### Hess' Law of Constant Heat SummationUsing four or more equations and their enthalpies

Some examples below (and some scattered in the problems) use five data equations. There's even one with six data equations.

Example #1: Calculate the value of ΔH° for the following reaction:

P4O10(s) + 6PCl5(g) ---> 10Cl3PO(g)

using the following four equations:

 a) P4(s) + 6Cl2(g) ---> 4PCl3(g) ΔH° = −1225.6 kJ b) P4(s) + 5O2(g) ---> P4O10(s) ΔH° = −2967.3 kJ c) PCl3(g) + Cl2(g) ---> PCl5(g) ΔH° = −84.2 kJ d) PCl3(g) + 1⁄2O2(g) ---> Cl3PO(g) ΔH° = −285.7 kJ

Solution:

1) We know that P4O10 MUST be on the left-hand side in the answer, so let's reverse (b):

 b) P4O10(s) ---> P4(s) + 5O2(g) ΔH° = +2967.3 kJ

2) We know that PCl5 MUST be on the left-hand side in the answer, so let's reverse (c) and multiply it by 6:

 c) 6PCl5(g) ---> 6PCl3(g) + 6Cl2(g) ΔH° = +505.2 kJ

3) We know that Cl3PO MUST have a 10 in front of it:

 d) 10PCl3(g) + 5O2(g) ---> 10Cl3PO(g) ΔH° = −2857 kJ

4) Now, write all four equations, but incorporate the revisions:

 a) P4(s) + 6Cl2(g) ---> 4PCl3(g) ΔH° = −1225.6 kJ b) P4O10(s) ---> P4(s) + 5O2(g) ΔH° = +2967.3 kJ c) 6PCl5(g) ---> 6PCl3(g) + 6Cl2(g) ΔH° = +505.2 kJ d) 10PCl3(g) + 5O2(g) ---> 10Cl3PO(g) ΔH° = −2857 kJ

5) Now, we will add all four equations as well as the ΔH° values. Notice the following:

a) P4(s) cancels out (see equations a and b)
b) Cl2 cancels out (see equations a and c)
c) O2 cancels out (see equations b and d)
d) PCl3 cancels out (see equations a+c and d)

6) The ΔH° values added together:

−1225.6 kJ + (+2967.3 kJ) + (+505.2 kJ) + (−2857 kJ) = −610.1 kJ

 P4O10(s) + 6PCl5(g) ---> 10Cl3PO(g) ΔH° = −610.1 kJ

Example #2: Calculate the reaction enthalpy for the formation of anhydrous aluminum chloride:

2Al(s) + 3Cl2(g) ---> 2AlCl3(s)

from the following data:

 2Al(s) + 6HCl(aq) ---> 2AlCl3(aq) + 3H2(g) ΔH° = −1049 kJ HCl(g) ---> HCl(aq) ΔH° = −74.8 kJ H2(g) + Cl2(g) ---> 2HCl(g) ΔH° = −185 kJ AlCl3(s) ---> AlCl3(aq) ΔH° = −323 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 ⇒ this one remains unchanged. It gives us 2Al(s), which is what we want. The other substances will cancel out, as described below.

eq. 2 ⇒ this one will get multiplied by six in order to cancel the 6HCl(aq).

eq. 3 ⇒ this one gets multiplied by three. This gives us 3Cl2(g), which is what we want, and cancels out the six HCl(g) that was in eq. 2. It also cancels the 3H2(g) from eq. 1.

eq. 4 ⇒ this one gets flipped (to put AlCl3(s) on the right) and it gets multiplied by two. It also cancels the AlCl3(aq) from eq. 1.

2) Rewrite the four equations with all applied changes:

 2Al(s) + 6HCl(aq) ---> 2AlCl3(aq) + 3H2(g) ΔH° = −1049 kJ 6HCl(g) ---> 6HCl(aq) ΔH° = −448.8 kJ 3H2(g) + 3Cl2(g) ---> 6HCl(g) ΔH° = -555 kJ 2AlCl3(aq) ---> 2AlCl3(s) ΔH° = +646 kJ

(−1049) + (−448.8) + (-555) + (+646) = −1406.8 kJ

 2Al(s) + 3Cl2(g) ---> 2AlCl3(s) ΔH° = −1406.8 kJ <--- NOT the final answer!!

This is not the enthalpy of formation for AlCl3(s). Remember that an enthalpy of formation equation is always for ONE mole of the target substance. In other words, this:
 Al(s) + 3⁄2Cl2(g) ---> AlCl3(s) ΔH°f = −703.4 kJ

The book value, by the way, is −705.63 kJ/mol.

