Solving Hess’ LawBy Dr. Robert D. Craig, Ph.D.

FINDING THE TARGET EQUATION

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• The value of the enthalpy change DH reported for a reaction is the amount of heat released or absorbed when reactants are converted to products at the same temperature and pressure and in the molar amounts represented by the coefficients

Pic of bomb calorimeter

P 233

• Step 1: arrange the given equations to get the reactants and products in the equation whose

• You wish to calculate on the correct sides of the equations

• (you may need to reverse some equations to do this!)

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Step 2: Get the correct formula amounts of the substances on each side.

(some adjustments might need! Like indicating (g) or (l)

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• Step 3: Make sure other substances in the equations cancel when the equations are added.

• (you may cancel out equal amounts on either side of equations)

CHM 141

• Would like to do some problems!!!

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1. Calculate the DHo rxn for

Ca(s) + 1/2 O2 (g) + CO2(g) -> CaCO3 (s) • The target! You have to use these to form itCa(s) + 1/2O2 (g) + CO2 -> CaO(s) DHo =-635 kJ

CaCO3 (s) -> CaO(s) + CO2 (g) DHo =+ 178.3 kJ

done

1. Calculate the DHo rxn for

2NOCl(g) -> N2 (g) + O2 (g) + Cl2 (g) • The target!1/2N2 (g) + 1/2O2 (g) -> NO(g) DHo =+90.3 kJ

NO(g) + 1/2Cl2 (g) -> NOCl(g) DHo =- 38.6 kJ

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• Need to eliminate NO species• Need 1 mol of N2 an O2 in products

• Need 1mol of Cl2 in products

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Equation 1Multiply by 2! And flip ! Then change + to –1/2N2 (g) + 1/2O2 (g) -> NO(g) DHo =+90.3 kJ

Now is 2NO(g) -> N2 (g) + O2 (g) DHo = (-)180.6 kJ

.Equation 2Need to start with the product so (Flip!)Need to multiply by 2 to address the target

concentration of 2NOClOriginalNO(g) + 1/2Cl2 (g) -> NOCl(g) DHo =- 38.6 kJ

Changed!NOCl(g) -> NO(g) + 1/2Cl2 (g) -> DHo =+38.6 kJ

2NOCl(g) -> 2NO(g) + Cl2 (g) -> DHo =+77.2 kJ

And rearranged equation and calculate the ethalpy

2NO(g) -> N2 (g) + O2 (g) DHo = (-)180.6 kJ

2NOCl(g) -> 2NO(g) + Cl2 (g) DHo =+77.2 kJ• Now add together and combine enthalpies!

Example 3

Calculate the enthalpy change for the reaction

P4O6 (s) + 2 O2 (g) P4O10 -Target

P4 (s) + 3O2 (s) P4O6 (s) DHo rxn = -1640.1

P4(s) + 5O2 (s) P4O10 (s) DHo rxn = -2940.1

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First:Just flip equation 1 and change enthalpyP4 (s) + 3O2 (s) P4O6 (s) DHo rxn = -1640.1

NowP4O6 (s) -> P4 (s) + 3O2 (s) DHo rxn = +1640.1

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Second combine (subtract from 2)• NowP4O6 (s) -> P4 (s) + 3O2 (s) DHo rxn = +1640.1

P4(s) + 5O2 (s) P4O10 (s) DHo rxn = -2940.1

-(P4O6 (s) -> P4 (s) + 3O2 (s) DHo rxn = +1640.1)

Target acquired!!!P4O6 (s) + 2 O2 (g) P4O10 -Target

What is enthalpy of target??

DHo rxn = -2940.1 +1640.1

= - 1300 kJ still exothermic

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• The standard enthalpies of formation of SO2 and SO3 are -297 and -396 kJ/mol respectively. Calculate the standard enthalpy of reaction for the reaction:

• SO2 + 1/2 O2 -> SO3.

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• Solution:In order to show how the chemical reactions take place, and for a better appreciation of the technique of problem solving, we write the equations according to the data given:

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From tabulated values

• which the heats of formation of all reactants and products are known:

solution

• SO2(g) -> S(s) + O2(g) DH = 297 kJS(s) + 3/2 O2 -> SO3 DH = -396 kJ

• Add the two equations to give • SO2(g) + 1/2 O2 -> SO3 DH = -99 kJ

Begin here

The enthalpy change can be measured by calorimetry for many, but not all , chemical

processes.

