Using four or more equations and their enthalpies

**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)

The ΔH° values added together:

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

6) 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:

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

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 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.eq. 4 ⇒ 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:

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

3) Add the four enthalpies for the answer:

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

2Al(s) + 3Cl _{2}(g) ---> 2AlCl_{3}(s)ΔH° = -1406.8 kJ

4) Comments:

(a) This 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)ΔH° _{f}= -703.4 kJThe book value, by the way, is -705.63 kJ/mol.

(b) This question was asked and answered on Yahoo Answers. The style of the answer is different than the way I answered the problem above. You may wish to take a look.

**Example #3:** Using only the equations below, calculate the molar heat of formation of nitrous acid HNO_{2}(aq).

NH _{4}NO_{2}(aq) ---> N_{2}(g) + 2H_{2}O(ℓ)ΔH° = -320.1 kJ NH _{3}(aq) + HNO_{2}(aq) ---> NH_{4}NO_{2}(aq)ΔH° = -37.7 kJ 2NH _{3}(aq) ---> N_{2}(g) + 3H_{2}(g)ΔH° = +169.9 kJ 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 very specific. 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)

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 HNO

_{2}(aq) on the product side. This forces eq. 1 to also be flipped to cancel out the NH_{4}NO_{2}.eq. 3 ⇒ this one gets divided by 2. The most obvious reason is in order to cancel the NH

_{3}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 2H

_{2}O that is in eq. 1.

3) Rewrite the four equations with all applied changes:

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

4) Some comments on substances cancelling:

nitrogen: eq. 1 and eq. 3 cancel to leave^{1}⁄_{2}N_{2}on the reactant side

hydrogen: eq. 3 and eq. 4 cancel to give^{1}⁄_{2}H_{2}on the reactant side. Think of the 2H_{2}in eq. 4 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):

ΔH°_{f, HNO2}= -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

The following two examples use more than four data equations.

**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)ΔH° _{f}= +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,

(1) H _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> H_{2}O(g)ΔH = -241.8 kJ (2) X(s) + 2Cl _{2}(g) ---> XCl_{4}(s)ΔH = +207.7 kJ (3) ^{1}⁄_{2}H_{2}(g) +^{1}⁄_{2}Cl_{2}(g) ---> HCl(g)ΔH = -92.3 kJ (4) X(s) + O _{2}(g) ---> XO_{2}(s)ΔH = -810.1 kJ (5) 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 equation two will have to be reversed.We know that 4HCl is a product. That means equation three will have to multiplied by 4.

XO

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

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

_{2}O(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) 2H _{2}O(g) ---> 2H_{2}(g) + O_{2}(g)ΔH = +483.6 kJ (2) XCl _{4}(s) ---> X(s) + 2Cl_{2}(g)ΔH = -207.7 kJ (3) 2H _{2}(g) + 2Cl_{2}(g) ---> 4HCl(g)ΔH = -369.2 kJ (4) X(s) + O _{2}(g) ---> XO_{2}(s)ΔH = -810.1 kJ (5) 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 (eq 1 and eq 5), as will H_{2} (1 and 3), O_{2} (1 and 4), Cl_{2} (2 and 3), and X(s) (2 and 4). Add up the five modified enthalpies for the final answer of -815.4 kJ.