Hess' Law of Constant Heat Summation
Using four or more equations and their enthalpies
Problems 1 - 10

Hess' Law: four or more equations and their enthalpies - Probs 11-20      Hess' Law: two equations and their enthalpies      Hess' Law: standard enthalpies of formation
Hess' Law - four or more equations and their enthalpies      Hess' Law: three equations and their enthalpies      Hess' Law: bond enthalpies
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Problem #1: Calculate the ΔH in kilojoules for the following reaction, the preparation of nitrous acid HNO2:

HCl(g) + NaNO2(s) ---> HNO2(ℓ) + NaCl(s)

Use the following thermochemical equations:

2NaCl(s) + H2O(ℓ) ---> 2HCl(g) + Na2O(s)ΔH = +507.31 kJ
NO(g) + NO2(g) + Na2O(s) ---> 2NaNO2(s)ΔH = -427.14 kJ
NO(g) + NO2(g) ---> N2O(g) + O2(g)ΔH = -42.68 kJ
2HNO2(ℓ) ---> N2O(g) + O2(g) + H2O(ℓ)ΔH = +34.35 kJ

Solution:

1) Let's examine each of the four equations in light of what needs to happen to it (in order to produce the target equation):

eq 1 ⇒ flip it (this puts NaCl on the right-hand side and HCl on the left-hand side)

eq 2 ⇒ flip it (this puts NaNO2 on the left-hand side)

eq 3 ⇒ leave untouched

eq 4 ⇒ flip it (this puts HNO2 on the right-hand side)

2) Rewrite all four equations with the above changes:

2HCl(g) + Na2O(s) ---> 2NaCl(s) + H2O(ℓ)ΔH = -507.31 kJ
2NaNO2(s) ---> NO(g) + NO2(g) + Na2O(s)ΔH = +427.14 kJ
NO(g) + NO2(g) ---> N2O(g) + O2(g)ΔH = -42.68 kJ
N2O(g) + O2(g) + H2O(ℓ) ---> 2HNO2(ℓ)ΔH = -34.35 kJ

Note the sign changes on the enthalpies of the three flipped reactions. The substances that get eliminated are:

Na2O(s) (eq 1 & 3); H2O(ℓ) (eq 1 & 4); NO(g) (eq 2 & 3); NO2(g) (eq 2 & 3); N2O(g) (eq 3 & 4); O2(g) (eq 3 & 4)

3) Add the four reactions to get this:

2HCl(g) + 2NaNO2(s) ---> 2HNO2(ℓ) + 2NaCl(s)ΔH = -157.2 kJ

The ΔH value came from this:

(-507.31) + (+427.14) + (-42.68) + (-34.35)

4) Divide everything by two for the final answer:

HCl(g) + NaNO2(s) ---> HNO2(ℓ) + NaCl(s)ΔH = -78.6 kJ

Problem #2: Determine the heat of reaction (in kJ) at 298 K for the reaction:

N2H4(ℓ) + O2(g) ---> N2(g) + 2H2O(ℓ)

given the following equations and ΔH values:

2NH3(g) + 3N2O(g) ---> 4N2(g) + 3H2O(ℓ)ΔH = -1013 kJ
N2O(g) + 3H2(g) ---> N2H4(ℓ) + H2O(ℓ)ΔH = -317 kJ
2NH3(g) + 12O2(g) ----> N2H4(ℓ) + H2O(ℓ)ΔH = -142.9 kJ
H2(g) + 12O2(g) ---> H2O(ℓ)ΔH = -285.8 kJ

Solution:

1) First, some discussion:

a) Equation 1 stays untouched. The main reason is because that's the only reaction that has N2 on the product side, which is where we need it. The 4 in front of the N2 is going to play a role. Suppose I divided through by 4 to get the one N2 in the final answer. That means I would wind up with 3/4 in front of the N2O and also in front of the H2O. Way too complicated. Keeping the 4 in front of the N2 means two things: (i) we will only deal with fractions that have a 2 in the denominator and (ii) the very least step will be to divide by 4.

