Using 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:

P_{4}O_{10}(s) + 6PCl_{5}(g) ---> 10Cl_{3}PO(g)

using the following four equations:

(a) P _{4}(s) + 6Cl_{2}(g) ---> 4PCl_{3}(g)ΔH° = −1225.6 kJ (b) P _{4}(s) + 5O_{2}(g) ---> P_{4}O_{10}(s)ΔH° = −2967.3 kJ (c) PCl _{3}(g) + Cl_{2}(g) ---> PCl_{5}(g)ΔH° = −84.2 kJ (d) PCl _{3}(g) +^{1}⁄_{2}O_{2}(g) ---> Cl_{3}PO(g)ΔH° = −285.7 kJ

**Solution:**

1) We know that P_{4}O_{10} **MUST** be on the left-hand side in the answer, so let's reverse (b):

(b) P _{4}O_{10}(s) ---> P_{4}(s) + 5O_{2}(g)ΔH° = +2967.3 kJ

2) We know that PCl_{5} **MUST** be on the left-hand side in the answer, so let's reverse (c) and multiply it by 6:

(c) 6PCl _{5}(g) ---> 6PCl_{3}(g) + 6Cl_{2}(g)ΔH° = +505.2 kJ

3) We know that Cl_{3}PO **MUST** have a 10 in front of it:

(d) 10PCl _{3}(g) + 5O_{2}(g) ---> 10Cl_{3}PO(g)ΔH° = −2857 kJ

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

(a) P _{4}(s) + 6Cl_{2}(g) ---> 4PCl_{3}(g)ΔH° = −1225.6 kJ (b) P _{4}O_{10}(s) ---> P_{4}(s) + 5O_{2}(g)ΔH° = +2967.3 kJ (c) 6PCl _{5}(g) ---> 6PCl_{3}(g) + 6Cl_{2}(g)ΔH° = +505.2 kJ (d) 10PCl _{3}(g) + 5O_{2}(g) ---> 10Cl_{3}PO(g)ΔH° = −2857 kJ

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

(a) P_{4}(s) cancels out (see equations a and b)

(b) Cl_{2}cancels out (see equations a and c)

(c) O_{2}cancels out (see equations b and d)

(d) PCl_{3}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

7) The answer:

P _{4}O_{10}(s) + 6PCl_{5}(g) ---> 10Cl_{3}PO(g)ΔH° = −610.1 kJ

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

2Al(s) + 3Cl_{2}(g) ---> 2AlCl_{3}(s)

from the following data:

(a) 2Al(s) + 6HCl(aq) ---> 2AlCl _{3}(aq) + 3H_{2}(g)ΔH° = −1049 kJ (b) HCl(g) ---> HCl(aq) ΔH° = −74.8 kJ (c) H _{2}(g) + Cl_{2}(g) ---> 2HCl(g)ΔH° = −185 kJ (d) AlCl _{3}(s) ---> AlCl_{3}(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):

(a) ---> this one remains unchanged. It gives us 2Al(s), which is what we want. The other substances will cancel out, as described below.(b) ---> this one will get multiplied by six in order to cancel the 6HCl(aq).

(c) ---> this one gets multiplied by three. This gives us 3Cl

_{2}(g), which is what we want, and cancels out the six HCl(g) that was in eq. 2. It also cancels the 3H_{2}(g) from eq. 1.(d) ---> this one gets flipped (to put AlCl

_{3}(s) on the right) and it gets multiplied by two. It also cancels the AlCl_{3}(aq) from eq. 1.

2) Rewrite the four equations with all applied changes:

(a) 2Al(s) + 6HCl(aq) ---> 2AlCl _{3}(aq) + 3H_{2}(g)ΔH° = −1049 kJ (b) 6HCl(g) ---> 6HCl(aq) ΔH° = −448.8 kJ (c) 3H _{2}(g) + 3Cl_{2}(g) ---> 6HCl(g)ΔH° = -555 kJ (d) 2AlCl _{3}(aq) ---> 2AlCl_{3}(s)ΔH° = +646 kJ

3) Add the four enthalpies for NOT the final answer:

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

2Al(s) + 3Cl _{2}(g) ---> 2AlCl_{3}(s)ΔH° = −1406.8 kJ <--- NOT the final answer!!

