Using two equations and their enthalpies

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Germain Henri Hess, in 1840, discovered a very useful principle which is named for him:

The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps.

Another way to state Hess' Law is:

If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations.

**An example of how Hess' Law is used**

What is the enthalpy of the following reaction?

C (s, graphite) ---> C (s, diamond) ΔH° = ??? kJ

By the way, notice the presence of the degree sign, °, on the enthalpy. This indicates that the reaction is happening under standard conditions. All the examples in this tutorial will be carried out under standard conditions.

In the common chemistry laboratory, this reaction cannot be examined directly. This is because, regardless of the low enthalpy, the reaction requires a very, very high activation energy to get the reaction started and, in this case, it means both high temperature and high pressure. The consequence is that the enthalpy value cannot be determined directly in almost all labs and, in the ones that can, the process is very, very difficult.

However, Hess' Law offers a way out. If we had two (or more) reactions that could be added together, then we can add the respective enthalpies of the reactions to get what we want. Here are the two reactions we need:

C (s, graphite) + O _{2}(g) ---> CO_{2}(g)ΔH° = -393.5 kJ C (s, diamond) + O _{2}(g) ---> CO_{2}(g)ΔH° = -395.4 kJ

What I am going to do is reverse the bottom equation. This will put the C (s, diamond) on the product side, where we need it. Also, when I add the two equations together, the oxygen and carbon dioxide will cancel out. This is, of course, what we want since those two substances are not in the final, desired equation. Here are the two equations again, with the second one reversed:

C (s, graphite) + O _{2}(g) ---> CO_{2}(g)ΔH° = -393.5 kJ CO _{2}(g) ---> C (s, diamond) + O_{2}(g)ΔH° = +395.4 kJ

I want you to notice the other change I made. Look at the enthalpy for the second equation, the one I reversed. Notice how the sign has changed also. This is an absolute requirement of using Hess' Law: reversing an equation means reversing the sign on the enthalpy value.

The reason? The first, unreversed equation is exothermic. We know this from the negative in front of the 395.4. That means that the opposite, reverse equation is endothermic. Putting in enthalpy (endothermic) is the reverse, the opposite of exothermic (giving off enthalpy). Hence, we change the sign EVERY time we reverse an equation.

Now I'm ready to add the equations together. When I do this I also add the enthalpies together. Here is the added equation without anything taken out:

CO _{2}(g) + C (s, graphite) + O_{2}(g) ---> CO_{2}(g) + C (s, diamond) + O_{2}(g)ΔH° = (-393.5 kJ) + (+395.4 kJ)

Notice the items which are the same on both sides and remove them:

C (s, graphite) ---> C (s, diamond) ΔH° = +1.9 kJ

We now have the answer we desire by using the indirect means of Hess' Law and two relatively easy experiments. Thus we avoid performing a tricky, expensive, possibly dangerous experiment. However, due to Hess' discovery, we know that our indirectly obtained answer is just as valid as if we had done the experiment directly.

Please go here for an answer to the above problem done on Yahoo Answers in 2006. The problem is asked using the term 'standard enthalpy of formation.' If you are not sure wat that term means, please make sure to find out. It will probably be tested! If you are not sure what it means, it won't affect your aility to understand the explanation given at the link.

**Problem #1:** Use the following data to determine the enthalpy (ΔH°) of reaction for:

NO _{2}(g) + (7/2) H_{2}(g) ---> 2H_{2}O(l) + NH_{3}(g)ΔH° = ??? kJ

Using the following two equations:

2NH _{3}(g) ---> N_{2}(g) + 3H_{2}(g)ΔH° = +92 kJ (1/2) N _{2}(g) + 2H_{2}O(l) ---> NO_{2}(g) + 2H_{2}(g)ΔH° = +170 kJ

**Solution:**

1) Notice how there is only one NH_{3} in the target equation and it's on the right-hand side. That means we have to flip our first equation and divide it by two. Like this:

(1/2)N _{2}(g) + (3/2)H_{2}(g) ---> NH_{3}(g)ΔH° = -46 kJ Notice that the sign changed on the ΔH and its numerical value was cut in half.

