1. Thermodynamics

2. 1st law of Thermodynamics
Based on the law of conservation of energy
States that energy can be converted from one form to another
Cannot calculate total energy but can calculate changes in energy

3. System Surroundings The rest of the universe outside the system
Specific part of the universe that is of interest to us
The rest of the universe outside the system
Heat transfer: the transfer of thermal energy between 2 bodies of different temperatures.
Heat is a measure of a process with a number and a unit.
(It is best thought of as a verb not a noun.)
open
closed
isolated
Exchange:
mass & energy
energy
nothing

4. Thermochemistry
Study of heat changes in a chemical or physical process
Enthalpy—used to quantify the heat content into or out of a system that occurs at constant volume in an open system
Symbol: H, cannot be measured
Symbol ΔH is used in heat changes
At constant volume q = ΔH
Heat of reaction (ΔHrxn) is the change in enthalpy with respect to a chemical reaction

5. DHrxn = H (products) – H (reactants)
Thermochemical equations
shows the enthalpy changes, ΔH
must specify states of matter
ΔHrxn (most general form, sometimes leave off the rxn)— used to describe changes in H.
More specific would be: ΔHcomb,ΔHsoln, ΔHvap, ΔHhydration ΔHfus
DHrxn = H (products) – H (reactants)
All chemical and physical changes release
or absorb heat
a. Endothermic: absorbs heat, q is positive, ∆H is positive
b. Exothermic: releases heat, q is negative, ∆H is negative

6. DHrxn = H (products) – H (reactants)
DHrxn = heat given off or absorbed during a reaction
at constant pressure, 1atm and 25C
Hproducts < Hreactants
Hproducts > Hreactants
DH > 0
DH < 0

7. Endothermic Exothermic releases heat to its surroundings (feels warm)
Energyproducts < Energyreactants
-q or -H
absorbs heat from the surroundings (feels cold)
Energyproducts > Energyreactants
+q or +H

8. Thermochemical Equations for Endothermic Reactions
Endothermic : H is positive; heat is on the reactants side of the equation (Heat absorbed)
Three ways to write the same thing:
NH4NO3(s)  NH4+(aq) + NO3-(aq) H = 27 kJ
NH4NO3(s) + Heat  NH4+(aq) + NO3-(aq)
NH4NO3(s) + 27kJ  NH4+(aq) + NO3-(aq)

9. Thermochemical Equations for Exothermic Reactions
Exothermic : H is negative; heat is on the products side of the equation (Heat released)
Three ways to write the same thing
CaO(s) + H2O(l)  Ca(OH)2(s) H = kJ
CaO(s) + H2O(l)  Ca(OH)2(s) + Heat
CaO(s) + H2O(l)  Ca(OH)2(s) kJ

12. 3. Heat of Combustion, DHcomb
Change the problem in NT
a. How much heat is released by the combustion of g of hydrogen gas?
DHcomb = -286 kJ 2
H O2  H2O 2
250.0 g H2
1 mole H2
-286 kJ
= -17,698 kJ
2.02g H2
2 mole H2
= x 104 kJ
Or x 104 kJ
released
How do you know when to use a table vs a formula?
If you have quantity (g or mol) you will use a table.
(When I say “grams,” you say, “table.”)

13. Standard Heat of Formation, DH f
Day 2

14. ii. Standard Heat of Formation, DH f
The circle is said “not,” which means standard
ii. Standard Heat of Formation, DH f
Take out your homework and find the C-13 table to use to look up values for DH f

15. ii. Standard Enthalpy, DH
The circle is said “not,” which means standard
ii. Standard Enthalpy, DH
What happens when the exchange occurs under normal conditions(1atm & 25C)?
Standard heat of Formation( Hf ) is
Forming one mole of a compound using only elements as the reactants
Example: C(s) + O2(g)  1CO2(g)
1 mole or molecule of CO2(g) Hf = kJ
The ΔHf of any element in its most stable form is 0kJ
ΔHf (O2) = 0 kJ
ΔHf (C, graphite) = 0 kJ
ΔHf (O3) = 142 kJ/mol
ΔHf (C, diamond) = 1.90 kJ/mol

