Using four or more equations and their enthalpies

Go to Hess' Law - using four or more equations and their enthalpies - Problems 11 - 20

Go to Hess' Law - using two equations and their enthalpies

Go to Hess' Law - using three equations and their enthalpies

Go to Hess' Law - using standard enthalpies of formation

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**Problem #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

**Problem #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

Comments:

(1) This is not the enthalpy of formation for AlCl_{3}(s). Remember that an enthalpy of formation equation is always for **ONE** mole of the target substance. In other words, this:

Al(s) + (3/2)Cl _{2}(g) ---> AlCl_{3}(s)ΔH° _{f}= -703.4 kJ

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

(2) 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.

**Problem #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

**Problem #4:** Calculate the ΔH in kilojoules for the following reaction, the preparation of nitrous acid HNO_{2}:

HCl(g) + NaNO_{2}(s) ---> HNO_{2}(l) + NaCl(s)

Use the following thermochemical equations:

2NaCl(s) + H _{2}O(l) ---> 2HCl(g) + Na_{2}O(s)ΔH = +507.31 kJ NO(g) + NO _{2}(g) + Na_{2}O(s) ---> 2NaNO_{2}(s)ΔH = -427.14 kJ NO(g) + NO _{2}(g) ---> N_{2}O(g) + O_{2}(g)ΔH = -42.68 kJ 2HNO _{2}(l) ---> N_{2}O(g) + O_{2}(g) + H_{2}O(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 NaNO

_{2}on the left-hand side)eq 3 ⇒ leave untouched

eq 4 ⇒ flip it (this puts HNO

_{2}on the right-hand side)

2) Rewrite all four equations with the above changes:

2HCl(g) + Na _{2}O(s) ---> 2NaCl(s) + H_{2}O(l)ΔH = -507.31 kJ 2NaNO _{2}(s) ---> NO(g) + NO_{2}(g) + Na_{2}O(s)ΔH = +427.14 kJ NO(g) + NO _{2}(g) ---> N_{2}O(g) + O_{2}(g)ΔH = -42.68 kJ N _{2}O(g) + O_{2}(g) + H_{2}O(l) ---> 2HNO_{2}(l)ΔH = -34.35 kJ

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

Na_{2}O(s) (eq 1 & 3); H_{2}O(l) (eq 1 & 4); NO(g) (eq 2 & 3); NO_{2}(g) (eq 2 & 3); N_{2}O(g) (eq 3 & 4); O_{2}(g) (eq 3 & 4)

3) Add the four reactions to get this:

2HCl(g) + 2NaNO _{2}(s) ---> 2HNO_{2}(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) + NaNO _{2}(s) ---> HNO_{2}(l) + NaCl(s)ΔH = -78.6 kJ

**Problem #5:** Determine the heat of reaction (in kJ) at 298 K for the reaction:

N_{2}H_{4}(l) + O_{2}(g) ---> N_{2}(g) + 2H_{2}O(l)

given the following equations and ΔH values:

2NH _{3}(g) + 3N_{2}O(g) ---> 4N_{2}(g) + 3H_{2}O(l)ΔH = -1013 kJ N _{2}O(g) + 3H_{2}(g) ---> N_{2}H_{4}(l) + H_{2}O(l)ΔH = -317 kJ 2NH _{3}(g) + 1/2O_{2}(g) ----> N_{2}H_{4}(l) + H_{2}O(l)ΔH = -142.9 kJ H _{2}(g) + 1/2O_{2}(g) ---> H_{2}O(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 N_{2}on the product side, which is where we need it. The 4 in front of the N_{2}is going to play a role. Suppose I divided through by 4 to get the one N_{2}in the final answer. That means I would wind up with 3/4 in front of the N_{2}O and also in front of the H_{2}O. Way too complicated. Keeping the 4 in front of the N_{2}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 N

_{2}O can be on the product side (to cancel with the 3N_{2}O in the first equation). I also have to multiply this reaction by 3, to give me my 3N_{2}O for cancelling purposes.c) This reaction needs to be flipped too. I must have the 2NH

_{3}be on the product side to cancel with the 2NH_{3}in equation 1.d) Notice that I have two equations (#2 and #3) that have N

_{2}H_{4}. When I add everything up, I'll have 4N_{2}H_{4}. 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 4O

_{2}(remember I will divide by 4 at the end), so I used 9 since I knew that would make 9/2O_{2}and the 1/2O_{2}in equation 3 would cancel, giving me 8/2O_{2}which is 4O_{2}.)f) I will not discuss the H

_{2}O, so that you may ponder how it works out.