2) The target equation has one NO_{2} and it's on the left-hand side, so we need to flip the second equation. Like this:

NO _{2}(g) + 2H_{2}(g) ---> (1/2) N_{2}(g) + 2H_{2}O(l)ΔH° = -170 kJ Notice that the sign changed on the ΔH.

3) The ΔH for the target equation is:

-46 + -170 = -216 kJ

**Problem #2:** Calculate the enthapy for the following reaction:

N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH° = ??? kJ

Using the following two equations:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH° = +180 kJ 2NO _{2}(g) ---> 2NO(g) + O_{2}(g)ΔH° = +112 kJ

**Solution:**

In order to solve this, we must reverse at least one equation and it turns out that the second one will require reversal. Here are both with the reversal to the second:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH° = +180 kJ 2NO(g) + O _{2}(g) ---> 2NO_{2}(g)ΔH° = -112 kJ

Notice that I have also changed the sign on the enthalpy from positive to negative.

Next, we add the two equations together and eliminate identical items. We also add the two enthalpies together.

N _{2}(g) + 2O_{2}(g) ---> 2NO_{2}(g)ΔH° = +68 kJ

The answer has been obtained: +68 kJ/mol.

**Problem #3:** Given the following data:

2NO(g) ---> N _{2}(g) + O_{2}(g)ΔH = -180.6 kJ N _{2}(g) + O_{2}(g) + Cl_{2}(g) ---> 2NOCl(g)ΔH = +103.4 kJ

Find the ΔH of the following reaction:

2NOCl(g) ---> 2NO(g) + Cl_{2}(g)

**Solution:**

1) Flip first reaction, flip second reaction:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH = +180.6 kJ 2NOCl(g) ---> N _{2}(g) + O_{2}(g) + Cl_{2}(g)ΔH = -103.4 kJ

2) Add the equations and the ΔH values:

+180.6 + (-103.4) = +77.2

2NOCl(g) ---> 2NO(g) + Cl _{2}(g)ΔH = +77.2 kJ

Comment: you could also just add up the two reactions without flipping them, then flip the answer (remembering to change the sign on the ΔH when you do so).

**Problem #4:** The compound carbon suboxide, C_{3}O_{2}, is a gas at room temperature. Use the data supplied to calculate the heat of formation of carbon suboxide.

2CO(g) + C(s) ---> C _{3}O_{2}(g)ΔH° = +127.3 kJ CO(g) ΔH° _{f}= -110.5 kJ

**Solution:**

1) The target equation is this:

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

2) Write the formation reaction for CO:

C(s) + (1/2)O _{2}(g) ---> CO(g)ΔH° _{f}= -110.5 kJthen multiply it by two (in order to cancel 2CO when we add the equations):

2C(s) + O _{2}(g) ---> 2CO(g)ΔH = -221.0 kJ

3) Write the two equations to be added:

2CO(g) + C(s) ---> C _{3}O_{2}(g)ΔH° = +127.3 kJ 2C(s) + O _{2}(g) ---> 2CO(g)ΔH = -221.0 kJ

4) The two equations above need only be added to obtain the desired answer:

+127.3 kJ + (-221.0 kJ) = -93.7 kJ

3C(s) + O _{2}(g) ---> C_{3}O_{2}(g)ΔH° _{f}= -93.7 kJ

Comment: notice that no chemical equation was given in the problem. That is because of the subscripted 'f' on the enthalpy. That indicates the enthalpy is a formation enthalpy and, as such, already has a chemical equation in the definition of the enthalpy of formation.

**Problem #5:** During discharge of a lead-acid storage battery, the following chemical reaction takes place:

Pb + PbO_{2}+ 2H_{2}SO_{4}---> 2PbSO_{4}+ 2H_{2}O

Using the following two reactions:

(1) Pb + PbO _{2}+ 2SO_{3}---> 2PbSO_{4}ΔH° = -775 kJ (2) SO _{3}+ H_{2}O ---> H_{2}SO_{4}ΔH° = -113 kJ

Determine the enthalpy of reaction for the discharge reaction above.

