1. Enthalpy and Hess’s Law
Chemistry I

2. Chemical Reactions at Constant Volume and Pressure
We know DE = q + w
DE = q – PDV
At constant volume, PDV = 0
So at constant volume, DE = qv
Subscript “v” reminds us that is at constant volume

3. Chemical Reactions at Constant Pressure
This equation already tells us at constant pressure, DE = qp – PDV
We can rearrange the equation as:
qp = DE – PDV
We can redefine “q” as H, or enthalpy
H = E + PV
If pressure is held constant, DH = DE + PDV
H is also a state function
Hint: most uppercase symbols are state functions

4. Enthalpy Enthalpy – heat absorbed or released in a reaction
Depends on the difference in quantity of heat
Represented by a H
When the pressure of a reaction remains constant, the heat absorbed or released during a chemical reaction is equal to the change in enthalpy for the reaction

5. Enthalpy Change
We can determine how energy was changed throughout a reaction
DH = Hproducts – Hreactants
Endothermic –
Heat absorbed ends up as “extra” enthalpy for the products compared to the reactants
DH = + (positive)
Exothermic –
Heat absorbed ends up as “less than” the enthalpy for the products compared to the reactants
DH = - (negative)

6. Enthalpy Diagram
Exothermic reaction

7. Enthalpy Sign of DH Process Heat Positive Endothermic Absorbed
Negative
Exothermic
released

8. Standard Enthalpy
Enthalpy measured when reactants are in their standard state (how they exist at room temperature) they are referred to as standard enthalpy
DH⁰

9. Using Enthalpy Changes
How much heat is released if 1.0 g of H2O2 decomposes in a beetle to produce a steam spray?
2 H2O2(l)  2 H2O (l) + O2(g) DH⁰ = -190 kJ
1.0 g H2O2 x (1 mol H2O2 / 34.0 g H2O2) = mol H2O2
Heat transferred = mol H2O2 x (-190 kJ/2 mol H2O2) = -2.8 kJ

10. Using Enthalpy Changes
How much heat is released if 6.44 g of S reacts with O2 according to the following equation?
2S +3 O2  2 SO3 DH⁰ = kJ
6.44 g S x (1 mol S / 32.1 g S) = mol S
Heat transferred = mol S x ( kJ/2 mol 3 O2) = 79.5 kJ

11. Relationship between DH and DE
Under constant volume, DE =qv
Under constant pressure, DH = qp
For most reactions not involving gases, DE is almost the same as DH since there (since there is no change in DV)
For gases, some of the energy (DE) is used to expand (or contract) the gas, therefore DH will be slightly smaller (DV is changing)

12. Relationship Cont. For gases, therefore, DE = DH – D(PV)
From ideal gas: PV = nRT
So DE = DH – RTDn
Dn = number of moles of product GAS – number of moles of reactant GAS

13. Example 6.4
Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25C.
2 CO(g) + O2(g)  2 CO2(g) DH = kJ/mol
DE = DH – RTDn
Dn = 2 -3 = -1
DE = kJ/mol – (8.314 J/K*mol)(1 kJ/1000 J)(298 K)(-1)
kJ/mol