+Room TemperatureReactantTrial NumberMass of reactant (from data)Moles of reactant (calculate)Initial time (ti) (from graph)Initial Temperature (Ti)(from the graph)Linear Curve fit slope (m)(from the graph)Linear Curve fit intercept (b)(from the graph)Final Temperature (T)(calculate from linear fit data)AT = T₁-TiawatercalorimeterarxaAHrxn26.0°CMg(s) + ½ O₂ (g) → MgO (s)Mg0.2578 g10.0106 mol33.5 s24.3°C37.23 °C37.06 °C12.8 °C5355.52 J(5.36 kJ)128 JMg-5483.52 J20.2551 g0.0105 mol41.05 s29.055 °C41.72 °C41.46 °C12.41 °C5192.344 J(5.192 kJ)-0.005191 °C/s -0.006452 °C/s -0.003932 °C/s -0.003809 °C/s124.1 JMgO-5316.4 J30.4939 g0.0123 mol46.52 s29.791 °C33.51 °C33.33 C3.54 °C1481.136 J(1.48 kJ)35.4 J-1516.5 JUse the following to determine the enthalpy of formation of magnesium oxide:Mg (s) + 2HCl (aq) → MgCl₂ (aq) + H₂ (g)MgO (s) + 2HCl (aq) → MgCl₂ (aq)+ H₂O (1)H₂(g) + ½O2 (g) → H₂O (1)0.5033 gMgO0.0125 mol32.47 s31.5714 °C35.27 °C35.15 °C3.59 °C1502.056 J(1.50 kJ)35.9 J-1538.0 J-517.3 kJ/mol -506.3 kJ/mol -123.3 kJ/mol -123.0 kJ/molAH = (calculated in the experiment)AH = (calculated in the experiment)AH=285.8 kJ/molAH = (determine using Hess's Law) Calculations:1. Using the enthalpies of reaction computed in the lab and the value given for the enthalpy of formation ofliquid water, show the Hess's Law determination of the enthalpy of formation of solid magnesium oxide.Show your work. If you are including a scan of your handwritten work, it must be legible.2. Compare the value you calculated with the actual value (-601.8 kJ/mol). Show your determination ofpercent error. Use the standard equation for this: |%error|actual value — experimental value||actual value|x 100%

Using the enthalpies of reaction computed in the lab and the value given for the enthalpy of formation of liquid water, show the Hess's Law determination of the enthalpy of formation of solid magnesium oxide. Show your work. If you are including a scan of your handwritten work, it must be legible.

Using the enthalpies of reaction computed in the lab and the value given for the enthalpy of formation of liquid water, show the Hess's Law determination of the enthalpy of formation of solid magnesium oxide. Show your work. If you are including a scan of your handwritten work, it must be legible.

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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Concept explainers

Thermochemistry

Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.

Exergonic Reaction

The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.

Question

+
Room Temperature
Reactant
Trial Number
Mass of reactant (from data)
Moles of reactant (calculate)
Initial time (ti) (from graph)
Initial Temperature (Ti)
(from the graph)
Linear Curve fit slope (m)
(from the graph)
Linear Curve fit intercept (b)
(from the graph)
Final Temperature (T)
(calculate from linear fit data)
AT = T₁-Ti
awater
calorimeter
arxa
AHrxn
26.0°C
Mg(s) + ½ O₂ (g) → MgO (s)
Mg
0.2578 g
1
0.0106 mol
33.5 s
24.3°C
37.23 °C
37.06 °C
12.8 °C
5355.52 J
(5.36 kJ)
128 J
Mg
-5483.52 J
2
0.2551 g
0.0105 mol
41.05 s
29.055 °C
41.72 °C
41.46 °C
12.41 °C
5192.344 J
(5.192 kJ)
-0.005191 °C/s -0.006452 °C/s -0.003932 °C/s -0.003809 °C/s
124.1 J
MgO
-5316.4 J
3
0.4939 g
0.0123 mol
46.52 s
29.791 °C
33.51 °C
33.33 C
3.54 °C
1481.136 J
(1.48 kJ)
35.4 J
-1516.5 J
Use the following to determine the enthalpy of formation of magnesium oxide:
Mg (s) + 2HCl (aq) → MgCl₂ (aq) + H₂ (g)
MgO (s) + 2HCl (aq) → MgCl₂ (aq)+ H₂O (1)
H₂(g) + ½O2 (g) → H₂O (1)
0.5033 g
MgO
0.0125 mol
32.47 s
31.5714 °C
35.27 °C
35.15 °C
3.59 °C
1502.056 J
(1.50 kJ)
35.9 J
-1538.0 J
-517.3 kJ/mol -506.3 kJ/mol -123.3 kJ/mol -123.0 kJ/mol
AH = (calculated in the experiment)
AH = (calculated in the experiment)
AH=285.8 kJ/mol
AH = (determine using Hess's Law)

Calculations:
1. Using the enthalpies of reaction computed in the lab and the value given for the enthalpy of formation of
liquid water, show the Hess's Law determination of the enthalpy of formation of solid magnesium oxide.
Show your work. If you are including a scan of your handwritten work, it must be legible.
2. Compare the value you calculated with the actual value (-601.8 kJ/mol). Show your determination of
percent error. Use the standard equation for this: |
%error
|actual value — experimental value|
|actual value|
x 100%

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The enthalpy change for a chemical reaction is the sum of the energy consumed in breaking bonds and the energy released during bond formation. One way to determine the overall energy change for a chemical reaction is to apply Hess’s law to add together a group of reactions which can be arranged such that the chemical equations, when combined, give the overall equation we are trying to characterize. a. Write a balanced chemical equation for the combustion of gaseous propane in gaseous oxygen to produce gaseous carbon dioxide and liquid water. C3H8(g)+ O2(g) ---->CO2(g)+ H2O(g) b. Combine the following equations to determine the enthalpy change for the combustion of 1 mole of propane. Assume that solid carbon is graphite. 3C(s,graphite)+4H2(g)--->C3H8(g) ΔHo = -103.8 kJ/mol C(s,graphite)+O2(g)--->CO2(g) ΔHo= -393.5 kJ/mol H2(g)+1/2 O2(g)--->H2O(g) ΔHo= -285.8 kJ/mol kJ/mol…