Given your calculated Enthalpies from Part I and Part II, as well as the given heat of formation for water, use Hess's law to solve for the Enthalpy of Formation for Magnesium Oxide Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) Mg0(s) + 2HCl(aq) → MgCl₂(aq) + H₂0 (1) H₂(g) + ½/20₂(g) → H₂O(1) ΔΗ = ΔΗ = AH-285.8 kJ/mol Part I Avg Part II Avg Given
Given your calculated Enthalpies from Part I and Part II, as well as the given heat of formation for water, use Hess's law to solve for the Enthalpy of Formation for Magnesium Oxide Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) Mg0(s) + 2HCl(aq) → MgCl₂(aq) + H₂0 (1) H₂(g) + ½/20₂(g) → H₂O(1) ΔΗ = ΔΗ = AH-285.8 kJ/mol Part I Avg Part II Avg Given
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The Question in the second picture I don't understand how to get Delta H for any of them. I'm supposed to use my calculations from part a & b which I included in the first pic.
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9:28
6.
Done 161_8_+Enthalpy_F2020
7.
Part I: Mg + HCI
Mass Mg (g)
Mass HCI Soln (g)
Initial Temp (°C)
Final Temp (°C)
molMg
msoln
AT
qsoln
1. molmgo
1. Calculate Moles Mg from mass Mg
2. Add Mg and HCl(aq) mass together
3. AT=Tr-Ti
4. See Equation 3 of the Background
Part I: Mg + HCI
qcal
2. msoln
qrxn
ΔΗ1,2
AH
AT
qsoln
qcal
qran
Trial 1
0.2536
99.9253
AH1.2
Trial 2
0.2532
100.1430
24.94
25
35.42 37.88
Moles Mg
Mass Mg + HCI Soln (g)
Change in Temp (°C)
Heat Absorbed by Soln (J)
[Ssotn = 4.184
Heat Absorbed by Cal (J)
[Ccal = 20.1
Heat Released by Reaction (J)
Enthalpy of Reaction (kJ)
Avg Enthalpy of Reaction (kJ)
Part II: MgO + HCI
Moles MgO
Mass MgO + HCI Soln (g)
Change in Temp (°C)
Heat Absorbed by Soln (J)
[Ssoln 4.184-
//
Heat Absorbed by Cal (J)
[Ccal = 20./1
Heat Released by Reaction (J)
Enthalpy of Reaction (kJ)
8. AHxn Avg Enthalpy of Reaction (kJ)
Part II: MgO + HCI
Mass MgO (g)
Mass HCI Soln (g)
Initial Temp (°C)
Final Temp (°C)
Trial 1
5. See Equation 2 of the Background
6.-qxn [qsoln+qcal]
7. See Equation 4 of the Background
Average AH for trials 1 and 2
8.
Trial 2
0.01043
100.1789
10.58
4434.59
211.6
-4646.19
-445.46
Trial 1
0.0248
100.8596
4.83
2038.24
96.6
2134.84
1
-86.08
1
#
Trial 1
Trial 2
0.9995
0.9990
99.8601 99.5041
26.07
30.77
25.97
30.80
-497.87
0.0103
100.3952
12.88
5410.28
257.6
-5667.88
-550.28
Trial 2
0.02478
84.81
100.4941
4.7
1976.20
94
-2070.2
-83.54
Show work for Part I Trial 1 for calculations 4, 5, 6, 7 (attach additional paper if necessary):](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd407bb69-65e0-4600-bfd9-f9e537a98b1a%2Fb98cfdcc-bdfe-42f6-9b27-7f34010a0223%2Fgd3or34_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1.
2.
3.
4.
5.
6.
7.
8.
3.
4.
5.
9:28
6.
Done 161_8_+Enthalpy_F2020
7.
Part I: Mg + HCI
Mass Mg (g)
Mass HCI Soln (g)
Initial Temp (°C)
Final Temp (°C)
molMg
msoln
AT
qsoln
1. molmgo
1. Calculate Moles Mg from mass Mg
2. Add Mg and HCl(aq) mass together
3. AT=Tr-Ti
4. See Equation 3 of the Background
Part I: Mg + HCI
qcal
2. msoln
qrxn
ΔΗ1,2
AH
AT
qsoln
qcal
qran
Trial 1
0.2536
99.9253
AH1.2
Trial 2
0.2532
100.1430
24.94
25
35.42 37.88
Moles Mg
Mass Mg + HCI Soln (g)
Change in Temp (°C)
Heat Absorbed by Soln (J)
[Ssotn = 4.184
Heat Absorbed by Cal (J)
[Ccal = 20.1
Heat Released by Reaction (J)
Enthalpy of Reaction (kJ)
Avg Enthalpy of Reaction (kJ)
Part II: MgO + HCI
Moles MgO
Mass MgO + HCI Soln (g)
Change in Temp (°C)
Heat Absorbed by Soln (J)
[Ssoln 4.184-
//
Heat Absorbed by Cal (J)
[Ccal = 20./1
Heat Released by Reaction (J)
Enthalpy of Reaction (kJ)
8. AHxn Avg Enthalpy of Reaction (kJ)
Part II: MgO + HCI
Mass MgO (g)
Mass HCI Soln (g)
Initial Temp (°C)
Final Temp (°C)
Trial 1
5. See Equation 2 of the Background
6.-qxn [qsoln+qcal]
7. See Equation 4 of the Background
Average AH for trials 1 and 2
8.
Trial 2
0.01043
100.1789
10.58
4434.59
211.6
-4646.19
-445.46
Trial 1
0.0248
100.8596
4.83
2038.24
96.6
2134.84
1
-86.08
1
#
Trial 1
Trial 2
0.9995
0.9990
99.8601 99.5041
26.07
30.77
25.97
30.80
-497.87
0.0103
100.3952
12.88
5410.28
257.6
-5667.88
-550.28
Trial 2
0.02478
84.81
100.4941
4.7
1976.20
94
-2070.2
-83.54
Show work for Part I Trial 1 for calculations 4, 5, 6, 7 (attach additional paper if necessary):
Transcribed Image Text:1.
2.
3.
4.
5.
6.
9:28
8.
Done 161_8_+Enthalpy_F2020
mol₂o
msoln
AT
qoln
qeal
gran
7. AH1,2
AH
Part II: MgO + HCI
Moles MgO
Mass MgO + HCI Soln (g)
Change in Temp (°C)
Heat Absorbed by Soln (J)
[Ssoln = 4.184
Heat Absorbed by Cal (J)
[Ccal = 20.
Heat Released by Reaction (J)
Enthalpy of Reaction (kJ)
Avg Enthalpy of Reaction (kJ)
M
Trial 1
0.0248
100.8596
4.83
2038.24
96.6
- 2134.84
-86.08
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)
Mg0(s) + 2HCl(aq) → MgCl₂(aq) + H₂0 (1)
-
H₂(g) + ½/20₂(g) → H₂0 (1)
Mg(s) + ½/20₂(g) → Mg0(s)
Trial 2
0.02478
Show work for Part I Trial 1 for calculations 4, 5, 6, 7 (attach additional paper if necessary):
4
100.4941
4.7
४५.81
8-Enthalpy
1976.20
94
-2070.2
-83.54
Given your calculated Enthalpies from Part I and Part II, as well as the given heat of formation for water,
use Hess's law to solve for the Enthalpy of Formation for Magnesium Oxide
ΔΗ =
ΔΗ =
AH = -285.8 kJ/mol
AH =
Part I Avg
Part II Avg
Given
Solve
...
8-8
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