1. Chapter 5 Thermochemistry Part B
Thermochemical Equations

2. First Law of Thermodynamics
The work, w, is positive if it is done on the system and negative if it is done by the system.
Positive q is heat gained by the system, negative q is heat lost by the system.
Change in Internal Energy U is due to work and heat
U = Ufinal – Uinitial = w + q
At constant volume and no other work w = o
U = Ufinal – Uinitial = qv
Heat at constant volume is equal to internal energy change, which depends only on the system initial and final states

3. Enthalpy (H) H = U + PV Similar to U, H is a state function
If there is only expansion work: w=– P V
Internal energy change at constant pressure
U= (qP + w)
Enthalpy change at constant pressure
H = U + PV
H = (qP + w ) -w = qP

4. Enthalpy Change (H) qp = H = Hfinal – Hinitial
Heat at constant pressure is equal to enthalpy change, which depends only on the initial and final states
qp = H = Hfinal – Hinitial
qp = H = Hproducts – Hreactants
Endothermic reaction. System absorbs energy from surroundings. H positive
Exothermic reaction. System loses energy to surroundings. H negative

5. Thermochemical Equation
Write H immediately after equation
N2(g) + 3H2(g)  2NH3(g) H = –92.38 kJ
Must give physical states of products and reactants
H different for different states
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l ) H ° rxn = –890.5 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H ° rxn = –802.3 kJ
Difference is equal to the energy to vaporize water

6. Multiple Paths; Same H °
Can often get from reactants to products by several different paths
Intermediate A
Products
Reactants
Intermediate B
Should get same H °
Enthalpy is state function and enthalpy change is path independent

7. Multiple Paths; Same H °
Path a: Single step
C(s) + O2(g)  CO2(g) H°rxn = –393.5 kJ
Path b: Two step
Step 1: C(s) + ½O2(g)  CO(g) H °rxn = –110.5 kJ
Step 2: CO(g) + ½O2(g)  CO2(g) H °rxn = –283.0 kJ
Net Rxn: C(s) + O2(g)  CO2(g) H °rxn = –393.5 kJ

8. Hess’s Law Hess’s Law of Heat Summation
Going from reactants to products
Enthalpy change is same whether reaction takes place in one step or many steps.
Chief Use
Calculation of H °rxn for reaction that can’t be measured directly
Thermochemical equations for individual steps of reaction sequence may be combined to obtain thermochemical equation of overall reaction

9. Hess’s Law of Heat Summation
Path a: N2(g) + 2O2(g)  2NO2(g) H °rxn = 68 kJ
Path b:
Step 1: N2(g) + O2(g)  2NO(g) H °rxn = kJ
Step 2: 2NO(g) + O2(g)  2NO2(g) H °rxn = –112 kJ
Net rxn: N2(g) + 2O2(g)  2NO2(g) H °rxn = kJ
For any reaction that can be written into steps, value of H °rxn for reactions = sum of H °rxn values of each individual step

10. Rules for Manipulating Thermochemical Equations
When equation is reversed, sign of H°rxn must also be reversed.
If all coefficients of equation are multiplied or divided by same factor, value of H°rxn must likewise be multiplied or divided by that factor
Formulas canceled from both sides of equation must be for substance in same physical states

11. Strategy for Adding Reactions Together:
Choose most complex compound in equation for one-step path
Choose equation in multi-step path that contains that compound
Write equation down so that compound
is on appropriate side of equation
has appropriate coefficient for our reaction
Repeat steps 1 – 3 for next most complex compound, etc.

12. Strategy for Adding Reactions (Cont.)
Choose equation that allows you to
cancel intermediates
multiply by appropriate coefficient
Add reactions together and cancel like terms
Add energies together, modifying enthalpy values in same way equation modified
If reversed equation, change sign on enthalpy
If doubled equation, double energy

13. Example Calculate H °rxn for 2 C (s, gr) + H2(g)  C2H2(g)
Given the following:
C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l ) H °rxn = – kJ
C(s, gr) + O2(g)  CO2(g) H °rxn = –393.5 kJ
H2(g) + ½O2(g)  H2O(l ) H °rxn = –285.8 kJ

14. 2C(s, gr) + H2(g)  C2H2(g)
2CO2(g) + H2O(l )  C2H2(g) + 5/2O2(g) H°rxn = – (– kJ) = kJ
2C(s, gr) + 2O2(g)  2CO2(g) H°rxn =(2–393.5 kJ) = –787.0 kJ
H2(g) + ½O2(g)  H2O(l ) H°rxn = –285.8 kJ
2CO2(g) + H2O(l ) + 2C(s, gr) + 2O2(g) + H2(g) + ½O2(g)  C2H2(g) + 5/2O2(g) + 2CO2(g) + H2O(l )
2C(s, gr) + H2(g)  C2H2(g) H°rxn = kJ

15. Standard Enthalpy of Formation, Hf°
A standard enthalpy of formation is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions.

16. Standard State
Most stable form and physical state of element at 1 atm and 25 °C (298 K)
Element
Standard state O
O2(g) C
C (s, gr) H
H2(g) Al
Al(s) Ne
Ne(g)
Note: All Hf° of elements in their standard states = 0

17. Standard Enthalpy of Formation, Hf°
Hf° of C2H5OH(l )
2C(s, gr) + 3H2(g) + ½O2(g)  C2H5OH(l ) Hf° = – kJ/mol
Hf° of Fe2O3(s)
2Fe(s) + 3/2O2(g)  Fe2O3(s) Hf° = –822.2 kJ/mol

18. Using Hf° H°reaction = – Sum of all H°f of all of the products
Sum of all H°f of all of the reactants

19. Calculate Horxn Using Hf°
Calculate H°rxn using Hf° data for the reaction
SO3(g)  SO2(g) + ½O2(g)
H°rxn = 99 kJ

20. Standard Enthalpy of combustion, Hc°
Standard enthalpy of combustion, Hc°, is the enthalpy change when 1 mole of a substance combines with oxygen under standard state conditions
C2H5OH(l) + 3O2(g)⟶2CO2 (g) + 3H2O(l)
Hc°= − kJ/mol

21. Using Hc° H°reaction = – Sum of all H°c of all of the reactants
Sum of all H°c of all of the products