is +2383.5 Jmol-1K. Besides, the absolute uncertainty of calculated enthalpy of formation of NH4Cl is ± 0.02, while the absolute uncertainty of experimental enthalpy of formation of NH4Cl is± 0.4. Similarly, the relative uncertainty of calculated enthalpy of formation of NH4Cl is ± 0.000036%, while the relative uncertainty of experimental enthalpy of formation of NH4Cl is ± 0.168%.
Suggest a reason why you are instructed to use ± 0.5mL as the uncertainty in the 50 mL volumes rather than ±0.05mL, which is the uncertainty of a 50 mL pipette or dispensator.
Other measurements, such as temperature, are to be recorded to one decimal place. For instance, according to the instructions, temperatures are to be measured to the nearest 0.1°C. For this reason, it is recommended for other readings to take the same form to enhance uniformity. Besides, the instruments may not be accurate because of their age; therefore, it is advisable for one to use the uncertainty of ± 0.5ml.
In conclusion, the experimental outcomes defied the calculated results because when ΔH1 for the two procedures are compared, different results are obtained. For instance, Hess’s Law is not obeyed in the experiment. The reason for the difference in the outcomes can be because of errors in volume measurements and temperature readings. Besides, it could be because of the rapid loss of heat into the surrounding, reducing the thermometer readings making inaccurate measurements to be used in the calculation.
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