4) Comments:

The above is not the enthalpy of formation for AlCl_{3}(s). Remember that an enthalpy of formation equation is always forONEmole of the target substance. In other words, this:

Al(s) + ^{3}⁄_{2}Cl_{2}(g) ---> AlCl_{3}(s)$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ = −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 HNO_{2}(aq).

(a) NH _{4}NO_{2}(aq) ---> N_{2}(g) + 2H_{2}O(ℓ)ΔH° = −320.1 kJ (b) NH _{3}(aq) + HNO_{2}(aq) ---> NH_{4}NO_{2}(aq)ΔH° = −37.7 kJ (c) 2NH _{3}(aq) ---> N_{2}(g) + 3H_{2}(g)ΔH° = +169.9 kJ (d) 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(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 ---> HNO_{2}(aq)the reactants must be elements in their standard states:

^{1}⁄_{2}H_{2}(g) +^{1}⁄_{2}N_{2}(g) + O_{2}(g) ---> HNO_{2}(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:

(a) ---> this will be flipped because (b) also gets flipped.(b) ---> this one gets flipped because we have to have HNO

_{2}(aq) on the product side. This forces (a) to also be flipped to cancel out the NH_{4}NO_{2}.(c) ---> this one gets divided by 2. The most obvious reason is in order to cancel the NH

_{3}from (b). The nitrogen and hydrogen will also cancel to give the final answer.(d) ---> this one is untouched. It will cancel the 2H

_{2}O that is in (a).

3) Rewrite the four equations with all applied changes:

(a) N _{2}(g) + 2H_{2}O(ℓ) ---> NH_{4}NO_{2}(aq)ΔH° = +320.1 kJ (b) NH _{4}NO_{2}(aq) ---> NH_{3}(aq) + HNO_{2}(aq)ΔH° = +37.7 kJ (c) NH _{3}(aq) --->^{1}⁄_{2}N_{2}(g) +^{3}⁄_{2}H_{2}(g)ΔH° = +84.95 kJ (d) 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(ℓ)ΔH° = −571.6 kJ

4) Some comments on substances cancelling:

nitrogen: (a) and (c) cancel to leave^{1}⁄_{2}N_{2}on the reactant side

hydrogen: (c) and (d) cancel to give^{1}⁄_{2}H_{2}on the reactant side. Think of the 2H_{2}in (d) as^{4}⁄_{2}H_{2}The only other substances that do not cancel are HNO

_{2}(aq) (product side) and O_{2}(g) (reactant side), which is exactly what we want.

5) Add the 4 enthalpies:

(+320.1) + (+37.7) + (+84.95) + (−571.6) = −128.85 kJDo not write −128.85 kJ/mol, write this (rounded to three sig figs):

$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ = −129 kJBy 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:

H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(ℓ)ΔH° = −285.8 kJ

**Example #4:** Determine ΔH for the reaction:

4CO + 8H_{2}---> 3CH_{4}+ CO_{2}+ 2H_{2}O

given the following data:

(a) C + ^{1}⁄_{2}O_{2}---> COΔH = −110.5 kJ (b) CO + ^{1}⁄_{2}O_{2}---> CO_{2}ΔH = −282.9 kJ (c) H _{2}+^{1}⁄_{2}O_{2}---> H_{2}OΔH = −285.8 kJ (d) C + 2H _{2}---> CH_{4}ΔH = −74.8 kJ (e) CH _{4}+ 2O_{2}---> CO_{2}+ 2H_{2}OΔH = −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 CO

_{2}. Notice that only one CO_{2}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 CH

_{4}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 H

_{2}and enough H_{2}O for the final equation.