16. ii. Standard Enthalpy, DH cont…
What happens when the exchange occurs under normal conditions (1atm & 25C)?
Standard heat of Reaction( Hrxn )
a. Direct method-equation (today’s lesson)
aA + bB  cC + dD
Hrxn = [c Hf(C) + dHf(D)] –[aHf(A) + bHf(B)]
Hrxn =  Hf(products )-  Hf(reactants)
b. Indirect method-Hess’s Law (tomorrow’s lesson)

17. ii. Standard Enthalpy DH f values

18. ii. Standard Enthalpy, DH cont… Direct method-equation examples
Calculate the standard heat of reaction for the combustion of methane.
Write the balanced
Chemical Equation:
Then use the Hf values to substitute into the equation. Be sure to use coefficients in the eqn!!!
aA + bB  cC + dD
Hrxn = [c Hf(c) + dHf(D)] –[aHf(A) + bHf(B)]
Hrxn =  Hf(products )-  Hf(reactants)
CH4(g) + 2O2(g)  CO2(g)+ 2H20(l)

19. ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants)
Use the standard enthalpies of formation to calculate ∆Hrxn
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
∆Hf(CO2(g))= -394kJ
∆Hf(H2O(l))= -286kJ
∆Hf(CH4(g))= -75kJ
∆Hf(O2(g))= 0kJ
ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants)
= [-394+(2×-286)] – [-75+0] =
= -891 kJ exothermic
reactants
products

20. ii. Standard Enthalpy, DH cont… Direct method-equation examples
2. Calculate the standard heat of reaction for the combustion of hydrogen sulfide gas. (hint: produces liquid water and sulfur dioxide gas) Write the balanced Chemical Equation:
2H2S(g) + 3O2(g)  2H2O (l)+ 2S02 (g)
Hrxn = [2( kJ )+2( kJ)] – [2(-20.63kJ)+ 3(0kJ)]
Hrxn = -1,124.06kJ or 1,124.06kJ evolved
Exothermic Reaction

21. ii. Standard Enthalpy, DH cont… Direct method-equation examples
3. Calculate the standard heat of reaction for the combustion of nitrogen monoxide gas and oxygen gas. (hint: yields nitrogen dioxide gas) Write the balanced Chemical Equation:
2NO(g) + O2(g)  2NO2 (g)
Hrxn = [2(33.18kJ)] – [2(90.25kJ)+ 0kJ]
Hrxn = kJ or kJ released
Exothermic Reaction

22. ii. Standard Enthalpy, DH cont… Direct method-equation examples
3. Calculate the standard heat of reaction for the combustion of carbon monoxide gas and oxygen gas. (hint: yields carbon dioxide gas) Write the balanced Chemical Equation:
2CO(g) + O2(g)  2CO2 (g)
Hrxn = [2( kJ )] – [2( kJ)+ 0kJ]
Hrxn = kJ or kJ evolved
Exothermic Reaction

23. Next Day-Indirect Method
Hess’s Law

24. Next Day-Indirect Method Hess’s Law
Hess’s Law is a lot like solving systems of equations but there are some differences.
You may use a multiple of an equation. This multiple may include positive, negative, and even fractions.
Do not try to subtract. Always add the equations together to avoid errors.

25. Hess’s Law-Example 1
2S(s) + 3O2(g)  2SO3(g) ∆H = ? S(s) + O2(g)  SO2(g) ∆H = -297 kJ 2SO3(g)  2SO2(g) + O2(g) ∆H = 198 kJ

26. Hess’s Law-Example 2
2NO(g) + O2(g)  2NO2(g) ∆H = ? N2(g) + O2(g)  2NO(g) ∆H = kJ N2(g) + 2O2(g)  2NO2(g) ∆H = 66.4 kJ

27. Hess’s Law-Example 3
2C (s) + O2(g)  2CO(g) ∆H = ? C(s) + O2(g)  CO2(g) ∆H = kJ 2CO(g) + O2(g)  2CO2(g) ∆H = -566kJ

28. Hess’s Law-Example 4
C(s) + 2H2(g)  CH4(g) ∆H = ? C(s) + O2(g)  CO2(g) ∆H = kJ 2H2(g) + O2(g)  2H2O(l) ∆H = kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H= kJ