th The law of conservation of energy energy cannot be or It can only be or T ransfer of energy always takes place from a substance at a ID: 932453
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Slide1
Bell Ringer May 11th
The law of conservation of energy: energy cannot be ________ or _______. It can only be ________ or __________. Transfer of energy always takes place from a substance at a _______ temperature to a substance at a _______ temperatureThree methods of heat energy transfer: _______ , __________, and __________.
1
Slide2Thermochemical Equations
2
Slide3Thermochemical Equations
A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH.Enthalpy (H) is the transfer of energy in a reaction (for chemical reactions it is in the form of heat) and ΔH is the change in enthalpy.By definition, ΔH = Hproducts – Hreactants Hproducts < Hreactants, ΔH is negativeHproducts > Hreactants, ΔH is positive3
Slide4Thermochemical & Endothermic/ Exothermic equations
Depending on the sign of ΔH°, the reaction can either be exothermic or endothermic. Exothermic reactions release heat from the system to the surroundings so the temperature will rise.ΔH° will be negative because the reaction loses heat.Endothermic reactions absorb heat from the surroundings into the system so the temperature will decrease.ΔH° will be positive because the reaction absorbs heat.4
Slide5Classify the following as endothermic or exothermic
Ice melting2 C4H10(g) + 13 O2(g) → 10 H2O(g) + 8 CO2(g) ΔHrx = -5506.2 kJ/mol 2 HCl (g) + 184.6 kJ → H2 (g) + Cl2 (g) Water vapor condensing5
Endothermic
Exothermic
Endothermic
Exothermic
Slide6Exothermic vs. Endothermic
ExothermicEndothermicA change in a chemical energy where energy/heat EXITS the chemical systemResults in a decrease in chemical potential energyΔH is negative
A change in chemical energy where energy/heat
ENTERS
the chemical system
Results in an increase in chemical potential energy
ΔH is positive
6
Slide7Thermochemical equations using Standard Heat of Formations - Solving for change in H
(ΔH)C2H2(g) + 2 H2(g) → C2H6(g)Remember By definition, ΔH = H products – H reactants
Substance
D
H
°
f
(kJ/
mol
)
C
2
H
2
(g)
226.7
C
2
H
6
(g)
-84.7
7
Standard Heat of Formations =
D
H
°
f
Slide8Thermochemical equations using Standard Heat of Formations
Write the equation for the heat of formation of C2H6(g)Substance
D
H
°
f
(kJ/mol)
C
2
H
2
(g)
226.7
C
2
H
6
(g)
-84.7
1
st
: Using our balanced chemical
equation, we see how many moles of each compound we have.
C
2
H
2
(g)
+
2
H
2
(g)
→
C2H6(g) [(H2) does not have a DH°f ]1 mol of C2H2(g) and 1 mol C2H6(g)
3rd: We solve for ∆H° ∆H° = [-84.7] – [226.7] = -331.4 kJ/mol
2nd: We plug in the ∆H°f for each of our compounds, remembering that ∆H° = [∆H°f products] – [∆H°f reactants]∆H° = [C2H6(g)] – [C2H2(g)] =
8
Slide9Practice Problems
Solve for the ΔHrx and write the following thermochemical equations.1. What is the ΔHrx for the process used to make lime (CaO)?CaCO3(s) → CaO(s) + CO2(g)
Substance
D
H
°
f
(kJ/mol)
CaCO
3
(s)
-1207.6
CaO
(s)
-634.9
C
4
H10 (g)
-30.0
H
2
O (g)
-241.82
CO
2
(g)
-393.5
2. What is the
ΔH
rx
for the combustion of
C
4H10(g)?2 C4H10 (g) + 13 O2 (g) → 10 H2O (g) + 8 CO2 (g)9
Slide10Practice Problems
Solve for the ΔHrx and write the following thermochemical equations.1. What is the ΔHrx for the process used to make lime (CaO)?CaCO3(s) → CaO(s) + CO2(g)
Substance
D
H
°
f
(kJ/mol)
CaCO
3
(s)
-1207.6
CaO
(s)
-634.9
C
4
H10 (g)
-30.0
H
2
O (g)
-241.82
CO
2
(g)
-393.5
10
ΔH
rx
= [ΔH°f (CaO) + ΔH°f (CO2)] – [ΔH°f (CaCO3)] ΔHrx = [(-634.9)+(-393.5)] – [(-1207.6)] ΔH
rx
= [ -1028.4]
– [-1207.6] = +179.2 kJ
CaCO
3(s) → CaO(s) + CO2(g) ΔHrx = 179.2 kJ/mol
Slide11Practice Problems
Solve for the ΔHrx and write the following thermochemical equations. SubstanceDH°f (kJ/mol)
CaCO
3
(s)
-1207.6
CaO
(s)
-634.9
C
4
H
10
(g)
-30.0
H
2
O (g)-241.82
CO
2
(g)
-393.5
11
2. What is the
ΔH
rx
for the
combustion of C
4
H
10
(g)?
2 C4H10 (g) + 13 O2 (g) → 10 H2O (g) + 8 CO2 (g) ΔHrx = [ΔH°f (H2O) + ΔH°f (CO2)] – [ΔH°f (C
4H
10
)]
(
We do not include O
2 because its ΔH°f is 0.) ΔHrx = [10(-241.82)+8(-393.5)] – [2(-30.0)] ΔHrx = [ -5566.2] – [-60.0] = -5506.2 kJ2 C4H10 (g) + 13 O2 (g) → 10 H2O (g) + 8 CO2
(g) ΔHrx = -5506.2 kJ/mol