3) The two equations with water in them:

One of them must be flipped. However, notice that we require eight H_{2}, 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 O_{2} 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 H_{2}on the left and also to get sufficient O_{2}on the left so as to cancel O_{2}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}⁄_{2}O_{2}ΔH = +110.5 kJ (b) 3CO + ^{3}⁄_{2}O_{2}---> 3CO_{2}ΔH = −848.7 kJ (c) 6H _{2}+^{6}⁄_{2}O_{2}---> 6H_{2}OΔH = −1714.8 kJ (d) C + 2H _{2}---> CH_{4}ΔH = −74.8 kJ (e) 2CO _{2}+ 4H_{2}O ---> 2CH_{4}+^{8}⁄_{2}O_{2}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}⁄_{2}O_{2}---> COΔH = −110.5 kJ <--- flip and mult. by 3 (b) CO + ^{1}⁄_{2}O_{2}---> CO_{2}ΔH = −282.9 kJ <--- unchanged (c) H _{2}+^{1}⁄_{2}O_{2}---> H_{2}OΔH = −285.8 kJ <--- mult. by 2 (d) C + 2H _{2}---> CH_{4}ΔH = −74.8 kJ <--- mult. by 3

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

**Example #5:** Acetylene, C_{2}H_{2}, is a gas commonly used in welding. It is formed in the reaction of calcium carbide, CaC_{2}, with water. Given the thermochemical equations below, calculate the value of ΔH°_{f} for acetylene in units of kilojoules per mole:

(a) CaO(s) + H _{2}O(ℓ) ---> Ca(OH)_{2}(s)ΔH° = −65.3 kJ (b) 2CaO(s) + 5C(s, gr) ---> 2CaC _{2}(s) + CO_{2}(g)ΔH° = +753 kJ (c) CaCO _{3}(s) ---> CaO(s) + CO_{2}(g)ΔH° = +178 kJ (d) CaC _{2}(s) + 2H_{2}O(ℓ) ---> Ca(OH)_{2}(s) + C_{2}H_{2}(g)ΔH° = −126 kJ (e) C(s, gr) + O _{2}(g) ---> CO_{2}(g)ΔH° = −393.5 kJ (f) 2H _{2}O(ℓ) ---> 2H_{2}(g) + O_{2}(g)ΔH° = +572 kJ

Here is the target equation:

2C(s, gr) + H_{2}(g) ---> C_{2}H_{2}(g)

Comment #1: the technique is to ignore simple things like CO_{2} and H_{2}O. 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 CaC

_{2}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 C

_{2}H_{2}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 2CaC

_{2}(f) flip, notice that it has 2H

_{2}, which is what we need.

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

(a) 2Ca(OH) _{2}(s) ---> 2CaO(s) + 2H_{2}O(ℓ)ΔH° = +130.6 kJ (b) 2CaO(s) + 5C(s, gr) ---> 2CaC _{2}(s) + CO_{2}(g)ΔH° = +753 kJ (c) not needed (d) 2CaC _{2}(s) + 4H_{2}O(ℓ) ---> 2Ca(OH)_{2}(s) + 2C_{2}H_{2}(g)ΔH° = −252 kJ (e) CO _{2}(g) ---> C(s, gr) + O_{2}(g)ΔH° = +393.5 kJ (f) 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(ℓ)ΔH° = −572 kJ

3) What cancels and where:

2Ca(OH)_{2}(s) ---> equations a and d2CaO(s) ---> equations a and b

4H

_{2}O(ℓ) ---> equations d with a and f5C(s) ---> cancels with C(s) in equation e to give 4C

2CaC

_{2}(s) ---> equations b and dCO

_{2}(g) ---> equations b and eO

_{2}(g) ---> equations e and f

4) Add up all the ΔH values:

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

4C(s, gr) + 2H _{2}(g) ---> 2C_{2}H_{2}(g)ΔH° = +453.1 kJ

5) Divide by 2:

2C(s, gr) + H _{2}(g) ---> C_{2}H_{2}(g)$\text{\Delta H}{\text{}}_{\mathrm{f}}^{\mathrm{o}}$ = +226.55 kJ

Note the addition of the subscripted f since we now have the correct formation reaction for C_{2}H_{2}(g).