Example #3: Using only the equations below, calculate the molar heat of formation of nitrous acid HNO2(aq).

 NH4NO2(aq) ---> N2(g) + 2H2O(ℓ) ΔH° = −320.1 kJ NH3(aq) + HNO2(aq) ---> NH4NO2(aq) ΔH° = −37.7 kJ 2NH3(aq) ---> N2(g) + 3H2(g) ΔH° = +169.9 kJ 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH° = −571.6 kJ

Solution:

1) Let's get the target equation:

a formation reaction is a very specific type of chemical reaction. The reactants produce one mole of the product in its standard state:
reactants ---> HNO2(aq)

the reactants must be elements in their standard states:

12H2(g) + 12N2(g) + O2(g) ---> HNO2(aq) <--- the coefficient of the product is always a 1 in a formation reaction

2) 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 ⇒ this will be flipped because eq. 2 also gets flipped.

eq. 2 ⇒ this one gets flipped because we have to have HNO2(aq) on the product side. This forces eq. 1 to also be flipped to cancel out the NH4NO2.

eq. 3 ⇒ this one gets divided by 2. The most obvious reason is in order to cancel the NH3 from eq. 2. The nitrogen and hydrogen will also cancel to give the final answer.

eq. 4 ⇒ this one is untouched. It will cancel the 2H2O that is in eq. 1.

3) Rewrite the four equations with all applied changes:

 N2(g) + 2H2O(ℓ) ---> NH4NO2(aq) ΔH° = +320.1 kJ NH4NO2(aq) ---> NH3(aq) + HNO2(aq) ΔH° = +37.7 kJ NH3(aq) ---> 1⁄2N2(g) + 3⁄2H2(g) ΔH° = +84.95 kJ 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH° = −571.6 kJ

4) Some comments on substances cancelling:

nitrogen: eq. 1 and eq. 3 cancel to leave 12N2 on the reactant side
hydrogen: eq. 3 and eq. 4 cancel to give 12H2 on the reactant side. Think of the 2H2 in eq. 4 as 42H2

The only other substances that do not cancel are HNO2(aq) (product side) and O2(g) (reactant side), which is exactly what we want.

(+320.1) + (+37.7) + (+84.95) + (−571.6) = −128.85 kJ

Do not write −128.85 kJ/mol, write this (rounded to three sig figs):

ΔH°f, HNO2 = −129 kJ

By the definition of formation, the amount is always for one mole of the target substance.

Note: for a variation on this question, use the fourth data equation like this:

 H2(g) + 1⁄2O2(g) ---> H2O(ℓ) ΔH° = −285.8 kJ

Example #4: Determine ΔH for the reaction:

4CO + 8H2 ---> 3CH4 + CO2 + 2H2O

given the following data:

 (a) C + 1⁄2O2 ---> CO ΔH = −110.5 kJ (b) CO + 1⁄2O2 ---> CO2 ΔH = −282.9 kJ (c) H2 + 1⁄2O2 ---> H2O ΔH = −285.8 kJ (d) C + 2H2 ---> CH4 ΔH = −74.8 kJ (e) CH4 + 2O2 ---> CO2 + 2H2O ΔH = −890.3 kJ

Solution:

When I solved this problem, I went through several combinations of flip/don't flip and what factor to use before getting the right answer. That's because, due to how the equations interweave (each substance in the final equation is in two data equations), there is lots of trial-and-error involved.

As best as I can, I'm going to describe some of my thinking that led to the correct solution below, but you might want to avoid the explanation and try this one on your own first. It's a very, very good problem and no, I did not write this problem!

The solution is described starting in step five of the explanation, if you want to stop your scrolling before seeing the solution.

1) The two equations with carbon monoxide in them:

Equation (a) is connected to equation (d) and one of them must be flipped. Also, whatever factor I choose to use must be applied to both equations.

Equation (b) is connected to equation (e) with respect to the CO2. Notice that only one CO2 will be required. That means that either equation (b) or (e) will wind up with a factor.

2) The two equations with methane in them:

If equation (d) gets flipped, then (e) must also be flipped.

If equation (d) gets flipped, then that means a possible factor of 4 for equation (e), since we required three CH4 in the final answer.

If (e) gets flipped, then that means we will need a pretty big factor in equation (c), in order to generate enough H2 and enough H2O for the final equation.

