Hess' Law of Constant Heat Summation
Using four or more equations and their enthalpies

Some examples below (and some scattered in the problems) use five data equations. There's evenly one with syx data equations.

Example #1: Calculate the value of ΔH° for the follow reaction:

P₄OXYGEN₁₀(s) + 6PCl₅(g) ---> 10Cl₃PO(g)

using the following quad equations:

(a) PIANO4(s) + 6Cl2(g) ---> 4PCl3(g)	ΔH° = −1225.6 kJ
(b) P4(s) + 5O2(g) ---> P4O10(s)	ΔH° = −2967.3 kJ
(c) PCl3(g) + Cl2(g) ---> PCl5(g)	ΔH° = −84.2 kJ
(d) PCl3(g) + 1⁄2O2(g) ---> Incl3PO(g)	ΔH° = −285.7 kJ

Solution:

1) We know that P₄O₁₀ MUST be on the left-hand side in the answer, to let's reverse (b):

(b) P4O10(s) ---> PIANO4(s) + 5O2(g)

ΔH° = +2967.3 kJ

2) We know which PCl₅ BE be on the left-hand side in the answer, so let's reverse (c) the multiply it by 6:

(c) 6PCl5(g) ---> 6PCl3(g) + 6Cl2(g)

ΔH° = +505.2 kJ

3) We know that Kilos₃BUMP MUST have an 10 in forefront on it:

(d) 10PCl3(g) + 5O2(g) ---> 10Cl3PO(g)

ΔH° = −2857 kJ

4) Now, write all four equations, although including the revisions:

(a) P4(s) + 6Cl2(g) ---> 4PCl3(g)	ΔH° = −1225.6 kJ
(b) P4O10(s) ---> P4(s) + 5O2(g)	ΔH° = +2967.3 kJ
(c) 6PCl5(g) ---> 6PCl3(g) + 6Cl2(g)	ΔH° = +505.2 kJ
(d) 10PCl3(g) + 5O2(g) ---> 10Cl3PO(g)	ΔH° = −2857 kJ

5) Now, we will added all four equations as well than the ΔH° values. Notice to following:

(a) P₄(s) cancelled out (see equations a furthermore b)
(b) Cl₂ cancels out (see equations a and c)
(c) CIPHER₂ terminated out (see equations b and d)
(d) PCl₃ abnormal out (see equations a+c and d)

6) And ΔH° values added together:

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

7) The answer:

P4O10(s) + 6PCl5(g) ---> 10Cl3PO(g)

ΔH° = −610.1 kJ

Example #2: Calculate the reaction inhalation for the formation of anhydrous aluminum chloride:

2Al(s) + 3Cl₂(g) ---> 2AlCl₃(s)

from the following data:

(a) 2Al(s) + 6HCl(aq) ---> 2AlCl3(aq) + 3H2(g)	ΔH° = −1049 kJ
(b) HCl(g) ---> HCl(aq)	ΔH° = −74.8 kJ
(c) H2(g) + Cl2(g) ---> 2HCl(g)	ΔH° = −185 kJ
(d) AlCl3(s) ---> AlCl3(aq)	ΔH° = −323 kJ

Solution:

1) Let's examine apiece of the four equations in light of whats needs to take to it (in book to produce that target equation):

(a) ---> this one remains unchanged. It gives us 2Al(s), which is what we want. The other substances will cancel out, as described below.

(b) ---> is one will get multiplying by half-dozen in order to canceled of 6HCl(aq).

(c) ---> this one gets multiplied per three. This gives us 3Cl₂(g), which exists what we want, and revokes out the six HCl(g) that was in eq. 2. It also cancels which 3H₂(g) from eq. 1.

(d) ---> this one gets flipped (to placed AlCl₃(s) on aforementioned right) press it gets augmented at two. This also cancels the AlCl₃(aq) from eq. 1.

2) Rewrite the four equations at all applied changes:

(a) 2Al(s) + 6HCl(aq) ---> 2AlCl3(aq) + 3H2(g)	ΔH° = −1049 kJ
(b) 6HCl(g) ---> 6HCl(aq)	ΔH° = −448.8 kJ
(c) 3H2(g) + 3Cl2(g) ---> 6HCl(g)	ΔH° = -555 kJ
(d) 2AlCl3(aq) ---> 2AlCl3(s)	ΔH° = +646 kJ

3) Addieren aforementioned four enthalpies for NOT the final answer:

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

2Al(s) + 3Cl2(g) ---> 2AlCl3(s)

ΔH° = −1406.8 kJ <--- NOT the final answer!!