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

(a) H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(g)ΔH = −241.8 kJ (b) X(s) + 2Cl _{2}(g) ---> XCl_{4}(s)ΔH = +207.7 kJ (c) ^{1}⁄_{2}H_{2}(g) +^{1}⁄_{2}Cl_{2}(g) ---> HCl(g)ΔH = −92.3 kJ (d) X(s) + O _{2}(g) ---> XO_{2}(s)ΔH = −810.1 kJ (e) H _{2}O(g) ---> H_{2}O(ℓ)ΔH = −44.0 kJ

What is the enthalpy for this reaction:

XCl_{4}(s) + 2H_{2}O(ℓ) ---> XO_{2}(s) + 4HCl(g)

**Solution:**

1) What we know and what it means:

We know that XCl_{4}is a reactant. That means (b) will have to be reversed.We know that 4HCl is a product. That means (c) will have to multiplied by 4.

XO

_{2}, in (d), is in the right place and the right amount. Leave that equation alone.We need 2H

_{2}O(ℓ) as a reactant. Reverse (e) and multiply by 2.2H

_{2}O(g), from our changed (e), needs to be canceled out because it's not in the final equation. We do that by reversing (a) and multiplying it by 2.

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

(a) 2H _{2}O(g) ---> 2H_{2}(g) + O_{2}(g)ΔH = +483.6 kJ (b) XCl _{4}(s) ---> X(s) + 2Cl_{2}(g)ΔH = −207.7 kJ (c) 2H _{2}(g) + 2Cl_{2}(g) ---> 4HCl(g)ΔH = −369.2 kJ (d) X(s) + O _{2}(g) ---> XO_{2}(s)ΔH = −810.1 kJ (e) 2H _{2}O(ℓ) ---> 2H_{2}O(g)ΔH = +88.0 kJ

3) Adding the five modified equations together will yield the target equation. H_{2}O(g) will cancel [(a) and (e)], as will H_{2} [(a) and (c)], O_{2} [(a) and (d)], Cl_{2} [(b) and (c)], and X(s) [(b) and (d)]. 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.

Br _{2}(ℓ) ---> Br_{2}(g)ΔH° = 30.91 kJ Br _{2}(g) ---> 2Br(g)ΔH° = 192.9 kJ Cl _{2}(g) ---> 2Cl(g)ΔH° = 243.4 kJ Br _{2}(ℓ) + Cl_{2}(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:

Br _{2}(ℓ) ---> Br_{2}(g)ΔH° = 30.91 kJ Br _{2}(g) ---> 2Br(g)ΔH° = 192.9 kJ Cl _{2}(g) ---> 2Cl(g)ΔH° = 243.4 kJ 2BrCl(g) ---> Br _{2}(ℓ) + Cl_{2}(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 Cl_{2}(g) cancels (as it should) between the third and fourth equations

(b) the Br_{2}(ℓ) cancels between the first and fourth equations

(c) the Br_{2}(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 kJThe 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:

Br _{2}(ℓ) + 5F_{2}(g) ---> 2BrF_{5}(ℓ)ΔH° = −918.0 kJ BrF _{3}(ℓ) + Br_{2}(ℓ) ---> 3BrF(g)ΔH° = 125.2 kJ 2NaBr(s) + F _{2}(g) ---> 2NaF(s) + Br_{2}(ℓ)ΔH° = −316.0 kJ NaBr(s) + F _{2}(g) ---> NaF(s) + BrF(g)ΔH° = −216.6 kJ

calculate ΔH° for the reaction:

BrF_{3}(ℓ) + F_{2}(g) ---> BrF_{5}(ℓ)

**Solution:**

1) Manipulate the four data equations:

^{1}⁄_{2}Br_{2}(ℓ) +^{5}⁄_{2}F_{2}(g) ---> BrF_{5}(ℓ)ΔH° = −459.0 kJ (divide by 2) BrF _{3}(ℓ) + Br_{2}(ℓ) ---> 3BrF(g)ΔH° = 125.2 kJ 3NaBr(s) + ^{3}⁄_{2}F_{2}(g) ---> 3NaF(s) +^{3}⁄_{2}Br_{2}(ℓ)ΔH° = −474.0 kJ (mult by ^{3}⁄_{2})3NaF(s) + 3BrF(g) ---> 3NaBr(s) + 3F _{2}(g)ΔH° = 649.8 kJ (flip, mult by 3)

