hess’s law 5.3.1. review q - what is the first law of thermodynamics?
TRANSCRIPT
HL5-3.PPT
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
○ ΔEuniverse = ΔEsystem + ΔEsurroundings = 0
Energy can only be transferred○ 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 (qp) that are calculated using this method have a special name, Enthalpy (ΔH).If ΔH is > 0 the reaction is endothermicIf ΔH is < 0 the reaction is exothermic
Enthalpies ΔH is also known as ΔHreaction because it is
often given for a particular reaction.Example: 2H2(g) + O2(g) 2H2O(g)
○ ΔHreaction = -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.)
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.
1940 - Germain Henri Hess
Hess’s Law Rules for using Hess's Law
1. 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:
○ N2(g) + 2O2(g) 2NO2 (g) Calculate the change in enthalpy for the above overall
reaction, given:○ N2(g) + O2(g) 2NO (g) ΔH = +181 kJ○ 2NO(g) + O2(g) 2NO2 (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:
○ N2(g) + 2O2(g) 2NO2 (g) Calculate the change in enthalpy for the above
overall reaction, given:○ N2(g) + O2(g) 2NO (g) ΔH = +181 kJ○ 2NO(g) + O2(g) 2NO2 (g) ΔH = -131 kJ
+ 50kJ
Hess’s Law
Another Easy Example Problem (# 2)
Calculate the value of ΔH⁰ for the reaction:
○ C (graphite) (s) C (diamond) (s)
Using the following equations:
1. C (graphite) (s) + O2(g) CO2(g) ΔH⁰ = ─ 394 kJ
2. C (diamond) (s) + O2(g) CO2(g) ΔH⁰ = ─ 396 kJ
Hess’s Law
ANSWER to Example Problem (# 2)
Using the following equations:
1. C (graphite) (s) + O2(g) CO2(g) ΔH⁰ = ─ 394 kJ
Use eqn #1 as is…
2. CO2(g) C (diamond) (s) + O2(g) ΔH⁰ = + 396 kJ
need to reverse eqn #2
-----------------------------------------------------------------------------C (graphite) (s) C (diamond) (s) ΔH⁰ = + 2 kJ
Hess’s Law
Harder Example Problem Calculate the value of ΔH for the reaction:
○ 2S(g) + 2OF2(g) SO2(g) + SF4(g)
Hess’s Law
Harder Example Problem Calculate the value of ΔH for the reaction:
○ 2S(g) + 2OF2(g) SO2(g) + SF4(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) + 2OF2(g) SO2(g) + SF4(g)
On the products side of the final reaction, there is 1 SF4 molecule, and the only possible source of the SF4 molecule is the second reaction.
However, the SF4 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 SF4 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 ΔH⁰ for the reaction:
○ 2N2 (g) + 5O2(g) 2N2O5(g)
Using the following equations:
1. H2 (g) + ½ O2(g) H2O(l) ΔH⁰ = ─285.5 kJ
2. N2O5(g) + H2O(l) 2 HNO3 (l) ΔH⁰ = ─ 76.6 kJ
3. ½ N2 (g) + 3/2 O2(g) + ½ H2 (g) HNO3 (l)
ΔH⁰ = ─ 174.1 kJ
ANSWER to Example Problem (# 3)
Using the following equations:
Multiply equation # 3 by 4
2 N2 (g) + 6 O2(g) + 2 H2 (g) 4 HNO3 (l)ΔH⁰ = ─ 696.4 kJ
Reverse equation # 2 and multiply by 2
4 HNO3 (l) 2 N2O5(g) + 2 H2O(l) ΔH⁰ = + 153.2 kJ
Reverse equation # 1 and multiply by 2
2 H2O(l) 2 H2 (g) + O2(g) ΔH⁰ = + 571 kJ
Cancel any substances and add eqn together
--------------------------------------------------------------------------
2N2 (g) + 5O2(g) 2N2O5(g) ΔH⁰ = + 27.8 kJ
HW (Due Wednesday)
Read Section 5.3 – if you haven’t already…
Do Exercise 5.3 – Due on Thursday 5/2P.143
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