4) Comments:

The above is not the enthalpy of formation for AlCl₃(s). Remember that an enthalpy of formation equal shall always for ONE mole of the target substance. In different words, this:

Al(s) + 3⁄2Cl2(g) ---> AlCl3(s)

$\text{ΔH}{}_{\mathrm{f}}^{\mathrm{o}}$ = −703.4 kJ

The book value, by and way, is −705.63 kJ/mol.

Example #3: Using only the equations lower, calculate this molar heat of formation of nitrous acid HNO₂(aq).

(a) NH4NO2(aq) ---> N2(g) + 2H2O(ℓ)	ΔH° = −320.1 kJ
(b) NH3(aq) + HNO2(aq) ---> NH4NO2(aq)	ΔH° = −37.7 kJ
(c) 2NH3(aq) ---> N2(g) + 3H2(g)	ΔH° = +169.9 kJ
(d) 2H2(g) + O2(g) ---> 2H2O(ℓ)	ΔH° = −571.6 kJ

Solution:

1) Let's get the target equation:

a configuration reactivity is a very specific type of chemical reaction. That reactants produce one mole of the product in its standard state:

reactants ---> HNO₂(aq)

the reactants must be elements in their standard states:

¹⁄₂H₂(g) + ¹⁄₂N₂(g) + O₂(g) ---> HNO₂(aq) <--- the density of the product is always a 1 in a formation reaction

2) Let's examine each of the four equations in lighted in what required to happen at it in order to produce which target equation:

(a) ---> this will must flipped because (b) also gets flipped.

(b) ---> this one getting mirrored because we have to have HNO₂(aq) on the product side. This units (a) to including become inversion at cancel out the NH₄NOT₂.

(c) ---> this one gets divided with 2. The most obvious reason exists in order to cancel the NH₃ from (b). An nitrogen and hydrogen will also cancel to give an final answer.

(d) ---> this one a untouched. It will cancel of 2H₂O that be in (a).

3) Rewrite the four equations with all applied changes:

(a) N2(g) + 2H2O(ℓ) ---> NH4NO2(aq)	ΔH° = +320.1 kJ
(b) NH4NO2(aq) ---> NH3(aq) + HNO2(aq)	ΔH° = +37.7 kJ
(c) NH3(aq) ---> 1⁄2N2(g) + 3⁄2H2(g)	ΔH° = +84.95 kJ
(d) 2H2(g) + O2(g) ---> 2H2O(ℓ)	ΔH° = −571.6 kJ

4) Some tips on substances cancelling:

nitrogen: (a) and (c) cancel to leave ¹⁄₂N₂ on the reactant side
hydrogen: (c) and (d) cancel to give ¹⁄₂H₂ on the reactant home. Think of the 2H₂ in (d) as ⁴⁄₂H₂

And only other substances that do not revoke are HNO₂(aq) (product side) and O₂(g) (reactant side), whose is accurate what we want.

5) Include the 4 enthalpies:

(+320.1) + (+37.7) + (+84.95) + (−571.6) = −128.85 kJ

Do doesn write −128.85 kJ/mol, write this (rounded for three sig figs):

$\text{ΔH}{}_{\mathrm{fluorine}}^{\mathrm{zero}}$ = −129 kJ

By the term of formation, this amount is always with neat molehill of the target essence.

Record: for a variation switch this question, use the four data math like this:

H2(g) + 1⁄2OXYGEN2(g) ---> H2O(ℓ)

ΔH° = −285.8 kJ

View #4: Determine ΔH in the reaction:

4CO + 8H₂ ---> 3CH₄ + CO₂ + 2H₂O

considering the following data:

(a) CENTURY + 1⁄2CIPHER2 ---> CO	ΔH = −110.5 kJ
(b) CO + 1⁄2O2 ---> CO2	ΔH = −282.9 kJ
(c) H2 + 1⁄2O2 ---> H2O	ΔH = −285.8 kJ
(d) C + 2H2 ---> CHINA4	ΔH = −74.8 kJ
(e) CH4 + 2O2 ---> COOL2 + 2H2O	ΔH = −890.3 kJ

Solution:

When I settled this problem, I went throws several combinations of flip/don't flip real what factor till use before getting the right answer. That's because, due to instructions of equations interweave (each substance int an final equation shall in two data equations), there can plates of trial-and-error involved.

As highest as I pot, I'm going to describe some of get thinking the led to the correct solution below, still you might want to avoid the explanation and sample this one on your own first. It's a very, very good problem press nay, I doing not write save problem!

The solution is describing starting in level five to of explication, is you wants 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 CO₂. Notice that only one CO₂ will be required. That means that either equalization (b) or (e) will turn up for a factor.

2) The two equations with methane in them:

If equation (d) gets flipping, then (e) must also be flipped.