**Solution:**

1) Multiply chemical equation (2) by 2:

2SO_{3}+ 2H_{2}O ---> 2H_{2}SO_{4}ΔH = -226 kJ

2) Switch the reactants and products in chemical reaction (2). Because of that, the sign of the change in enthalpy becomes positive. Let's number the following chemical equation as (3):

(3) 2H_{2}SO_{4}---> 2SO_{3}+ 2H_{2}O ΔH° = 226 kJ

3) Add chemical equations (1) and (3):

Pb + PbO_{2}+ 2SO_{3}+ 2H_{2}SO_{4}---> 2PbSO_{4}+ 2SO_{3}+ 2H_{2}O

4) Then, add the enthalpy changes of equations (1) and (3):

Pb + PbO_{2}+ 2H_{2}SO_{4}---> 2PbSO_{4}+ 2H_{2}O ΔH° = -549 kJ

Remember to cancel out 2SO_{3} because it appears on both the reactant and product side, leaving you with the desired chemical reaction.

**Problem #6:** Consider the following reaction:

N _{2}H_{4}(l ) + O_{2}(g) ---> N_{2}(g) + 2 H_{2}O(l )ΔH = -622.2 kJ

Given the following data, calculate the heat of reaction for the same reaction where water is a gaseous product instead of a liquid:

ΔH_{f}for H_{2}O(g) = -285.83 kJ/mol

ΔH_{f}for H_{2}O(l) = -241.83 kJ/mol

**Solution:**

1) We need to find ΔH for this reaction:

H_{2}O(l) ---> H_{2}O(g)ΔH = -285.83 - (-241.83) = -44

2) Now, we use the two reactions we have:

N _{2}H_{4}(l) + O_{2}(g) ---> N_{2}(g) + 2H_{2}O(l)ΔH = -622.2 kJ H _{2}O(l) ---> H_{2}O(g)ΔH = -44 kJ

3) Multiply second equation by 2:

N _{2}H_{4}(l) + O_{2}(g) ---> N_{2}(g) + 2H_{2}O(l)ΔH = -622.2 kJ 2H _{2}O(l) ---> 2H_{2}O(g)ΔH = -88 kJ

4) Add the two reactions (and eliminate 2H_{2}O(l)) and add the enthalpies to get the final answer:

ΔH = -622.2 kJ + -88 kJ = -710.2 kJ

**Problem #7:** When one mole of sulfur burns to form SO_{2}, 1300 calories are released. When one mole of sulfur burns to form SO_{3}, 3600 calories are released. What is the ΔH when one mole of SO_{2} is burned to form SO_{3}?

**Solution:**

1) Let's write out our information in a chemical way:

S + O _{2}---> SO_{2}ΔH = -1.3 kcal S + (3/2)O _{2}---> SO_{3}ΔH = -3.6 kcal

2) The target equation we want is this:

SO_{2}+ (1/2)O_{2}---> SO_{3}

3) To obtain the target equation, what I will do is flip the first equation, remembering to also change the sign on the enthalpy. Here are the equations with the first one flipped:

SO _{2}---> S + O_{2}ΔH = +1.3 kcal S + (3/2)O _{2}---> SO_{3}ΔH = -3.6 kcal

4) When those two equations are added, the S and an O_{2} will cancel out:

SO_{2}+ (1/2)O_{2}---> SO_{3}

5) We add the enthalpies:

+1.3 plus -3.6 = -2.3 kcal or -2300 cal

Comment: note that calories were used rather than Joules. This in no way affects the solution technique.

**Problem #8:** Calculate the enthalpy of formation for sulfur dioxide, SO_{2}. Use the following information:

S(s) + (3/2)O _{2}(g) ----> SO_{3}(g)ΔH = -395.8 kJ 2SO _{2}(g) + O_{2}(g) ---> 2SO_{3}(g)ΔH = -198.2 kJ

**Solution:**

1) This is the formation reaction for SO_{2}:

S(s) + O_{2}(g) ---> SO_{2}(g)

2) What should be done to the two data equations:

eq 1 ---> nothing

eq 2 ---> reverse it and divide by 2 (this puts one SO_{2}on the product side, which is where we want it)

3) The result:

S(s) + (3/2)O _{2}(g) ---> SO_{3}(g)ΔH = -395.8 kJ SO _{3}(g) ---> SO_{2}(g) + (1/2)O_{2}(g)ΔH = +99.1 kJ