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

2NH _{3}(g) + 3N_{2}O(g) ---> 4N_{2}(g) + 3H_{2}O(l)ΔH = -1013 kJ 3N _{2}H_{4}(l) + 3H_{2}O(l) ---> 3N_{2}O(g) + 9H_{2}(g)ΔH = +951 kJ N _{2}H_{4}(l) + H_{2}O(l)---> 2NH_{3}(g) + 1/2O_{2}(g)ΔH = +142.9 kJ 9H _{2}(g) + 9/2O_{2}(g) ---> 9H_{2}O(l)ΔH = -2572.2 kJ

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

4N_{2}H_{4}(l) + 4O_{2}(g) ---> 4N_{2}(g) + 8H_{2}O(l)The 2NH

_{3}, the 3N_{2}O and the 9H_{2}cancel completely.1/2O

_{2}cancels and four of the 12 H_{2}O 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 kJb) divide by 4 for the final answer: -623 kJ (to three sig figs)

**Problem #6:** Calculate the standard enthalpy change for H_{2}O(g) ---> H_{2}O(l) at 298.15 K, given the following reaction enthalpies:

CH _{4}(g)+ 2O_{2}(g) ---> CO_{2}(g) + 2H_{2}O(g)ΔH = -802.30 kJ C(s, graphite) + O _{2}(g) ---> CO_{2}(g)ΔH = -393.51 kJ C(s, graphite) + 2H _{2}(g) ---> CH_{4}(g)ΔH = -74.81 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(l)ΔH = -571.66 kJ

**Solution:**

1) Do these:

a) flip first equation (puts H_{2}O(g) on the reactant side)

b) leave second equation alone (need to cancel the CO_{2}from the first equation)

c) flip third equation (to cancel the CH_{4}in the first equation)

d) leave fourth equation alone (H_{2}O(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:

CO _{2}(g) + 2H_{2}O(g) ---> CH_{4}(g)+ 2O_{2}(g)ΔH = +802.30 kJ C(s, graphite) + O _{2}(g) ---> CO_{2}(g)ΔH = -393.51 kJ CH _{4}(g) ---> C(s, graphite) + 2H_{2}(g)ΔH = +74.81 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(l)ΔH = -571.66 kJ

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

2H_{2}O(g) ---> 2H_{2}O(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

H _{2}O(g) ---> H_{2}O(l)ΔH = -44.03 kJ

**Problem #7:** Using these reactions, where M represents a generic metal:

2M(s) + 6HCl(aq) ---> 2MCl _{3}(aq) + 3H_{2}(g)ΔH = -881 kJ HCl(g) ---> HCl(aq) ΔH = -74.8 kJ H _{2}(g) + Cl_{2}(g) ---> 2HCl(g)ΔH = -1845.0 kJ MCl _{3}(s) ---> MCl_{3}(aq)ΔH = -316 kJ

Determine the enthalpy of:

2M(s) + 3Cl_{2}(g) ---> 2MCl_{3}(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 H_{2}in eq 1, to cancel 6HCl(g), to get 3Cl_{2}in the final answer)

4) flip and mult by 2 (put 2MCl_{3}(s) on the product side)

2) Result:

2M(s) + 6HCl(aq) ---> 2MCl _{3}(aq) + 3H_{2}(g)ΔH = -881 kJ 6HCl(g) ---> 6HCl(aq) ΔH = -448.8 kJ 3H _{2}(g) + 3Cl_{2}(g) ---> 6HCl(g)ΔH = -5535 kJ 2MCl _{3}(aq) ---> 2MCl_{3}(s)ΔH = +632 kJ

3) Add the four enthalpies for the answer:

2M(s) + 3Cl _{2}(g) ---> 2MCl_{3}(s)ΔH = -6232.8 kJ

The following three problems use more than four data equations.

