1. Calculating Heats of Reaction
Hess’s Law and Heats of Formation

2. Introduction
If we want to measure the heat of reaction of methane, we just burn a known amount of methane gas in oxygen and measure the heat evolved by using a calorimeter.
But, sometimes we don’t want consume the material when we are interested in the heat of reaction.
What is the heat of formation of a diamond?

3. Hess’s Law
There is a way to measure the heat of reaction of a compound in a way that does not require destroying the compound.
We use Hess’s Law:
If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

4. Hess’s Law The law is easier to use than it is to say. For example:
Find the heat of reaction to go from diamond to graphite.
C(diamond) → C(graphite)

5. Hess’s Law C(diamond) → C(graphite)
We can’t measure this directly, because the process is too slow.
But we do know the following information:
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

6. Hess’s Law C(diamond) → C(graphite)
If we rearrange the first equation ...
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

7. Hess’s Law C(diamond) → C(graphite)
If we rearrange the first equation ...
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

8. Hess’s Law C(diamond) → C(graphite) ... add them together ...
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

9. Hess’s Law C(diamond) → C(graphite) ... add them together ...
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)

10. Hess’s Law C(diamond) → C(graphite) ... and simplify ...
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)

11. Hess’s Law C(diamond) → C(graphite) ... and simplify ...
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)

12. Hess’s Law C(diamond) → C(graphite) ... and simplify ...
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

13. Hess’s Law C(diamond) → C(graphite) ... we get the desired equation.
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

14. Hess’s Law C(diamond) → C(graphite) Now, we add the heats of reaction.
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

15. Hess’s Law C(diamond) → C(graphite) Now, we add the heats of reaction.
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
C(diamond) → C(graphite) ∆H = kJ − kJ

16. Hess’s Law C(diamond) → C(graphite) Now, we add the heats of reaction.
CO2(g) → C(graphite) + O2(g) ∆H = kJ
C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ
C(diamond) → C(graphite) ∆H = − 1.9 kJ
This is an exothermic process.

17. Hess’s Law Let’s try another one. Find ∆H for the reaction:
C(graphite) + ½ O2(g) → CO(g)
given:
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

18. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
Notice that C(graphite) is on the left of both equations.
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

19. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
This means that we will write the equation with C(graphite) the same way in our final equations.
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

20. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
Notice that CO(g) is on the right above and on the left of below.
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

21. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
This means that we will write the equation with CO(g) the opposite way in our final equations.
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

22. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
This means that we will write the equation with CO(g) the opposite way in our final equations.
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ

23. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
Next, we add everything together ... spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ

24. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
Next, we add everything together ... spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ
C(graphite) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g)

25. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
... and simplify ... spacer spacer spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ
C(graphite) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g)
½ O2(g)

26. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
... and simplify ... spacer spacer spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ

27. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
... to get the equation that we want. spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ

28. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
Now, we add the heats of reaction together. spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ

29. Hess’s Law C(graphite) + ½ O2(g) → CO(g)
Now, we add the heats of reaction together. spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ
C(graphite) + ½ O2(g) → CO(g) ∆H = −393.5 kJ kJ

30. Hess’s Law This is an exothermic process.
C(graphite) + ½ O2(g) → CO(g)
Now, we add the heats of reaction together. spacer spacer spacer
C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ
CO2(g) → CO(g) + ½ O2(g) ∆H = kJ
C(graphite) + ½ O2(g) → CO(g) ∆H = −110.5 kJ
This is an exothermic process.

31. Hess’s Law Let’s try another one. Find ∆H for the reaction:
2 P(s) + 5 Cl2(g) → 2 PCl5(s)
given:
PCl5(s) → PCl3(g) + Cl2(g) ∆H = kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

32. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Notice that P(s) is on the left of both equations. spacer spacer
PCl5(s) → PCl3(g) + Cl2(g) ∆H = kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

33. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
This means that we will write the equation with P(s) the same way in our final equations.
PCl5(s) → PCl3(g) + Cl2(g) ∆H = kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

34. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Notice that PCl5(g) is on the right above and on the left of below.
PCl5(s) → PCl3(g) + Cl2(g) ∆H = kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

35. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
This means that we will write the equation with CO(g) the opposite way in our final equations.
PCl5(s) → PCl3(g) + Cl2(g) ∆H = kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

36. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
This means that we will write the equation with CO(g) the opposite way in our final equations.
PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

37. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Also notice that PCl5(g) is has a coefficient of “2” above and “1” of below.
PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

38. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
This means that we need to multiply the equation below by 2.
PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

39. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
This means that we need to multiply the equation below by 2.
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

40. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Next, we add everything together ... spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

41. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Next, we add everything together ... spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ
2 PCl3(s) + 2 Cl2(g) + 2 P(s) + 3 Cl2(g) → 2 PCl5(s) + 2 PCl3(g)

42. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
... and simplify ... spacer spacer spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ
2 PCl3(s) + 2 Cl2(g) + 2 P(s) + 3 Cl2(g) → 2 PCl5(s) + 2 PCl3(g)

43. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
... and simplify ... spacer spacer spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

44. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
... to get the equation that we want. spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

45. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Now, we add the heats of reaction together. spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

46. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Now, we add the heats of reaction together. spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ
2 P(s) + 5 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ − 574 kJ

47. Hess’s Law 2 P(s) + 5 Cl2(g) → 2 PCl5(s)
Now, we add the heats of reaction together. spacer spacer spacer
2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ
2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ
2 P(s) + 5 Cl2(g) → 2 PCl5(s) ∆H = −749.8 kJ = −750. kJ
This is a highly exothermic reaction.

48. Standard Heats of Reaction
Changes in enthalpy (∆H) are usually specified at a set of standard conditions.
Standard temperature = 25°C = 298 K
Standard pressure = 1 atm = kPa
The change in enthalpy that accompanies the formation of one mole of a compound from its elements under standard conditions is called the standard heat of formation, ∆Hfo.

49. Standard Heats of Reaction
For example:
H2(g) + ½ O2(g) → H2O(l) ∆Hfo = − kJ
Ca(s) + ½ O2(g) → CaO(s) ∆Hfo = −635 kJ
C(s) + 2 H2(g) → CH4(g) ∆Hfo = −74.86 kJ
½ N2(g) + ½ O2(g) → NO(g) ∆Hfo = kJ

50. Standard Heats of Reaction
For a reaction occurring at standard conditions, you can calculate the heat of reaction by using the standard heats of formation of the reactants and the products.
∆Ho = ∆Hfo(products) − ∆Hfo(reactants)

51. Standard Heats of Reaction
For example, to calculate the standard heat of reaction of methane burned in air.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
∆Hfo(CH4, g) = −74.86 kJ
∆Hfo(O2, g) = 0.0 kJ
∆Hfo(CO2, g) = −393.5 kJ
∆Hfo(H2O, l) = −282.8 kJ

52. Standard Heats of Reaction
For example, to calculate the standard heat of reaction of methane burned in air.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
∆Ho = ∆Hfo(products) − ∆Hfo(reactants)
∆Hfo(products) = ∆Hfo(CO2) + 2 ∆Hfo(H2O)
= −393.5 kJ + 2(−285.8 kJ) = −965.1 kJ
∆Hfo(reactants) = ∆Hfo(CH4) + 2 ∆Hfo(O2)
= −74.86 kJ + 2(0.0 kJ) = −74.86 kJ

53. Standard Heats of Reaction
For example, to calculate the standard heat of reaction of methane burned in air.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
∆Ho = −965.1 kJ − (−74.86 kJ) = −890.2 kJ
This is a highly exothermic reaction.

54. Standard Heats of Reaction
Let’s do another one.
Calculate the standard heat of reaction of :
SO2(g) + ½ O2(g) → SO3(g)
∆Hfo(SO2, g) = −296.8 kJ
∆Hfo(O2, g) = 0.0 kJ
∆Hfo(SO3, g) = −395.7 kJ

55. Standard Heats of Reaction
Calculate the standard heat of reaction of :
SO2(g) + ½ O2(g) → SO3(g)
∆Ho = ∆Hfo(products) − ∆Hfo(reactants)
∆Hfo(products) = ∆Hfo(SO3)
= −395.7 kJ
∆Hfo(reactants) = ∆Hfo(SO2) + ½ ∆Hfo(O2)
= −296.8 kJ + ½(0.0 kJ) = −296.8 kJ

56. Standard Heats of Reaction
Calculate the standard heat of reaction of :
SO2(g) + ½ O2(g) → SO3(g)
∆Ho = −395.7 kJ − (−296.8 kJ) = −100.9 kJ
This is a highly exothermic reaction.