b) Equation 2 needs to be flipped, so the N2O can be on the product side (to cancel with the 3N2O in the first equation). I also have to multiply this reaction by 3, to give me my 3N2O for cancelling purposes.

c) This reaction needs to be flipped too. I must have the 2NH3 be on the product side to cancel with the 2NH3 in equation 1.

d) Notice that I have two equations (#2 and #3) that have N2H4. When I add everything up, I'll have 4N2H4. Remember: my very last step will be to divide everything by 4.

e) Equation 4 gets multiplied by 9. Look at the oxygens. I know I need 4O2 (remember I will divide by 4 at the end), so I used 9 since I knew that would make 92O2 and the 12O2 in equation 3 would cancel, giving me 82O2 which is 4O2.)

f) I will not discuss the H2O, so that you may ponder how it works out.

2) Here's the result of everything I described:

2NH3(g) + 3N2O(g) ---> 4N2(g) + 3H2O(ℓ)ΔH = -1013 kJ
3N2H4(ℓ) + 3H2O(ℓ) ---> 3N2O(g) + 9H2(g)ΔH = +951 kJ
N2H4(ℓ) + H2O(ℓ)---> 2NH3(g) + 12O2(g)ΔH = +142.9 kJ
9H2(g) + 92O2(g) ---> 9H2O(ℓ)ΔH = -2572.2 kJ

3) When we add the four chemical reactions together, here is what results:

4N2H4(ℓ) + 4O2(g) ---> 4N2(g) + 8H2O(ℓ)

The 2NH3, the 3N2O and the 9H2 cancel completely.

12O2 cancels and four of the 12H2O on the right cancel.

4) Calculating the enthalpy:

a) add the four enthalpies: -1013 kJ + +951 kJ + +142.9 kJ + -2572.2 kJ = -2491.3 kJ

b) divide by 4 for the final answer: -623 kJ (to three sig figs)


Problem #3: Calculate the standard enthalpy change for H2O(g) ---> H2O(ℓ) at 298.15 K, given the following reaction enthalpies:

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(g)ΔH = -802.30 kJ
C(s, graphite) + O2(g) ---> CO2(g)ΔH = -393.51 kJ
C(s, graphite) + 2H2(g) ---> CH4(g)ΔH = -74.81 kJ
2H2(g) + O2(g) ---> 2H2O(ℓ)ΔH = -571.66 kJ

Solution:

1) Do these:

a) flip first equation (puts H2O(g) on the reactant side)
b) leave second equation alone (need to cancel the CO2 from the first equation)
c) flip third equation (to cancel the CH4 in the first equation)
d) leave fourth equation alone (H2O(ℓ) is on the product side, which is where we want it)

I ignored the other items which will, if I did it right, take care of themselves.

2) The result:

CO2(g) + 2H2O(g) ---> CH4(g) + 2O2(g)ΔH = +802.30 kJ
C(s, graphite) + O2(g) ---> CO2(g)ΔH = -393.51 kJ
CH4(g) ---> C(s, graphite) + 2H2(g)ΔH = +74.81 kJ
2H2(g) + O2(g) ---> 2H2O(ℓ)ΔH = -571.66 kJ

3) When the four modified equations are added together, we get this:

2H2O(g) ---> 2H2O(ℓ)

4) Add up the four enthalpies above and then divide that answer by 2 for the final answer.

[+802.30 + (-393.51) + (+74.81) + (-571.66)] / 2 = -44.03 kJ

H2O(g) ---> H2O(ℓ)ΔH = -44.03 kJ


Problem #4: Using these reactions, where M represents a generic metal:

2M(s) + 6HCl(aq) ---> 2MCl3(aq) + 3H2(g)ΔH = -881 kJ
HCl(g) ---> HCl(aq)ΔH = -74.8 kJ
H2(g) + Cl2(g) ---> 2HCl(g)ΔH = -1845.0 kJ
MCl3(s) ---> MCl3(aq)ΔH = -316 kJ

Determine the enthalpy of:

