Hess’s Law
5.3.1

Review
 Q - What is the first Law of
Thermodynamics?

The Laws of Thermodynamics
 The First Law
 The total energy of the universe (which hates you) is constant.
○ This is similar to the law of conservation of energy .
 It can be written as
○ ΔE universe
= ΔE system
+ ΔE surroundings
 Energy can only be transferred
= 0
○ The flow of heat is considered one such transfer

Review
 Q - What kind of calorimeter is this?
 Q - What is held constant in this type of calorimeter?

Calorimeters
Cup Calorimeter
* Constant P
Bomb Calorimeter
* Constant V

Enthalpies
 Calorimetry under constant pressure is so common that the heat values (q p
) that are calculated using this method have a special name, Enthalpy ( ΔH).
 If ΔH is > 0 the reaction is endothermic
 If ΔH is < 0 the reaction is exothermic

Enthalpies

ΔH is also known as ΔH reaction because it is often given for a particular reaction.
 Example: 2H
2
( g ) + O
2
( g )  2H
2
O( g )
○ ΔH reaction
= -483.6 kJ
 Q - Is this reaction endothermic or exothermic?

Enthalpies

ΔH is dependent on the state of matter so be very careful to note which state you are talking about when working problems.
 Making liquid water does not release the same amount of energy as the formation of steam…
 The same is true for nearly all variables in thermo chemistry (c, S, etc.)

1940 - Germain Henri Hess
Hess’s Law


You have already measured enthalpy using a calorimeter.
You can also calculate it using Hess’s

Law which states:
The heat transferred, ΔH, in a reaction is the same regardless whether the reaction occurs in a single step or in several steps.
 If a series of reactions are added together, the net change in the heat of the reaction is the sum of the enthalpy changes for each step.


Hess’s Law
1.
Rules for using Hess's Law
If the reaction is multiplied (or divided) by some factor, H must also be multiplied (or divided) by that same factor.
2.
If the reaction is reversed (flipped), the sign of H must also be reversed.


Hess’s Law


Easy Example Problem
○
Nitrogen and oxygen gas combine to form nitrogen dioxide according to the following reaction:
N
2
( g ) + 2O
2
( g )  2NO
2
( g )
○
○
Calculate the change in enthalpy for the above overall reaction, given:
N
2
( g ) + O
2
( g )  2NO ( g ) ΔH = +181 kJ
2NO( g ) + O
2
( g )  2NO
2
( g ) ΔH = -131 kJ


Hess’s Law


Easy Example Problem - ANSWER
○
Nitrogen and oxygen gas combine to form nitrogen dioxide according to the following reaction:
N
2
( g ) + 2O
2
( g )  2NO
2
( g )
○
○
Calculate the change in enthalpy for the above overall reaction, given:
N
2
( g ) + O
2
( g )  2NO ( g ) ΔH = +181 kJ
2NO( g ) + O
2
( g )  2NO
2
( g ) ΔH = -131 kJ
+ 50kJ

Hess’s Law


Another Easy Example Problem (# 2)
Calculate the value of ΔH ⁰ for the reaction:
○ C (graphite) (s)

C (diamond) (s)
 Using the following equations:
1.
2.
C
(graphite)
(s) + O
2
(g)

CO
2
(g) ΔH ⁰ = ─ 394 kJ
C
(diamond)
(s) + O
2
(g)

CO
2
(g) ΔH ⁰ = ─ 396 kJ

Hess’s Law
 ANSWER to Example Problem (# 2)
Using the following equations:
1.
C
(graphite)
(s) + O
2
(g)

CO
2
(g) ΔH ⁰ = ─ 394 kJ
Use eqn #1 as is…
2.
CO
2
(g)

C
(diamond)
(s) + O
2
(g) ΔH ⁰ = + 396 kJ need to reverse eqn #2
-----------------------------------------------------------------------------
C (graphite) (s)

C (diamond) (s) ΔH ⁰ = + 2 kJ

Hess’s Law


Harder Example Problem
○
Calculate the value of ΔH for the reaction:
2S(g) + 2OF
2
(g)

SO
2
(g) + SF
4
(g)

Hess’s Law


Harder Example Problem
○
Calculate the value of ΔH for the reaction:
2S(g) + 2OF
2
(g)

SO
2
(g) + SF
4
(g)
 If we look at the final reaction, we see that we need 2
S atoms on the reactants side.
○ Q - Which step reaction has just S?
 The only reaction with S atoms is the third reaction, and in order to get 2 S atoms, we need to multiply the whole reaction by a factor of 2. ( put on board)

○
The next reactant in the final reaction is 2 OF molecules.
Q - Which step reaction has OF?
 The only reaction with an OF molecule is the first reaction, and in order to get 2 OF molecules, we need to multiply the whole reaction by a factor of 2. ( p.o.b.)

Hess’s

Law
Harder Example Problem

○
Calculate the value of ΔH for the reaction:
2S(g) + 2OF
2
(g)

SO
2
(g) + SF
4
(g)


On the products side of the final reaction, there is 1 SF
4 molecule, and the only possible source of the SF
4 molecule is the second reaction.
However, the SF
4 molecule is on the reactants side, which is not the side we need it on.


So we'll have to FLIP the second reaction to get the SF
4 molecule where we need it.
Now if we total up the reactions, we should end up with the given overall reaction:

Hess’s Law



Another Example Problem (# 3)
Calculate the value of ΔH ⁰ for the reaction:
○ 2N
2
(g) + 5O
2
(g)

2N
2
O
5
(g)
Using the following equations:
1.
2.
3.
H
2
(g) + ½ O
2
(g)

H
2
O(l) ΔH ⁰ = ─285.5 kJ
N
2
O
5
(g) + H
2
O(l)

2 HNO
3
(l) ΔH ⁰ = ─ 76.6 kJ
½ N
2
(g) + 3/2 O
2
(g) + ½ H
2
(g)

HNO
3
(l)
ΔH ⁰ = ─ 174.1 kJ



ANSWER to Example Problem (# 3)
Using the following equations:


Multiply equation # 3 by 4

2 N
2
(g) + 6 O
2
(g) + 2 H
2
(g)

4 HNO
3
(l)
ΔH ⁰ = ─ 696.4 kJ
Reverse equation # 2 and multiply by 2
4 HNO
3
(l)

2 N
2
O
5
(g) + 2 H
2
O(l) ΔH ⁰ = + 153.2 kJ
Reverse equation # 1 and multiply by 2

2 H
2
O(l)

2 H
2
(g) + O
2
(g) ΔH ⁰ = + 571 kJ
Cancel any substances and add eqn together
--------------------------------------------------------------------------
2N
2
(g) + 5O
2
(g)

2N
2
O
5
(g) ΔH ⁰ = + 27.8 kJ

HW (Due Wednesday)
 Read Section 5.3 – if you haven’t already…
 Do Exercise 5.3 – Due on Thursday 5/2
 P.143
○ #1-5