**Problem #8:** Determine ΔH for the reaction:

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

given the following data:

(a) C (s, gr) + (1/2)O _{2}(g) ---> CO (g)ΔH = -110.5 kJ (b) CO + (1/2)O _{2}---> CO_{2}ΔH = -282.9 kJ (c) H _{2}+ (1/2)O_{2}---> H_{2}O(l)ΔH = -285.8 kJ (d) C (s, gr) + 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, in early August 2011, 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 (g) ---> C (s, gr) + (1/2)O _{2}(g)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}O(l)ΔH = -1714.8 kJ (d) C (s, gr) + 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

**Problem #9:** 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).

**Problem #10:** From a consideration of the following reactions, calculate ΔH°_{f} for ethane, C_{2}H_{6}(g).

CH _{3}CHO(g) + 2 H_{2}(g) ---> C_{2}H_{6}(g) + H_{2}O(l)ΔH° = -204 kJ 2 H _{2}(g) + O_{2}(g) ---> 2 H_{2}O(g)ΔH° = -484 kJ 2 C _{2}H_{5}OH(l) + O_{2}(g) ---> 2 CH_{3}CHO(g) + 2 H_{2}O(l)ΔH° = -348 kJ H _{2}O(l) ---> H_{2}O(g)ΔH° = 44 kJ 2 C _{2}H_{5}OH(l) ---> 4 C(s) + 6 H_{2}(g) + O_{2}(g)ΔH° = 555 kJ

**Solution:**

1) The formation reaction for ethane is this:

2C(s) + 3H_{2}(g) ---> C_{2}H_{6}

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

a) equation 1 of the 5 has C_{2}H_{6}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 CH_{3}CHO on the product side, so it will cancel the CH_{3}CHO 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 C_{2}H_{5}OH to cancel the one in eq 3.

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

CH _{3}CHO(g) + 2 H_{2}(g) ---> C_{2}H_{6}(g) + H_{2}O(l)ΔH° = -204 kJ 2 H _{2}(g) + O_{2}(g) ---> 2 H_{2}O(g)ΔH° = -484 kJ C _{2}H_{5}OH(l) + (1/2)O_{2}(g) ---> CH_{3}CHO(g) + H_{2}O(l)ΔH° = -174 kJ H _{2}O(l) ---> H_{2}O(g)ΔH° = 44 kJ 2 C(s) + 3 H _{2}(g) + (1/2)O_{2}(g) ---> C_{2}H_{5}OH(l)ΔH° = -277.5 kJ

4) Flip equation 2 and multiply equation 4 by 2. This will cancel all of the H_{2}O(g) and H_{2}O(l):

CH _{3}CHO(g) + 2 H_{2}(g) ---> C_{2}H_{6}(g) + H_{2}O(l)ΔH° = -204 kJ 2 H _{2}O(g) ---> 2 H_{2}(g) + O_{2}(g)ΔH° = 484 kJ C _{2}H_{5}OH(l) + (1/2)O_{2}(g) ---> CH_{3}CHO(g) + H_{2}O(l)ΔH° = -174 kJ 2 H _{2}O(l) ---> 2 H_{2}O(g)ΔH° = 88 kJ 2 C(s) + 3 H _{2}(g) + (1/2)O_{2}(g) ---> C_{2}H_{5}OH(l)ΔH° = -277.5 kJ There are 5H

_{2}on the reactant side and 2H_{2}on the product side, leaving 3H_{2}on the reactant side, which is what we want. The one O_{2}on each side will cancel.

5) Adding up the five enthalpies gives

ΔH° = -83.5 kJ

**Bonus Problem:**

**Solution:**

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

Reaction Enthalpy Value A ---> B ΔH _{AB}B ---> C ΔH _{BC}A ---> E ΔH _{AE}E ---> D ΔH _{ED}

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:

Reaction Enthalpy Value B ---> A -ΔH _{AB}C ---> B -ΔH _{BC}A ---> E ΔH _{AE}E ---> D ΔH _{ED}

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

4) ΔH_{CD} is arrived at by adding the four data enthalpies:

(-ΔH_{AB}) + (-ΔH_{BC}) + ΔH_{AE}+ ΔH_{ED}

Go to Hess' Law - using four or more equations and their enthalpies - Problems 11 - 20

Go to Hess' Law - using two equations and their enthalpies

Go to Hess' Law - using three equations and their enthalpies

Go to Hess' Law - using standard enthalpies of formation