4) The answer:

When the two equations are added, the SO_{3}and (1/2)O_{2}will cancel out, leaving the desired equation. Add the two enthalpies for the answer:

S(s) + O _{2}(g) ---> SO_{2}(g)ΔH = -296.7 kJ

5) Comment: try the exact same problem as above, but with slightly different data equations:

2SO _{3}(g) ---> 2S(s) + 3O_{2}(g)ΔH = +791.6 kJ 2SO _{2}(g) + O_{2}(g) ---> 2SO_{3}(g)ΔH = -198.2 kJ

Your assignment is to get to the answer, the -296.7 kJ.

**Problem #9:** Given the following:

I _{2}(g) + 3Cl_{2}(g) ---> 2ICl_{3}ΔH = -214 kJ I _{2}(s) ----> I_{2}(g)ΔH = +38 kJ

Calculate the enthalpy of formation for ICl_{3}.

**Solution #1:**

1) The target equation is this:

(1/2)I_{2}(s) + (3/2)Cl_{2}(g) ---> ICl_{3}(s)

2) What needs to be done to the two data equations

eq 1 ---> divide by 2

eq 2 ---> divide by 2

3) The result:

(1/2)I _{2}(g) + (3/2)Cl_{2}(g) ---> ICl_{3}ΔH = -107 kJ (1/2)I _{2}(s) ----> (1/2)I_{2}(g)ΔH = +19 kJ When you add the two equations just above, the (1/2)I

_{2}(g) will cancel out, leaving only the desired target equation.The enthalpy of the target equation is this:

-107 + 19 = -88 kJ

**Solution #2:**

Keeping mind that the enthalpy of formation is always for 1 mole of the product and in standard states, can simply add both the data equation's enthalpies:-214 +38 = -176Since this is for 2 moles of ICl

_{3}, diving by 2 yields the answer of -88 kJ.

**Problem #10:** Given the follow two reactions:

X _{2}+ 5 Y_{2}---> 2 XY_{5}ΔH _{1}3 X _{2}+ Z_{2}---> 2 X_{3}ZΔH _{2}

Calculate ΔH_{rxn} for the following reaction:

15 Y_{2}+ 2 X_{3}Z ---> 6 XY_{5}+ Z_{2}

**Solution:**

Comment: A completely made-up problem? Doesn't matter. The technique for solving is the same.

1) Changes to be applied:

eq 1 ---> multiply by 3 in order to get 15Y_{2}

eq 2 ---> flip in order to get X_{3}Z onto the reactant side

2) The result:

3 X _{2}+ 15 Y_{2}---> 6 XY_{5}3ΔH _{1}2 X _{3}Z ---> 3 X_{2}+ Z_{2}-ΔH _{2}

3) Solve for the enthalpy of the desired reaction:

When the two equations above are added, the 3X_{2}will cancel out and the enthalpy is this:ΔH

_{rxn}= 3ΔH_{1}+ (-ΔH_{2})

**Bonus Problem:** Given the following two reactions at 298 K and 1 atm pressure:

N _{2}(g) + O_{2}(g) ---> 2NO(g)ΔH _{1}NO(g) + (1/2)O _{2}(g) ---> NO_{2}(g)ΔH _{2}

which of the statements below correctly describes the ΔH_{f}° for NO_{2}(g)?

(A) ΔH_{1}

(B) ΔH_{2}

(C) ΔH_{2}+ (1/2)ΔH_{1}

(D) ΔH_{2}- (1/2)ΔH_{1}

(E) ΔH_{2}+ ΔH_{1}

**Solution:**

1) The target equation we want is the formation reaction for NO_{2}(g):

(1/2)N_{2}(g) + O_{2}(g) ---> NO_{2}(g)

2) We get that by doing this:

(1/2)N _{2}(g) + (1/2)O_{2}(g) ---> NO(g)(1/2)ΔH _{1}NO(g) + (1/2)O _{2}(g) ---> NO_{2}(g)ΔH _{2}

3) Adding the two above equations yields the target equation. When we add the equations together, we also add the enthalpies together:

(1/2)ΔH_{1}+ ΔH_{2}answer choice C

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