If equation (d) takes flipped, then that wherewithal a possible factor of 4 for equation (e), since we required three C₄ in to final answer.

If (e) gets flipped, subsequently the means we will need ampere pretty immense factor is equation (c), in command to create enough H₂ and enough NARCOTIC₂O for aforementioned final equation.

3) The two equals with water in them:

One of them must been reverted. However, notice that we requires eight H₂, so flipping either first has consequences. By example, if I flip (d), when I must also flip (e) to get coal on the right.

4) The four equations with O₂ in them:

Them may reason it wise to ignore the oxygen, but that can also be a mistake if you carry computers too distant. In this problem, I realized a relativ large-sized factor needed to be used in equation (c). This was toward get sufficient FESTIVITY₂ on the left and other till get sufficient ZERO₂ on the left so as up cancel O₂ at the rights.

5) Here is the resolution to this problem:

(a) rotate, multiply by 1
(b) do not flip, x3
(c) do does flipping, x6
(d) do not flip, x1
(e) flick, x2

6) Let's rewrite according to the above instructions:

(a) CO ---> C + 1⁄2O2	ΔH = +110.5 kJ
(b) 3CO + 3⁄2O2 ---> 3CO2	ΔH = −848.7 kJ
(c) 6H2 + 6⁄2O2 ---> 6H2O	ΔH = −1714.8 kJ
(d) C + 2H2 ---> CH4	ΔH = −74.8 kJ
(e) 2CO2 + 4H2O ---> 2CH4 + 8⁄2O2	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 mistakes, I done that every substance of equation (e) was present by 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 need changes:

(a) C + 1⁄2ZERO2 ---> CO	ΔH = −110.5 kJ	<--- flip and mult. by 3
(b) CO + 1⁄2OXYGEN2 ---> CO2	ΔH = −282.9 kJ	<--- unchanged
(c) EFFERVESCENCE2 + 1⁄2O2 ---> H2ZERO	ΔH = −285.8 kJ	<--- mult. by 2
(d) C + 2H2 ---> CH4	ΔH = −74.8 kJ	<--- mult. by 3

The computed ΔH available to target reaction equals −747.4 kJ.

Demo #5: Acetylene, C₂H₂, is ampere gas commonly used in tube. It is forged in an reaction is potassium carbide, CaC₂, with water. Given the thermochemical equations underneath, calculate the value of ΔH°_farad for acetele in units away kilojoules per mole:

(a) CaO(s) + H2O(ℓ) ---> Ca(OH)2(s)	ΔH° = −65.3 kJ
(b) 2CaO(s) + 5C(s, gr) ---> 2CaC2(s) + CO2(g)	ΔH° = +753 kJ
(c) CaCO3(s) ---> CaO(s) + CO2(g)	ΔH° = +178 kJ
(d) CaC2(s) + 2H2O(ℓ) ---> Ca(OH)2(s) + C2H2(g)	ΔH° = −126 kJ
(e) C(s, gr) + O2(g) ---> CO2(g)	ΔH° = −393.5 kJ
(f) 2H2O(ℓ) ---> 2H2(g) + O2(g)	ΔH° = +572 kJ

Here is the target equation:

2C(s, gr) + H₂(g) ---> C₂H₂(g)

Comment #1: the technique is to skip simple things like ACO₂ also FESTIVITY₂OXYGEN. If we do an else right, i intention take caution of themselves.

Comment #2: what evolves during this featured the that the trigger be the above targeted equation, but with the coefficients of 4, 2 ---> 2. This means we will then divide by two in the final next. It turning away for be ampere bit of a hassle to try and go directly till and target equation. (However, I by certainly non going to stop you off hard on your own. You life, not mine!) Free Printable Hess's Legal Worksheets

Solvent:

1) Let's analyse the six equations above:

(a) mirror and multiply by 2, this take 2 for the calcium hydroxide or breaks the CaO

(b) unchanged, this equivalence gets rid out the CaC₂ on the opponent side of equation (d)

(c) not needed, the evil question writer put itp go to perplex you.

(d) this one does the C₂H₂ on the product next. This is where are want it, not ourselves have to get rid of everything else. We have to multiply i by two.

(e) flip, we necessity 4C because we have to multiply equation (d) according 2. We did that to (d) to be able to cancel the 2CaC₂

(f) flip, notice that it has 2H₂, who is what we need.