2M(s) + 3Cl2(g) ---> 2MCl3(s)

Solution:

1) These changes to the four data equations:

1) untouched (because it has 2M on the reactant side, where we want it)
2) mult by 6 (to cancel the 6HCl(aq) in equation 1)
3) mult by 3 (to cancel the H2 in eq 1, to cancel 6HCl(g), to get 3Cl2 in the final answer)
4) flip and mult by 2 (put 2MCl3(s) on the product side)

2) Result:

2M(s) + 6HCl(aq) ---> 2MCl3(aq) + 3H2(g)ΔH = -881 kJ
6HCl(g) ---> 6HCl(aq)ΔH = -448.8 kJ
3H2(g) + 3Cl2(g) ---> 6HCl(g)ΔH = -5535 kJ
2MCl3(aq) ---> 2MCl3(s)ΔH = +632 kJ

3) Add the four enthalpies for the answer:

2M(s) + 3Cl2(g) ---> 2MCl3(s) ΔH = -6232.8 kJ

Problem #5: From a consideration of the following reactions, calculate ΔH°f for ethane, C2H6(g).

CH3CHO(g) + 2H2(g) ---> C2H6(g) + H2O(ℓ)ΔH° = -204 kJ
2H2(g) + O2(g) ---> 2H2O(g)ΔH° = -484 kJ
2C2H5OH(ℓ) + O2(g) ---> 2CH3CHO(g) + 2H2O(ℓ)ΔH° = -348 kJ
H2O(ℓ) ---> H2O(g)ΔH° = 44 kJ
2C2H5OH(ℓ) ---> 4C(s, gr) + 6H2(g) + O2(g)ΔH° = 555 kJ

Solution:

1) The formation reaction for ethane is this:

2C(s) + 3H2(g) ---> C2H6

2) We will need to manipulate the 5 data equation to get what we want.

a) equation 1 of the 5 has C2H6 as the product. That's where we want it, so leave eq 1 untouched.
b) equation 3 must be divided by 2. This is to get one CH3CHO on the product side, so it will cancel the CH3CHO in eq 1.
c) equation 5 must be flipped and divided by 2. This will put 2C on the reactant side and give us one C2H5OH to cancel the one in eq 3.

3) Let's apply the above changes and then look at equations 2 and 4.

CH3CHO(g) + 2H2(g) ---> C2H6(g) + H2O(ℓ)ΔH° = -204 kJ
2H2(g) + O2(g) ---> 2H2O(g)ΔH° = -484 kJ
C2H5OH(ℓ) + 12O2(g) ---> CH3CHO(g) + H2O(ℓ)ΔH° = -174 kJ
H2O(ℓ) ---> H2O(g)ΔH° = 44 kJ
2C(s, gr) + 3H2(g) + 12O2(g) ---> C2H5OH(ℓ)ΔH° = -277.5 kJ

4) Flip equation 2 and multiply equation 4 by 2. This will cancel all of the H2O(g) and H2O(ℓ):

CH3CHO(g) + 2H2(g) ---> C2H6(g) + H2O(ℓ)ΔH° = -204 kJ
2H2O(g) ---> 2H2(g) + O2(g)ΔH° = 484 kJ
C2H5OH(ℓ) + 12O2(g) ---> CH3CHO(g) + H2O(ℓ)ΔH° = -174 kJ
2H2O(ℓ) ---> 2H2O(g)ΔH° = 88 kJ
2C(s, gr) + 3H2(g) + 12O2(g) ---> C2H5OH(ℓ)ΔH° = -277.5 kJ

There are 5H2 on the reactant side and 2H2 on the product side, leaving 3H2 on the reactant side, which is what we want. The one O2 on each side will cancel.

5) Adding up the five enthalpies gives

ΔH° = -83.5 kJ

Compare that to the data given here.