3) The two equations with water in them:

One of them must be flipped. However, notice that we require eight H2, so flipping either one has consequences. For example, if I flip (d), then I must also flip (e) to get methane on the right.

4) The four equations with O2 in them:

You might think it wise to ignore the oxygen, but that can also be a mistake if you carry it too far. In this problem, I realized a relatively large factor needed to be used in equation (c). This was to get sufficient H2 on the left and also to get sufficient O2 on the left so as to cancel O2 on the right.

5) Here is the solution to this problem:

(a) flip, multiply by 1
(b) do not flip, x3
(c) do not flip, x6
(d) do not flip, x1
(e) flip, x2

6) Let's rewrite according to the above instructions:

 (a) CO ---> C + 1⁄2O2 ΔH = +110.5 kJ (b) 3CO + 3⁄2O2 ---> 3CO2 ΔH = −848.7 kJ (c) 6H2 + 6⁄2O2 ---> 6H2O ΔH = −1714.8 kJ (d) C + 2H2 ---> CH4 ΔH = −74.8 kJ (e) 2CO2 + 4H2O ---> 2CH4 + 8⁄2O2 ΔH = +1780.6 kJ

7) The enthalpy is:

(+110.5) + (−848.7) + (−1714.8) + (−74.8) + (+1780.6) = −747.2 kJ

Postscript: one day, while checking for errors, I realized that every substance of equation (e) was present in one of the other data equations. Perhaps (e) wasn't required for the solution to the problem. Here are the four data equations with the required changes:

 (a) C + 1⁄2O2 ---> CO ΔH = −110.5 kJ <--- flip and mult. by 3 (b) CO + 1⁄2O2 ---> CO2 ΔH = −282.9 kJ <--- unchanged (c) H2 + 1⁄2O2 ---> H2O ΔH = −285.8 kJ <--- mult. by 2 (d) C + 2H2 ---> CH4 ΔH = −74.8 kJ <--- mult. by 3

The computed ΔH for the target reaction equals −747.4 kJ.

Example #5: Acetylene, C2H2, is a gas commonly used in welding. It is formed in the reaction of calcium carbide, CaC2, with water. Given the thermochemical equations below, calculate the value of ΔH°f for acetylene in units of kilojoules per mole:

 (a) CaO(s) + H2O(ℓ) ---> Ca(OH)2(s) ΔH° = −65.3 kJ (b) 2CaO(s) + 5C(s, gr) ---> 2CaC2(s) + CO2(g) ΔH° = +753 kJ (c) CaCO3(s) ---> CaO(s) + CO2(g) ΔH° = +178 kJ (d) CaC2(s) + 2H2O(ℓ) ---> Ca(OH)2(s) + C2H2(g) ΔH° = −126 kJ (e) C(s, gr) + O2(g) ---> CO2(g) ΔH° = −393.5 kJ (f) 2H2O(ℓ) ---> 2H2(g) + O2(g) ΔH° = +572 kJ

Here is the target equation:

2C(s, gr) + H2(g) ---> C2H2(g)

Comment #1: the technique is to ignore simple things like CO2 and H2O. If we do the others right, they will take care of themselves.

Comment #2: what evolves during the solution is that the answer is the above target equation, but with the coefficients of 4, 2 ---> 2. This means we will then divide by two in the final step. It turns out to be a bit of a hassle to try and go directly to the target equation. (However, I am certainly not going to stop you from trying on your own. Your life, not mine!)

Solution:

1) Let's analyse the six equations above:

(a) flip and multiply by 2, this gets 2 for the calcium hydroxide and cancels the CaO

(b) unchanged, this equation gets rid of the CaC2 on the reactant side of equation (d)

(c) not needed, the evil question writer put it there to confuse you.

(d) this one has the C2H2 on the product side. This is where we want it, but we have to get rid of everything else. We have to multiply it by two.

(e) flip, we need 4C because we have to multiply equation (d) by 2. We did that to (d) to be able to cancel the 2CaC2

(f) flip, notice that it has 2H2, which is what we need.