Consider, The example , the oxidation of carbon to form carbon monoxide.

C(graphite) + 1/2O2(g) -> CO(g)

C(graphite) + 1/2O2(g) -> CO(g)

• Hess's law can be applied to calculate enthalpies of reactions that are difficult to measure. In this example, it is very difficult to control the oxidation of graphite to give pure CO.

• However, enthalpy for the oxidation of graphite to CO2 can easily be measured. So can the enthalpy of oxidation of CO to CO2.

• The application of Hess's law enables us to estimate the enthalpy of formation of CO

skip this slide!!can use this!!

However, enthalpy for the oxidation of graphite to CO2 can easily be measured. So can the enthalpy of oxidation of CO to CO2. The application of Hess's law enables us to estimate the enthalpy of formation of CO.

Problem is never see this : C(graphite) + 1/2O2(g) -> CO(g)

• Even if a deficiency of oxygen is used, the primary product of the reaction of carbon and oxygen is CO .

• As soon as CO is formed, it reacts with O2 to form CO2, Because the reaction cannot be carried out in a way that allows CO to be the sole product, it is not possible to measure the change in enthalpy for this reaction by calorimetry.

C(graphite) + 1/2O2(g) -> CO(g)

• The enthalpy change for the reaction forming CO(g) from C(s) and 02 (g) will and can be determined indirectly, from enthalpy changes for other reactions that can be measured.

One use of Hess law!!!

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• The calculation is based on Hess's law, which states that if a reaction is the sum of two or more other re actions, D H o for the overall process is the sum of the D H o rxn values of those reactions.

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• The oxidation of C(s) to CO2(g ) can be viewed as occurring in two steps:

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First the oxidation of C(s) to CO (g) (Equation 1) , and then the oxidation of CO (g) to CO2 (g) (Equation 2) .

• Adding these two equations gives the equation for the oxidation of C( s) to CO, (g) (Equation 3) ,

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Eq 1: C(graphite) + 1/2O2(g) -> CO(g)

Eq 2: CO(g) + 1/2O2(g) -> CO2(g)• ._____________________________________Eq 3: C(s) + O2(g) -> CO2(g)

D H o1 = ?

D H o2 = - 283.0 kJjmol-rxn

D H o3= - 393 .5 kJjmol-rxn

P 233

• Hess's law tells us that the enthalpy change for the overall reaction (D H o

3) will equal the sum of the enthalpy changes for reactions I and 2 (D H o

1+ D H o2). Both D H o

1 and D H o2

can be measured, and these values are then used to calculate the enthalpy change for reaction 1.

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D H o3 = (D H o

1+ D H o2).

- 393.5 kJ/mol-rxn = (D H o1 -283.0 kJ/mol-rxn)

D H o1 = - 110.5 kJ/ mol-rxn

Energy level diagram 0 kJ ------------ C(graphite) + O2

| |-110 kJ | |• V | CO + 0.5 O2 ----- |

| | -393 kJ | | -283 kJ | | | | V V ------------ CO2

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There is another

• This formulation is based on steps• Can also use a bond energy table

For tabulated data

• Estimate the change in enthalpy, ΔH, for the following reaction:

• H2 (g) + Cl2 (g) → 2 HCl (g)

These problems are much easier

• Solution • To work this problem, think of the reaction in

terms of simple steps: • Step 1 The reactant molecules, H2 and Cl2,

break down into their atoms • H2(g) → 2 H(g)

Cl2(g) → 2 Cl(g)

Step 2 These atoms combine to form HCl molecules

• 2 H (g) + 2 Cl (g) → 2 HCl (g) • In the first step, the H-H and Cl-Cl bonds are

broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:

• ΔH1 = +(436 kJ + 243 kJ) = +679 kJ

Attempt to Target this !!!C(s) + 2H2 (g) -> CH4 (g)

The enthalpy of combustion D H ocombustion for

H2, C(graphite) and CH4 are -285.8, -393.5, and -890.4 kJ/mol respectively.

