Using four or more equations and their enthalpies

Problems 1 - 10

**Problem #1:** 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

4) The ΔH value came from this:

(−507.31) + (+427.14) + (−42.68) + (−34.35)

5) Divide everything by two for the final answer:

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

**Problem #2:** 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}⁄_{2}O_{2}(g) ----> N_{2}H_{4}(ℓ) + H_{2}O(ℓ)ΔH = −142.9 kJ H _{2}(g) +^{1}⁄_{2}O_{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}⁄_{2}O_{2}and the^{1}⁄_{2}O_{2}in equation 3 would cancel, giving me^{8}⁄_{2}O_{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}⁄_{2}O_{2}(g)ΔH = +142.9 kJ 9H _{2}(g) +^{9}⁄_{2}O_{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}⁄_{2}O_{2}cancels and four of the 12H_{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 kJ(b) divide by 4 for the final answer: −623 kJ (to three sig figs)

**Problem #3:** 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 #4:** 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

**Problem #5:** From a consideration of the following reactions, calculate ΔH°_{f} for ethane, C_{2}H_{6}(g).

CH _{3}CHO(g) + 2H_{2}(g) ---> C_{2}H_{6}(g) + H_{2}O(ℓ)ΔH° = −204 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(g)ΔH° = −484 kJ 2C _{2}H_{5}OH(ℓ) + O_{2}(g) ---> 2CH_{3}CHO(g) + 2H_{2}O(ℓ)ΔH° = −348 kJ H _{2}O(ℓ) ---> H_{2}O(g)ΔH° = 44 kJ 2C _{2}H_{5}OH(ℓ) ---> 4C(s, gr) + 6H_{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) + 2H_{2}(g) ---> C_{2}H_{6}(g) + H_{2}O(ℓ)ΔH° = −204 kJ 2H _{2}(g) + O_{2}(g) ---> 2H_{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 2C(s, gr) + 3H _{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) + 2H_{2}(g) ---> C_{2}H_{6}(g) + H_{2}O(ℓ)ΔH° = −204 kJ 2H _{2}O(g) ---> 2H_{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 2H _{2}O(ℓ) ---> 2H_{2}O(g)ΔH° = 88 kJ 2C(s, gr) + 3H _{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

**Problem #6:** Find the heat of reaction for:

CO + 2H_{2}---> CH_{3}OH

using these data:

(I) 2C(s) + O _{2}(g) ---> 2CO(g)ΔH = −220.9 kJ (II) C(s) + O _{2}(g) ---> 2CO_{2}(g)ΔH = −391.1 kJ (III) 2CH _{3}OH(ℓ) + 3O_{2}(g) ---> 2CO_{2}(g) + 4H_{2}O(ℓ)ΔH = −1451.9 kJ (IV) 2H _{2}(g) + O_{2}(g) ---> 2H_{2}O(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) + O _{2}(g)ΔH = +220.9 kJ (II) 2C(s) + 2O _{2}(g) ---> 4CO_{2}(g)ΔH = −782.2 kJ (III) 2CO _{2}(g) + 4H_{2}O(ℓ) ---> 2CH_{3}OH(ℓ) + 3O_{2}(g)ΔH = +1451.9 kJ (IV) 4H _{2}(g) + O_{2}(g) ---> 4H_{2}O(ℓ)ΔH = −1142.4 kJ

3) Add the four equations:

2CO + 4H_{2}---> 2CH_{3}OH ΔH = −251.8 kJ

4) Divide the above equation by 2:

CO + 2H_{2}---> CH_{3}OH ΔH = −125.9 kJ

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

NGiven the following reactions and enthalpy changes_{2}O_{3}(g) + N_{2}O_{5}(s) ---> 2N_{2}O_{4}(g)

NO(g) + NO _{2}(g) ---> N_{2}O_{3}(g)ΔH = −39.8 kJ NO(g) + NO _{2}(g) + O_{2}(g) ---> N_{2}O_{5}(g)ΔH = −112.5 kJ 2NO _{2}(g) ---> N_{2}O_{4}(g)ΔH = −57.2 kJ 2NO(g) + O _{2}(g) ---> 2NO_{2}(g)ΔH = −114.2 kJ N _{2}O_{5}(s) ---> N_{2}O_{5}(g)ΔH = +54.4 kJ