Problem #6: Find the heat of reaction for:

CO + 2H2 ---> CH3OH

using these data:

(I) 2C(s) + O2(g) ---> 2CO(g)ΔH = -220.9 kJ
(II) C(s) + O2(g) ---> 2CO2(g)ΔH = -391.1 kJ
(III) 2CH3OH(ℓ) + 3O2(g) ---> 2CO2(g) + 4H2O(ℓ)ΔH = -1451.9 kJ
(IV) 2H2(g) + O2(g) ---> 2H2O(ℓ)ΔH = -571.2 kJ

Solution:

1) Apply the following changes:

Reverse (I), change the sign of ΔH
Multiply (II) by two.
Reverse (III) and change the sign of ΔH
Multiply equation (IV) by two.

2) The result of the changes:

(I) 2CO(g) ---> 2C(s) + O2(g)ΔH = +220.9 kJ
(II) 2C(s) + 2O2(g) ---> 4CO2(g)ΔH = -782.2 kJ
(III) 2CO2(g) + 4H2O(ℓ) ---> 2CH3OH(ℓ) + 3O2(g)ΔH = +1451.9 kJ
(IV) 4H2(g) + O2(g) ---> 4H2O(ℓ)ΔH = -1142.4 kJ

3) Add the four equations:

2CO + 4H2 ---> 2CH3OH ΔH = -251.8 kJ

4) Divide the above equation by 2:

CO + 2H2 ---> CH3OH ΔH = -125.9 kJ

Problem #7: Find the heat of reaction for the equation:

N2O3(g) + N2O5(s) ---> 2N2O4(g)
Given the following reactions and enthalpy changes
NO(g) + NO2(g) ---> N2O3(g)ΔH = -39.8 kJ
NO(g) + NO2(g) + O2(g) ---> N2O5(g)ΔH = -112.5 kJ
2NO2(g) ---> N2O4(g)ΔH = -57.2 kJ
2NO(g) + O2(g) ---> 2NO2(g)ΔH = -114.2 kJ
N2O5(s) ---> N2O5(g)ΔH = +54.4 kJ

Solution:

1) Manipulate the data equations as follows:

eq 1 - flip (puts N2O3 as a reactant)
eq 2 - flip (puts N2O5(g) as a reactant, to cancel with eq 5)
eq 3 - multiply by 2 (need 2N2O4 as a product)
eq 4 - untouched
eq 5 - untouched

2) When you add the five data equations together, you will have the following cancel:

4NO2 (four on the right from eqs 1, 2, and 4; four on the left from eq 3
2NO (two on the right from eq 1 and 2; two on the left from eq 4)
O2 (from eq 2 and 4)
N2O5(g) (from eq 2 and 5)

3) The answer:

39.8 + 112.5 + (-114.4) + (-114.2) + 54.4 = -21.9 kJ

Problem #8: This information is given:

Cl2(g) + 5F2(g) ---> 2ClF5(g)ΔH° = -510.0 kJ
ClF3(g) + Cl2(g) ---> 3ClF(g)ΔH° = -5.5 kJ
2NaCl(s) + F2(g) ---> 2NaF(s) + Cl2(g)ΔH° = -316.0 kJ
NaCl(s) + F2(g) ---> NaF(s) + ClF(g)ΔH° = -214.5 kJ

Please calculate the ΔH° for this reaction:

ClF3(g) + F2(g) ---> ClF5(g)

Solution:

1) I'm going to modify the first and the fourth data equations:

12Cl2(g) + 52F2(g) ---> ClF5(g)ΔH° = -255.0 kJ
ClF3(g) + Cl2(g) ---> 3ClF(g)ΔH° = -5.5 kJ
2NaCl(s) + F2(g) ---> 2NaF(s) + Cl2(g)ΔH° = -316.0 kJ
3NaF(s) + 3ClF(g) ---> 3NaCl(s) + 3F2(g)ΔH° = +643.5 kJ

Modifying the first equation gives me one ClF5 as a product, which is what I want in my target equation.

Modifying the fourth equation gives me 3ClF on the left-hand side, which I need to cancel the 3ClF in the second equation.