Calculate the standard enthalpy of formation D H o

f for CH4.

From this

(1) H2(g) + 1/2 O2(g) -> H2O(l) -285.8

(2) C(graphite) + O2(g) -> CO2(g) -293.5

(3) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) -890.4

Move 1

• Strategy:• The three reactions (1, 2, and 3), as they are

written , cannot be added together to obtain the equation for the formation of CH4 from its elements.

• Methane, CH4, is a production the reaction for which we wish to calculate D H o rxn but it is a reactant in Equation 3.

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• Water appears in two of these equations although it is not a component of the reaction forming CH 4 from carbon and hydrogen

Move 2

• To use Hess's law to solve this problem , we will first have to manipulate the equations and adjust the D H o values accordingly before adding equations together.

• Writing an equation in the reverse direction changes the sign of D H o and that doubling the amount of reactants and products doubles the value of D H o.

• Adjustments to Equations 2 and 3 will produce new equations that along with Equation 1

.asking DHf From the D H o rxn

(1) H2(g) + 1/2 O2(g) -> H2O(l) -285.8

(2) C(graphite) + O2(g) -> CO2(g) -293.5

(3) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) -890.4

From the above equations, deriveC(s) + 2H2 (g) -> CH4 (g)

Answer: C + 2H2 -> CH4 -74.7

Hint: 2*(1) + (2) - (3), Thus,DHf = 2 * (-285.8) + (-393.5) - (-890.4) = ?

construct an energy level diagram

===C(graphite) + 2 H2(g) + 2 O2(g)===• - 74.7 kJ | |• == CH4 (g) + 2 O2(g)== |• | |• | |• | |• | |-965.1 kJ• -890.4 kJ | | [(-2*285.8-393.5) kJ]• | |• | |• | |• | |• V V======CO2(g) + 2 H2O(l)==========

Let’s burn methane in air!!

• CH4 + 2 O2(g) -> CO2(g) + 2 H2O(g) dHo =???

Go to original handout From the following enthalpies of reactions• 2 O(g) -> O2(g) dHo = -249 kJ/mol

• H2O(l) -> H2O(g) dHo = 44 kJ/mol at 298 K

• 2 H(g) + O(g) -> H2O(g) dHo = -803 kJ/mol

• C(graphite) + 2 O(g) -> CO2(g) dHo = -643 kJ/mol

• C(graphite) + O2(g) -> CO2(g) dHo = -394 kJ/mol

• C(graphite) + 2 H2(g) -> CH4(g) dHo = -75 kJ/mol

• 2 H(g) -> H2(g) dHo = -436 kJ/mol

• H2O(l) -> H2O(g) dH = 41 kJ/mol at 373 K, non-standard condition

• Calculate the heat of combustion of methane into gaseous H2O.

What is the target?

• CH4 + 2 O2(g) -> CO2(g) + 2 H2O(g) dHo =???

• Would you like to try first???

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Solution:-2(1) + 2(3) + (4) - (6) - 2(7) gives

CH4(g) + 2 O2(g) -> CO2(g) + H2O(g), and therefore,dH = -2*(-249) + 2*(-803) + (-643) - (-75) - 2(-436) = -804 kJ/mol

.a possible start

-2(1) means 2 O(g) -> O2(g) dHo = -249 kJ/mol• Now 2 O2(g) -> 4 O(g) dHo = +498 kJ/mol

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2(3) equals =

2 H(g) + O(g) -> H2O(g) dHo = -803 kJ/mol

4H(g) + 2 O(g) -> 2 H2O(g) dHo = -1606kJ/mol

Then need to combine!!!

Discussion:

•Work out the details yourself and check the result. The calculation is rather complicated. Reading it will not be able to master the technique. Data from equations 2, 5 and 8 are not required.

• Often, you have to select suitable data from a Table of Standard Enthalpy of Formation in problem solving.

CRC handbook or cancelation

• One method is to re-write the key equations as follows and then add them to cancel out undesirable compound on both sides.

Practice the cancellation of the formula yourself.