2) Rewrite all the equations, with all changes applied:

 (a) 2Ca(OH)2(s) ---> 2CaO(s) + 2H2O(ℓ) ΔH° = +130.6 kJ (b) 2CaO(s) + 5C(s, gr) ---> 2CaC2(s) + CO2(g) ΔH° = +753 kJ (c) not needed (d) 2CaC2(s) + 4H2O(ℓ) ---> 2Ca(OH)2(s) + 2C2H2(g) ΔH° = −252 kJ (e) CO2(g) ---> C(s, gr) + O2(g) ΔH° = +393.5 kJ (f) 2H2(g) + O2(g) ---> 2H2O(ℓ) ΔH° = −572 kJ

3) What cancels and where:

2Ca(OH)2(s) ⇒ equations a and d

2CaO(s) ⇒ equations a and b

4H2O(ℓ) ⇒ equations d with a and f

5C(s) ⇒ cancels with C(s) in equation e to give 4C

2CaC2(s) ⇒ equations b and d

CO2(g) ⇒ equations b and e

O2(g) ⇒ equations e and f

4) Add up all the ΔH values:

+130.6 + (+753) + (−252) + (+393.5) + (−572) = +453.1

 4C(s, gr) + 2H2(g) ---> 2C2H2(g) ΔH° = +453.1 kJ

5) Divide by 2:

 2C(s, gr) + H2(g) ---> C2H2(g) ΔH°f = +226.55 kJ

Note the addition of the subscripted f since we now have the correct formation reaction for C2H2(g).

Example #6: Given the following reactions where X represents a generic metal or metalloid,

 (1) H2(g) + 1⁄2O2(g) ---> H2O(g) ΔH = −241.8 kJ (2) X(s) + 2Cl2(g) ---> XCl4(s) ΔH = +207.7 kJ (3) 1⁄2H2(g) + 1⁄2Cl2(g) ---> HCl(g) ΔH = −92.3 kJ (4) X(s) + O2(g) ---> XO2(s) ΔH = −810.1 kJ (5) H2O(g) ---> H2O(ℓ) ΔH = −44.0 kJ

What is the enthalpy for this reaction:

XCl4(s) + 2H2O(ℓ) ---> XO2(s) + 4HCl(g)

Solution:

1) What we know and what it means:

We know that XCl4 is a reactant. That means equation two will have to be reversed.

We know that 4HCl is a product. That means equation three will have to multiplied by 4.

XO2, in equation four, is in the right place and the right amount. Leave that equation alone.

We need 2H2O(ℓ) as a reactant. Reverse equation five and multiply by 2.

2H2O(g), from our changed equation five, needs to be canceled out because it's not in the final equation. We do that by reversing equation one and multiplying it by 2.

2) We modify the data equations according to the above:

 (1) 2H2O(g) ---> 2H2(g) + O2(g) ΔH = +483.6 kJ (2) XCl4(s) ---> X(s) + 2Cl2(g) ΔH = −207.7 kJ (3) 2H2(g) + 2Cl2(g) ---> 4HCl(g) ΔH = −369.2 kJ (4) X(s) + O2(g) ---> XO2(s) ΔH = −810.1 kJ (5) 2H2O(ℓ) ---> 2H2O(g) ΔH = +88.0 kJ

3) Adding the five modified equations together will yield the target equation. H2O(g) will cancel (eq 1 and eq 5), as will H2 (1 and 3), O2 (1 and 4), Cl2 (2 and 3), and X(s) (2 and 4). Add up the five modified enthalpies for the final answer of −815.4 kJ.

Example #7: The bond enthalpy of the Br-Cl bond is equal to ΔH° for the reaction

BrCl(g) ---> Br(g) + Cl(g)

Use the following data to find the bond enthalpy of the Br-Cl bond.

 Br2(ℓ) ---> Br2(g) ΔH° = 30.91 kJ Br2(g) ---> 2Br(g) ΔH° = 192.9 kJ Cl2(g) ---> 2Cl(g) ΔH° = 243.4 kJ Br2(ℓ) + Cl2(g) ---> 2BrCl(g) ΔH° = 29.2 kJ

Solution:

1) It seems pretty obvious that the fourth equation will be reversed. Here are all four with that reversal applied:

 Br2(ℓ) ---> Br2(g) ΔH° = 30.91 kJ Br2(g) ---> 2Br(g) ΔH° = 192.9 kJ Cl2(g) ---> 2Cl(g) ΔH° = 243.4 kJ 2BrCl(g) ---> Br2(ℓ) + Cl2(g) ΔH° = −29.2 kJ

I could have also divided through by 2 to get BrCl instead of 2BrCl. I'll wait on that.

2) Notice several things:

(a) the Cl2(g) cancels (as it should) between the third and fourth equations
(b) the Br2(ℓ) cancels between the first and fourth equations
(c) the Br2(g) cancels between the first and second equations
(d) 2Cl(g) and 2Br(g) will remain after the above cancelling.