2) The reasons:

(a) get BrF_{5}with a coefficient of 1

(b) leave unchanged, BrF_{3}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 equationNotice that there are now 4F

_{2}on the left (from^{5}⁄_{2}+^{3}⁄_{2}). When the equations are added, that will cancel with the 3F_{2}on the right, giving us one F_{2}on the left, which is what we want.Notice also that there will be

^{3}⁄_{2}Br_{2}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:

$\text{\Delta H}{\text{}}_{\mathrm{f,\; KCl}}^{\mathrm{o}}$ = −436.7 kJ/mol $\text{\Delta H}{\text{}}_{\mathrm{f,\; Cl(g)}}^{\mathrm{o}}$ = +121.3 kJ/mol $\text{\Delta H}{\text{}}_{\mathrm{lattice\; energy,\; KCl}}^{\mathrm{}}$ = −715 kJ/mol $\text{\Delta H}{\text{}}_{\mathrm{electron\; affinity,\; Cl}}^{\mathrm{}}$ = −349 kJ/mol $\text{\Delta H}{\text{}}_{\mathrm{first\; ionization\; energy,\; K}}^{\mathrm{}}$ = +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}⁄_{2}Cl_{2}(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}⁄_{2}Cl_{2}(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

^{1}⁄_{2}Cl_{2}(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}⁄_{2}Cl_{2}(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}⁄_{2}Cl_{2}(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

**Example #10:** Determine the enthalpy of reaction (in kJ) at 298 K for the reaction:

2SO_{2}(g) +^{1}⁄_{2}P_{4}(s) + 5Cl_{2}(g) ---> 2SOCl_{2}(ℓ) + 2POCl_{3}(ℓ)

Given the following equations and ΔH° values,

(a) SOCl _{2}(ℓ) + H_{2}O(ℓ) ---> SO_{2}(g) + 2HCl(g)ΔH° = +10.3 kJ (b) PCl _{3}(ℓ) +^{1}⁄_{2}O_{2}(g) ---> POCl_{3}(ℓ)ΔH° = −325.7 kJ (c) ^{1}⁄_{4}P_{4}(s) +^{3}⁄_{2}Cl_{2}(g) ---> PCl_{3}(ℓ)ΔH° = −306.7 kJ (d) 4HCl(g) + O _{2}(g) ---> 2Cl_{2}(g) + 2H_{2}O(ℓ)ΔH° = −202.6 kJ

**Solution:**

1) Let's examine each data equation for what must be done in order to generate the target equation:

(a) ---> the equation must be flipped (to get SO_{2}on the reactant side) and multiplied by 2 (to get 2SO_{2})(b) ---> multiply by 2 because POCl

_{3}has a 2 in front of it in the target equation(c) ---> multiply by 2 because P

_{4}needs a^{1}⁄_{2}in front of it(d) ---> flip it so as to cancel 4HCl in addition to creating 5Cl

_{2}on the reactant side

2) Apply the above changes:

(a) 2SO _{2}(g) + 4HCl(g) ---> 2SOCl_{2}(ℓ) + 2H_{2}O(ℓ)ΔH° = −20.6 kJ (b) 2PCl _{3}(ℓ) + O_{2}(g) ---> 2POCl_{3}(ℓ)ΔH° = −651.4 kJ (c) ^{1}⁄_{2}P_{4}(s) + 3Cl_{2}(g) ---> 2PCl_{3}(ℓ)ΔH° = −613.4 kJ (d) 2Cl _{2}(g) + 2H_{2}O(ℓ) ---> 4HCl(g) + O_{2}(g)ΔH° = +202.6 kJ

3) Add the for equations. 4HCl, 2H_{2}O, O_{2}, and PCl_{3} all cancel out, leaving the target equation.

4) Add the four enthalpies:

(−20.6 kJ) + (−651.4 kJ) + (−613.4 kJ) + (+202.6 kJ) = −1082.8 kJ

5)I found this problem on a Hess' Law worksheet and a wrong answer (−1041.6 kJ) was provided. That wrong answer is obtained by not changing the sign of the first data equation, this using +20.6 when you should be using −20.6.