lecture 4: hess’s law reading: zumdahl 9.5 outline: definition of hess’ law using hess’ law...
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Lecture 4: Hess’s Law
Reading: Zumdahl 9.5
Outline:
Definition of Hess’ Law
Using Hess’ Law (examples)
Q: What is Hess’s Law ?
• Recall (lecture 3) Enthalpy is a state function. As such, H for going from some initial state to some final state is pathway independent.
• Hess’s Law: H for a process involving the transformation of reactants into products is not dependent on pathway.
(This means we can calculate H for a reaction by a single step, or by multiple steps)
Using Hess’s Law
When calculating H for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine H for our “single step” reaction.
2NO2 (g)
N2 (g) + 2O2 (g)
q
2NO2 (g)N2 (g) + 2O2 (g)
The reaction of interest is:
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
This reaction can also be carried out in two steps:
N2 (g) + O2 (g) 2NO(g) H = +180 kJ
2NO (g) + O2(g) 2NO2(g) H = -112 kJ
• If we take the previous two reactions and add them, we get the original reaction of interest:
N2 (g) + O2 (g) 2NO(g) H = +180 kJ 2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
N2 (g) + 2O2 (g) 2NO2(g) H = + 68 kJ
Note: the important things about this example is that the sum of H for the two reaction steps is equal to the H for the reaction of interest.
Big idea: We can combine reactions of known
H to determine the H for the overall
“combined” reaction.
Hess’s Law: An Important Detail
One can always reverse the direction of a
reaction when making a combined reaction.
When you do this, the sign of H changes.
N2(g) + 2O2(g) 2NO2(g) H = +68 kJ
2NO2(g) N2(g) + 2O2(g) H = - 68 kJ
One more detail:
• The magnitude of H is directly proportional to the quantities involved. (This means H is an “extensive” quantity).
• So, if the coefficients of a reaction are multiplied by a number, the value of H is also multiplied by the same number.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
N2(g) + 4O2(g) 4NO2(g) H = 136 kJ
Using Hess’s Law: tips
• When trying to combine reactions to form a reaction of interest, it is usually best to work backwards from the reaction of interest.
• Example:
What is H for the following reaction?
3C (gr) + 4H2 (g) C3H8 (g)
3C (gr) + 4H2 (g) C3H8 (g) H = ?
You’re given the following reactions:
C (gr) + O2 (g) CO2 (g) H = -394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ
Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation.
(x3) C (gr) + O2 (g) CO2 (g) H = -394 kJ
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
Step 2: To get C3H8 on the product side of the reaction, we need to reverse reaction 2, and change
the sign of H.
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 kJ
Step 3: Add two “new” reactions together to see what
remains:
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
• Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ
3C (gr) + 4H2 (g) C3H8 (g)
Need to multiply second reaction by 4
Example (cont.)
• Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g)
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g) H = -106 kJ
Which is the one step reaction of interest
Another Example:
• Calculate H for the following reaction:
H2(g) + Cl2(g) 2HCl(g)
Given the following:
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
N2 (g) + 3H2 (g) 2NH3 (g) H = - 92 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = - 629 kJ
Step 1: Only the first reaction contains the product of interest (HCl), but as a reactant. Therefore, reverse this reaction and multiply by 2 to get stoichiometry correct.
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = +352 kJ
Step 2: Need Cl2 as a reactant, therefore, add reaction 3 to result from step 1 and see what is left.
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ
N2(g) + 4H2(g) + Cl2(g) 2NH4Cl(s) H = -629 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)
H = -277 kJ
Step 3: Use remaining known reaction in combination with the result from Step 2 to get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
? ( N2 (g) + 3H2(g) 2NH3(g) H = -92 kJ)
H2(g) + Cl2(g) 2HCl(g) H = ?
Key: need to reverse the middle reaction
• Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
2NH3(g) 3H2 (g) + N2 (g) H = +92 kJ
H2(g) + Cl2(g) 2HCl(g) H = -185 kJ
1
This is the desired reaction and resultant H!