Transcript Chemistry 1011 - Marine Institute of Memorial University

Chemistry 1011
TOPIC
Thermochemistry
TEXT REFERENCE
Masterton and Hurley Chapter 8
Chemistry 1011 Slot 5
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8.4 Thermochemical Equations
YOU ARE EXPECTED TO BE ABLE TO:
– Define molar enthalpy of reaction, molar heat of fusion
and molar heat of vaporization.
– Carry out calculations relating heat absorbed or
released in a chemical reaction, the quantity of a
reactant or product involved, and DH for the reaction.
– Use Hess's Law to determine the heat of reaction given
appropriate equations and thermochemical data.
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Molar Heat of Fusion and of
Vaporization
• The latent heat of fusion (vaporization) of a
substance is the heat absorbed or released
when 1 gram of the substance changes
phase.
• The molar heat of fusion (vaporization) of a
substance is the heat absorbed or released
when 1mole of the substance changes
phase.
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Thermochemical Equations
• Balanced chemical equations that show the
enthalpy relation between products and reactants
H2(g) + Cl2(g)  2HCl(g); DH= -185 kJ
Exothermic reaction: 185 kJ of heat is evolved when 2
moles of HCl are formed.
2HgO(s)  2Hg(l) + O2(g); DH= +182 kJ
Endothermic reaction: 182 kJ of heat must be absorbed to
decompose 2 moles HgO.
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Thermochemical Equations
• The sign of DH indicates whether the reaction is
endothermic (+) or exothermic (–)
• Coefficients in the equations represent the
numbers of moles of reactants and products
• Phases must be specified (s), (l), (g), (aq)
• The value quoted for DH applies when products
and reactants are at the same temperature, usually
25oC
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Thermochemical Equations
Rule #1
• DH is directly proportional to the amount of
reactants and products.
When one mole of ice melts, DH = +6.00 kJ.
When one gram of ice melts, DH = +0.333 kJ.g-1
H2(g) + Cl2(g)  2HCl(g); DH = -185kJ
1/ H
1
2 2(g) + /2Cl2(g)  HCl(g); DH = -92.5kJ
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Thermochemical Equations
Rule #2
• DH for a reaction is equal in magnitude but
opposite in sign to DH for the reverse
reaction.
2HgO(s)  2Hg(l) + O2(g); DH= +182 kJ
2Hg(l) + O2(g)  2HgO(s) ; DH= -182 kJ
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Thermochemical Equations Rule #3
DH is independent of the path of the reaction –
it depends only on the initial and final states
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Hess’ law
DH = DH1 + DH2
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Using Hess’ Law
•
Hess’ Law is used to calculate DH for reactions
that are difficult to carry out directly.
In order to apply Hess’ Law we need to know
DH for other related reactions.
For example, we can calculate DH for
1.
C(s) + 1/2O2(g)  CO(g);
•
•
DH1 = ?
If we know DH for
2.
3.
CO(g) + 1/2O2 (g)  CO2(g); DH2 = -283.0 kJ
C(s) + O2(g)  CO2(g); DH3 = -393.5 kJ
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Calculating the Unknown DH
•
Write equation #3
C(s) + O2(g)  CO2(g); DH3 = -393.5 kJ
•
Reverse equation #2
CO2(g)  CO(g) + 1/2O2 (g); -DH2 = +283.0 kJ
Add the two equations
C(s) + 1/2O2(g)  CO(g);
DH1 = -393.5kJ + 283.0kJ = -110.5kJ
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Enthalpy Diagram
C(s) + O2(g)
DH1-110.5kJ
DH3 -393.5kJ
CO(g) + 1/2O2(g)
DH2 +283.0kJ
CO2(g)
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Some Other Examples
•
Given
1.
2.
+ 3/2H2(g)  NH3(g); DH = -46.1kJ
C(s) + 2H2(g)  CH4(g); DH = -74.7kJ
1/ N
2 2(g)
3. C(s) + 1/2H2(g) + 1/2 N2(g)  HCN(g); DH = +135.2kJ
•
Find DH for the reaction:
4. CH4(g) + NH3(g)  HCN(g) + 3H2(g)
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Some Other Examples
•
Given
+ 3/2H2(g)  NH3(g); DH = -46.1kJ
C(s) + 2H2(g)  CH4(g); DH = -74.7kJ
•
•
1/ N
2 2(g)
•
C(s) + 1/2H2(g) + 1/2 N2(g)  HCN(g); DH = +135.2kJ
Find DH for the reaction:
•
•
CH4(g) + NH3(g)  HCN(g) + 3H2(g)
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Find DH for the reaction:
CH4(g) + NH3(g)  HCN(g) + 3H2(g
• Add Reverse Eq #2 to Reverse Eq #1
CH4(g)  C(s) + 2H2(g); DH = +74.7kJ
NH3(g)  1/2N2(g) + 3/2H2(g); DH = +46.1kJ
CH4(g) + NH3(g)  C(s) + 2H2(g) + 1/2N2(g) + 3/2H2(g); DH = +120.8kJ
• Add Eq #3
C(s) + 1/2H2(g) + 1/2 N2(g)  HCN(g); DH = +135.2kJ
CH4(g) + NH3(g)  HCN(g) + 3H2(g; DH = +256.0kJ
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Some Other Examples
• A 1.00g sample of table sugar (sucrose),
C12H22O11, is burned in a bomb calorimeter
containing 1.50 x 103g water.
– The temperature of the calorimeter and water rises from
25.00oC to 27.32oC.
– The heat capacity of the metal components of the bomb
is 837J.K-1.
– The specific heat of water is 4.184J.g-1.K-1.
• Calculate
– (a) the heat evolved by the 1.00g of sucrose, and
– (b) the heat evolved per mole of sucrose.
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Solution
• Heat absorbed by the water = m.c.Dt
= 1.50 x 103 g x 4.184 J.g-1.K-1 x 2.32K
= 14.6 kJ
• Heat absorbed by the calorimeter = C.Dt
= 837 J.K-1 x 2.32K = 2.03 kJ
• Total heat absorbed by the calorimeter and
contents
= - total heat released by 1.00g sucrose
= - 16.6kJ
• Heat evolved per mole of sucrose
= - 16.6 kJ x 342 g.mol-1 = - 5680 kJ
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