**Solution:**

1) Manipulate the data equations as follows:

eq 1 - flip (puts N_{2}O_{3}as a reactant)

eq 2 - flip (puts N_{2}O_{5}(g) as a reactant, to cancel with eq 5)

eq 3 - multiply by 2 (need 2N_{2}O_{4}as a product)

eq 4 - untouched

eq 5 - untouched

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

4NO_{2}(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)

O_{2}(from eq 2 and 4)

N_{2}O_{5}(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:

Cl _{2}(g) + 5F_{2}(g) ---> 2ClF_{5}(g)ΔH° = −510.0 kJ ClF _{3}(g) + Cl_{2}(g) ---> 3ClF(g)ΔH° = −5.5 kJ 2NaCl(s) + F _{2}(g) ---> 2NaF(s) + Cl_{2}(g)ΔH° = −316.0 kJ NaCl(s) + F _{2}(g) ---> NaF(s) + ClF(g)ΔH° = −214.5 kJ

Please calculate the ΔH° for this reaction:

ClF_{3}(g) + F_{2}(g) ---> ClF_{5}(g)

**Solution:**

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

^{1}⁄_{2}Cl_{2}(g) +^{5}⁄_{2}F_{2}(g) ---> ClF_{5}(g)ΔH° = −255.0 kJ ClF _{3}(g) + Cl_{2}(g) ---> 3ClF(g)ΔH° = −5.5 kJ 2NaCl(s) + F _{2}(g) ---> 2NaF(s) + Cl_{2}(g)ΔH° = −316.0 kJ 3NaF(s) + 3ClF(g) ---> 3NaCl(s) + 3F _{2}(g)ΔH° = +643.5 kJ Modifying the first equation gives me one ClF

_{5}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 ^{3}⁄_{2}:

^{1}⁄_{2}Cl_{2}(g) + (5/2)F_{2}(g) ---> ClF_{5}(g)ΔH° = −255.0 kJ ClF _{3}(g) + Cl_{2}(g) ---> 3ClF(g)ΔH° = −5.5 kJ 3NaCl(s) + ^{3}⁄_{2}F_{2}(g) ---> 3NaF(s) +^{3}⁄_{2}Cl_{2}(g)ΔH° = −474.0 kJ 3NaF(s) + 3ClF(g) ---> 3NaCl(s) + 3F _{2}(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) +^{1}⁄_{8}S_{8}(s) + 2O_{2}(g) ---> ZnSO_{4}(s)

given the following data:

Zn(s) + ^{1}⁄_{8}S_{8}(s) ---> ZnS(s)ΔH = −183.92 kJ 2ZnS(s) + 3O _{2}(g) ---> 2ZnO(s) + 2SO_{2}(g)ΔH = −927.54 kJ 2SO _{2}(g) + O_{2}(g) ---> 2SO_{3}(g)ΔH = −196.04 kJ ZnO(s) + SO _{3}(g) ---> ZnSO_{4}(s)ΔH = −230.32 kJ

**Solution:**

1) The first data equation has Zn and S_{8} 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) + ^{3}⁄_{2}O_{2}(g) ---> ZnO(s) + SO_{2}(g)ΔH = −463.77 kJ

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

SO _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> SO_{3}(g)ΔH = −98.02 kJ

3) I still need to make the ZnO go away as well as the SO_{3}. Both of those are accomplished by the fourth data equation being left untouched. It also introduces the ZnSO_{4} in its desired place as a product.

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

Zn(s) + ^{1}⁄_{8}S_{8}(s) ---> ZnS(s)ΔH = −183.92 kJ ZnS(s) + ^{3}⁄_{2}O_{2}(g) ---> ZnO(s) + SO_{2}(g)ΔH = −463.77 kJ SO _{2}(g) +^{1}⁄_{2}O_{2}(g) ---> SO_{3}(g)ΔH = −98.02 kJ ZnO(s) + SO _{3}(g) ---> ZnSO_{4}(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) + ^{1}⁄_{8}S_{8}(s) + 2O_{2}(g) ---> ZnSO_{4}(s)ΔH = −976.03 kJ The 2O

_{2}came from the O_{2}in data equations 2 and 3.

**Problem #10:**

**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}