2) We have more to do. The NaCl/NaF amounts between equations three and four will not cancel when the four equations are added. The answer is to multiply equation three by 32:

12Cl2(g) + (5/2)F2(g) ---> ClF5(g)ΔH° = -255.0 kJ
ClF3(g) + Cl2(g) ---> 3ClF(g)ΔH° = -5.5 kJ
3NaCl(s) + 32F2(g) ---> 3NaF(s) + 32Cl2(g)ΔH° = -474.0 kJ
3NaF(s) + 3ClF(g) ---> 3NaCl(s) + 3F2(g) ---> ΔH° = +643.5 kJ

3) When you add the four modified equations together, everything will cancel except the components of the target equation. Add the four enthalpies for the enthalpy of the target equation.

By the way, I made a strategic decision to break out the modifications needed into two parts. I wanted to highlight the issue in getting the NaCl/NaF equations to balance out so the NaCl and the NaF would cancel.


Problem #9: Determine the enthalpy for this reaction:

Zn(s) + 18S8(s) + 2O2(g) ---> ZnSO4(s)

given the following data:

Zn(s) + 18S8(s) ---> ZnS(s)ΔH = -183.92 kJ
2ZnS(s) + 3O2(g) ---> 2ZnO(s) + 2SO2(g)ΔH = -927.54 kJ
2SO2(g) + O2(g) ---> 2SO3(g)ΔH = -196.04 kJ
ZnO(s) + SO3(g) ---> ZnSO4(s)ΔH = -230.32 kJ

Solution:

1) The first data equation has Zn and S8 the way we want them. How to get rid of the ZnS in data equation #1? We modify the second data equation as follows:

ZnS(s) + 32O2(g) ---> ZnO(s) + SO2(g)ΔH = -463.77 kJ

2) Now, we need to eliminate the SO2 in the second data equation. We do that by modifying the third data equation:

SO2(g) + 12O2(g) ---> SO3(g)ΔH = -98.02 kJ

3) I still need to make the ZnO go away as well as the SO3. Both of those are accomplished by the fourth data equation being left untouched. It also introduces the ZnSO4 in its desired place as a product.

4) Let's rewrite all four data equations with the modifications:

Zn(s) + 19S8(s) ---> ZnS(s)ΔH = -183.92 kJ
ZnS(s) + 32O2(g) ---> ZnO(s) + SO2(g)ΔH = -463.77 kJ
SO2(g) + 12O2(g) ---> SO3(g)ΔH = -98.02 kJ
ZnO(s) + SO3(g) ---> ZnSO4(s)ΔH = -230.32 kJ

5) Adding the four above chemical equations will give us our target equation. Add the four enthalpies for the final answer.

Zn(s) + 18S8(s) + 2O2(g) ---> ZnSO4(s)ΔH = -976.03 kJ

The 2O2 came from the O2 in data equations 2 and 3.


Problem #10:

Solution:

1) Let us write the chemical equations associated with the four enthalpies given:

ReactionEnthalpy Value
A ---> BΔHAB
B ---> CΔHBC
A ---> EΔHAE
E ---> DΔHED

2) Rearrange the four data equations as follows:

first ⇒ flip, in order to cancel the B in the second equation
second ⇒ flip, because this puts C as the reactant
third ⇒ leave untouched, A will cancel with the A in the first equation
fourth ⇒ leave untouched, E cancels with the third equation and D is the product (which is what we want)

3) The results of the above-described modifications:

ReactionEnthalpy Value
B ---> A-ΔHAB
C ---> B-ΔHBC
A ---> EΔHAE
E ---> DΔHED

Note how the signs for the first two enthalpies have changed.

4) ΔHCD is arrived at by adding the four data enthalpies:

(-ΔHAB) + (-ΔHBC) + ΔHAE + ΔHED

Hess' Law: four or more equations and their enthalpies - Problems 11-20      Hess' Law: two equations and their enthalpies      Hess' Law: standard enthalpies of formation
Hess' Law - four or more equations and their enthalpies      Hess' Law: three equations and their enthalpies      Hess' Law: bond enthalpies
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