1. Hess’s Law

2. After today you will be able to…
Add, multiply, divide, or reverse chemical equations
Calculate the enthalpy changes for an overall reaction using Hess’s Law

3. Enthalpy
Enthalpy: The total energy or total heat of a system. Its symbol is H, and ∆H shows the change in energy of a reaction.

4. Hess’s Law
Hess’s Law: If you add two or more chemical equations to get an overall equation, then you can also add the heat changes (ΔHs) to get the overall heat change.

5. How to add chemical equations:
If two identical substances are on opposite sides of the arrow, they will cancel.
If two identical substances are on same side of the arrow, add the coefficients together.
Keep substances on the same side of the arrow in the final equation.

6. Helpful hint: Align arrows underneath each other!
Example: Add the following equations. C + O2  CO2 CO2  CO + ½ O2
Helpful hint: Align arrows underneath each other!
C + ½O2  CO
If the same chemical appears on both sides of the equation, nothing changes, that’s why it cancels.

7. Example: Add the following equations
Example: Add the following equations. 2 Cu + O2  2 CuO 4 Cu + O2  2 Cu2O
6Cu + 2O2  2CuO + 2Cu2O
In this example, Cu and O2 are both reactants, and they are not recreated on the products side in the same form. So instead of cancelling, these chemicals are added together.

8. How to reverse chemical equations:
To reverse an equation, the products and reactants change sides. The sign of the ΔH also reverses. If the forward reaction is exothermic, the reverse reaction is endothermic (ex: - to +).

9. Example: Reverse this reaction. 2 P + 3 Cl2  2 PCl3 H = -574 kJ
Switch sides!
Reverse the sign!
2 PCl3  2 P + 3 Cl2
H = +574 kJ
The forward reaction was exothermic, so the reverse reaction is endothermic. The value of H is still the same.

10. How to multiply/divide a chemical equation:
Multiply or divide all coefficients by the same number. Multiply or divide the H by the same number. If more or less chemicals are reacting, H will change by the same factor.

11. Example: Multiply this equation by 2. Ca + 2 C  CaC2 H = -62.8kJ [ ] [ ] x2 x2
2 Ca + 4 C  2 CaC2
H = kJ
If there are twice as many reactants and products, twice as much heat is going to be released (because this particular reaction is exothermic).

12. These rearrangements can be combined…
Example: Reverse and multiply by 3. 2Fe + 3/2 O2  Fe2O3 H = kJ [ [ ] ] x3 x3
Reverse the sign!
Switch sides!
3 Fe2O3  6 Fe + 9/2 O2
H = kJ
The reaction is reversed, so it goes from exothermic to endothermic. The reaction is also increased by a factor of three times as many chemicals, so the heat involved is increased by a factor of three.

13. Now it’s time to apply Hess’s law to some examples!

14. Example: Find the H for C2H4 + H2  C2H6 using the following information: 2 C + 2 H2  C2H4 H = +52kJ 2 C + 3 H2  C2H6 H = -85kJ
Helpful hint: Pick one substance that appears only in the overall equation!
(Reverse)
(Same)
C2H4  2 C + 2 H2
ΔH= -52kJ
2 C + 3 H2  C2H6
ΔH= -85kJ
C2H4 + H2  C2H6
ΔH= -137kJ

15. 3 H2 + 3 Cl2  6HCl ΔH= -552kJ 2 NH3  N2 + 3 H2 ΔH= +92kJ
Example: Find the H for 2 NH3 + 3 Cl2  6 HCl + N2 using the following information: ½ H2 + ½ Cl2  HCl H = -92kJ ½ N2 + 3/2 H2  NH3 H = -46kJ
(x 6)
(Rev, x2)
3 H Cl2  6HCl
ΔH= -552kJ
2 NH3  N2 + 3 H2
ΔH= +92kJ
3 Cl2 + 2 NH3  6HCl + N2
ΔH= -460kJ

16. Questions?