1. Energy Lecture 4
Hess’s Law & Review

2. Calculating Enthalpy Change
A theoretical way to determine ∆H for a chemical reaction is provided by Hess’s law, which states that if two or more thermochemical equations can be added to produce a final equation for a reaction, then the enthalpy change for the final reaction equals the sum of the enthalpy changes for the individual reactions.

3. Hess’s Law
There is an amount of heat associated with every chemical reaction.

4. Hess’s Law
Often you know the heat for parts of the reaction, and you must add them together to find the heat for the total reaction.

5. Applying Hess’s Law
Use 2 thermochemical reactions to determine ∆H for the oxidation of ethanol (C2H5OH) to form acetaldehyde (C2H4O) and water.
Here is the overall reaction:
ethanol + oxygen gas acetaldehyde + water
Here are the two component reactions:

6. Applying Hess’s Law
For the overall reaction, Acetaldehyde should be on the right side of the equation, so reverse equation a.
Note that you must change the sign of ∆H.
The desired equation has two moles of ethanol, so double equation b and its ∆H.

7. Applying Hess’s Law
Add these two equations, and cancel any terms common to both sides of the combined equation. 1 2
∆ H = -349kJ
Note that ∆H is negative, which means the reaction is exothermic. (releasing energy)

8. Basic Assessment Questions
Practice Hess’s Law!
Use reactions a and b to determine ∆H for this single-displacement reaction.
Cl2(g) + 2HBr(g) HCl (g) + Br2(g)
a. H2(g) + Cl2(g) HCl (g) ∆H= -185
b. H2(g) + Br2(g) HBr (g) ∆H= -73

9. Practice Hess’s Law a. H2(g) + Cl2(g) 2HCl (g) ∆H= -185kJ
Keep equation “a” as written because HCl is on the right in the total reaction:
a. H2(g) + Cl2(g) HCl (g) ∆H= -185kJ
Flip equation “b” because HBr needs to be on the left in the overall equation
b. H2(g) + Br2(g) HBr (g) ∆H= -73kJ
b HBr (g) H2(g) + Br2(g) ∆H=73kJ

10. a. H2(g) + Cl2(g) 2HCl (g) ∆H= -185 kJ
Now add the two equations together
a. H2(g) + Cl2(g) HCl (g) ∆H= -185 kJ
b HBr (g) H2(g) + Br2(g) ∆H= 73kJ
Cl2(g) + 2HBr(g) HCl (g) + Br2(g)
∆H = -112
= -112kJ

11. Answer
∆Hrxn = –112 kJ

12. Practice using q=cm∆T
A 15.6-g sample of ethanol absorbs 868 J as it is heated. If the initial temperature of the ethanol was 21.5°C, what is the final temperature of the ethanol?
Hint:
solve for ∆T then add 21.5!

13. A 15. 6-g sample of ethanol absorbs 868 J as it is heated
A 15.6-g sample of ethanol absorbs 868 J as it is heated. If the initial temperature of the ethanol was 21.5°C, what is the final temperature of the ethanol?
868 J
(2.44J/gºC ) (15.6g)
∆T =
q .
c m =
Remember this is not your final answer.
You are looking for the final temp… so add the initial temp to this number
∆T= 22.8ºC
22.8ºC ºC = 44.3ºC = final temp

14. Practice q=mc∆T again!
If 335 g water at 65.5°C loses 9750 J of heat, what is the final temperature of the water?

15. Answer
58.5°C

16. Practice using q=mol x ∆H
How much heat is evolved when 24.9 g of propanol (C3H7OH) is burned? ∆Hcomb = –2010 kJ/mol
Molar mass of C3H7OH is 60.1g/mol

17. Answer
833 kJ