2 O2(g) -> 4 O(g) dHo = 498 kJ/mol4 H(g) + 2 O(g) -> 2 H2O(g) dHo = -1606 kJ/mol2 H2(g) -> 4 H(g) dHo = 872 kJ/mol

CH4(g) -> C(graphite) + 2 H2(g) dHo = 75 kJ/mol

C(graphite) + 2 O(g) -> CO2(g) dHo = -643 kJ/mol add all equations ---------------------------add all dHs

CH4 + 2 O2(g) -> CO2(g) + 2 H2O(g) dHo = -804 kJ/mol

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• Skill:Apply Hess's law to calculate standard enthalpy of reaction from standard enthalpies of formation. There are more than one way to solve the problem

Target: 2 H2(g) + O2(g) -> 2 H2O(g) dHo = ?

• From the following enthalpies of reactions: • 2 O(g) -> O2(g) dHo = -249 kJ/mol

• 2 H(g) -> H2(g) dHo = -436 kJ/mol

• H2O(l) -> H2O(g) dH = 41 kJ/mol at 373 K, non-standard condition

• H2O(l) -> H2O(g) dHo = 44 kJ/mol at 298 K

• 2 H(g) + O(g) -> H2O(g) dHo = -803 kJ/mol

• C(graphite) + 2 O(g) -> CO2(g) dHo = -643 kJ/mol

• C(graphite) + O2(g) -> CO2(g) dHo = -394 kJ/mol

• C(graphite) + 2 H2(g) -> CH4(g) dHo = -75 kJ/mol

2 H2(g) + O2(g) -> 2 H2O(g) dHo = ?

• Skill:From the result in the previous problem, only one step is required to solve this problem

2 H2(g) + O2(g) -> 2 H2O(g) dHo = ?

• 2 H(g) + O(g) -> H2O(g) dHo = -803 kJ/mol

• 2[2 H(g) + O(g) -> H2O(g) dHo = -803 kJ/mol

• 4H(g) + 2O(g) -> 2H2O(g) -1603

• (-1) 2 O(g) -> O2(g) dHo = -249 kJ/mol

• (-2) 2 H(g) -> H2(g) dHo = -436 kJ/mol

• Adjustments to Equations 2 and 3 will produce newequations that alongwith Equation 1,

• can be combined to give the desired net reaction.• SoLution To have CH4 appear as a product in the overall reaction,

we reverse Equation 3, which changes• the sign of .1.,W.• Equation 3': CO,(g) + 2 H,O(f) --> CH,(g) + 20,(g)• j" H]' = - j" H'l = + 890.3 kJjmo l-rxn• Next, wesee that 2 mol of H2( g) is on the reactant side in our

desired equation. Equation 2 is written for• only 1 molof H2(g) as a reactant. Therefore we multiply the

stoichiometric coefficients in Equation 2 by 2

• and multiply the value of j" H'" by 2.• Equation 2': 2 H,(g) + O, (g) --> 2 H,O(f)• ~ ,H r = 2 j, rH't = 2 ( - 285.8 kJjmol-rxn) = - 571.6

kJjmol-rxn• We now have three equations that when added

toget her, will give the targeted equation for the formation

• of methane from carbon and hydrogen. In this summation process, 02(g). H20(!'). and COAg) aU

• cancel.• Equation 1:• Equation 2':• Equation 3':• C(s) + O,(g ) --> CO,(g)• 2 H,(g) + O, (g) --> 2 H,O(f)• CO,(g) + 2 H,O(t) --> CH,(g) + 2 0, (g)• j" Hi = - 393.5 kf/rncl-rxn• j" H'r = 2 .1.,H'2 = - 571.6 kl/mcl-rxn• .1,H3' = - ~,Hl = + 890.3 kJjmol-rxn• Net Equation: C(s) + 2 H, (g) --> CH, (g) .l ,H:" ~ .l ,H; + 2 .l ,Hl + (- .l ,H;)• .l ,H:" ~ (-393 .5 kf/mcl-rxn) + (-571.6 kJ/mol-rxn) + (+890.3 kf/mcl-rxn)• = - 74.8 kJj mol-rxn• Thus, for the formatio n of 1 mol of CH 4(g) from the elements, we find j" H'" = - 74.8

kJjmol-rxn.