3) Add the four equations (and the four entalpies) to obtain:

2BrCl(g) ---> 2Br(g) + 2Cl(g) ΔH° = 438.01 kJ

4) Since our answer is double the requested equation, we divide through by 2:

BrCl(g) ---> Br(g) + Cl(g) ΔH° = 219 kJ

The reason I waited on the division is that I knew I'd have to divide the other equations by 2, making lots of one-halves appear in the data equations. Since waiting was possible in this problem, I elected to follow that path.

Example #8: Given:

 Br2(ℓ) + 5F2(g) ---> 2BrF5(ℓ) ΔH° = −918.0 kJ BrF3(ℓ) + Br2(ℓ) ---> 3BrF(g) ΔH° = 125.2 kJ 2NaBr(s) + F2(g) ---> 2NaF(s) + Br2(ℓ) ΔH° = −316.0 kJ NaBr(s) + F2(g) ---> NaF(s) + BrF(g) ΔH° = −216.6 kJ

calculate ΔH° for the reaction:

BrF3(ℓ) + F2(g) ---> BrF5(ℓ)

Solution:

1) Manipulate the four data equations:

 1⁄2Br2(ℓ) + 5⁄2F2(g) ---> BrF5(ℓ) ΔH° = −459.0 kJ (divide by 2) BrF3(ℓ) + Br2(ℓ) ---> 3BrF(g) ΔH° = 125.2 kJ 3NaBr(s) + 3⁄2F2(g) ---> 3NaF(s) + 3⁄2Br2(ℓ) ΔH° = −474.0 kJ (mult by 3⁄2) 3NaF(s) + 3BrF(g) ---> 3NaBr(s) + 3F2(g) ΔH° = 649.8 kJ (flip, mult by 3)

2) The reasons:

(a) get BrF5 with a coefficient of 1
(b) leave unchanged, BrF3 is on correct side and with coefficient of 1
(c) make 3NaF to cancel with 4th equation
(d) cancel the 3BrF in the second data equation

Notice that there are now 4F2 on the left (from 52 + 32). When the equations are added, that will cancel with the 3F2 on the right, giving us one F2 on the left, which is what we want.

Notice also that there will be 32Br2 on each side.

3) Add the four changed enthalpies for the final answer of −158 kJ.

Example #9: Determine the enthalpy of sublimation for solid potassium, given the following data:

 $\text{ΔH}{\text{}}_{f, KCl}^{o}$ = −436.7 kJ/mol $\text{ΔH}{\text{}}_{f, Cl\left(g\right)}^{o}$ = +121.3 kJ/mol $\text{ΔH}{\text{}}_{lattice energy, KCl}^{}$ = −715 kJ/mol $\text{ΔH}{\text{}}_{electron affinity, Cl}^{}$ = −349 kJ/mol $\text{ΔH}{\text{}}_{first ionization energy, K}^{}$ = +418.7 kJ/mol

Solution:

1) Write the target equation:

 K(s) ---> K(g) ΔH = ???

2) Write all the above data with the chemical equations rather than word descriptions:

 (1) K(s) + 1⁄2Cl2(g) ---> KCl(s) ΔH = −436.7 kJ (2) K+(g) + Cl¯(g) ---> KCl(s) ΔH = −715 kJ (3) K(g) ---> K+(g) + e¯ ΔH = +418.7 kJ (4) 1⁄2Cl2(g) ---> Cl(g) ΔH = +121.3 kJ (5) Cl(g) + e¯ ---> Cl¯(g) ΔH = −349 kJ

3) Analyze the equations in view of what is needed for the target equation.

(1) Stays the same in order to keep K(s) on the reactant side.

(2) Flip this equation to cancel out the KCl in equation (1)(s).

(3) Flip to put K(g) on the product side.

(4) Flip so as to cancel the 12Cl2(g) in equation (1).

(5) Flip so as to cancel the Cl(g) in equation (4).

Note that K+(g) and e¯ are not mentioned. They will, however, cancel also.

4) Apply the changes:

 K(s) + 1⁄2Cl2(g) ---> KCl(s) ΔH = −436.7 kJ KCl(s) ---> K+(g) + Cl¯(g) ΔH = +715 kJ K+(g) + e¯ ---> K(g) ΔH = −418.7 kJ Cl(g) ---> 1⁄2Cl2(g) ΔH = −121.3 kJ Cl¯(g) ---> Cl(g) + e¯ ΔH = +349 kJ

5) Add the equations and the enthalpies to obtain:

 K(s) ---> K(g) ΔH = +87.3 kJ