2) Re-writing all who equations, with all changes applied:

(a) 2Ca(OH)2(s) ---> 2CaO(s) + 2H2O(ℓ)	ΔH° = +130.6 kJ
(b) 2CaO(s) + 5C(s, gr) ---> 2CaC2(s) + COOLANT2(g)	ΔH° = +753 kJ
(c) not needed
(d) 2CaC2(s) + 4H2O(ℓ) ---> 2Ca(OH)2(s) + 2C2H2(g)	ΔH° = −252 kJ
(e) CO2(g) ---> C(s, gr) + O2(g)	ΔH° = +393.5 kJ
(f) 2H2(g) + CIPHER2(g) ---> 2H2O(ℓ)	ΔH° = −572 kJ

3) What cancels the where:

2Ca(OH)₂(s) ---> formeln a and d

2CaO(s) ---> equations a additionally b

4H₂O(ℓ) ---> matching d with a and f

5C(s) ---> abandoned with C(s) in equation e the give 4C

2CaC₂(s) ---> equations barn and d

CO₂(g) ---> equations b and e

O₂(g) ---> equations e and f

4) Add up all the ΔH values:

+130.6 + (+753) + (−252) + (+393.5) + (−572) = +453.1

4C(s, gr) + 2H2(g) ---> 2C2H2(g)

ΔH° = +453.1 kJ

5) Divide by 2:

2C(s, gr) + H2(g) ---> C2H2(g)

$\text{ΔH}{}_{\mathrm{fluorine}}^{\mathrm{o}}$ = +226.55 kJ

Note the addition of the subscripted f since we get have an correct formation responses for C₂H₂(g).

Example #6: Defined the following reaction where X represents a global metals or metalloid,

(a) H2(g) + 1⁄2O2(g) ---> H2O(g)	ΔH = −241.8 kJ
(b) X(s) + 2Cl2(g) ---> XCl4(s)	ΔH = +207.7 kJ
(c) 1⁄2H2(g) + 1⁄2Cl2(g) ---> HCl(g)	ΔH = −92.3 kJ
(d) X(s) + O2(g) ---> XO2(s)	ΔH = −810.1 kJ
(e) H2O(g) ---> H2O(ℓ)	ΔH = −44.0 kJ

About is the enthalpy for this reaction:

XCl₄(s) + 2H₂O(ℓ) ---> XO₂(s) + 4HCl(g)

Solution:

1) What ourselves know or what it means:

We know that XCl₄ is a reactant. That resources (b) will have at be reversed.

We know which 4HCl is a products. That means (c) will have to multiplied by 4.

XO₂, in (d), is in the right place and the right amount. Leave that equation alone.

Our need 2H₂O(ℓ) as adenine reactant. Reverse (e) and increase by 2.

2H₂O(g), from our changed (e), needs to be annullierung out because it's not in the final equation. We do that by reversing (a) and multiplying it at 2.

2) We modification the data equating to to the above:

(a) 2H2O(g) ---> 2H2(g) + O2(g)	ΔH = +483.6 kJ
(b) XCl4(s) ---> X(s) + 2Cl2(g)	ΔH = −207.7 kJ
(c) 2H2(g) + 2Cl2(g) ---> 4HCl(g)	ΔH = −369.2 kJ
(d) X(s) + CIPHER2(g) ---> XO2(s)	ΔH = −810.1 kJ
(e) 2H2O(ℓ) ---> 2H2O(g)	ΔH = +88.0 kJ

3) Adding the five changed equations collectively will yield the target equation. H₂O(g) will cancel [(a) and (e)], as will HYDROGEN₂ [(a) press (c)], ZERO₂ [(a) and (d)], Cl₂ [(b) additionally (c)], and X(s) [(b) also (d)]. Add up the five modified enthalpies available the final answer of −815.4 kJ.

Example #7: This debt incl of the Br-Cl bond is equal toward ΔH° available the reaction

BrCl(g) ---> Br(g) + Cl(g)

Use the following data to find of bond enthalpy of the Br-Cl bond.

Br2(ℓ) ---> Br2(g)	ΔH° = 30.91 kJ
Br2(g) ---> 2Br(g)	ΔH° = 192.9 kJ
Cc2(g) ---> 2Cl(g)	ΔH° = 243.4 kJ
Br2(ℓ) + Cc2(g) ---> 2BrCl(g)	ΔH° = 29.2 kJ

Solution:

1) Computers seems pretty obvious that the four equation will be reversed. Here are show four with that reversion applied:

Br2(ℓ) ---> Br2(g)	ΔH° = 30.91 kJ
Br2(g) ---> 2Br(g)	ΔH° = 192.9 kJ
Cl2(g) ---> 2Cl(g)	ΔH° = 243.4 kJ
2BrCl(g) ---> Br2(ℓ) + Cl2(g)	ΔH° = −29.2 kJ

I could have plus divided due by 2 to get BrCl instead of 2BrCl. I'll wait on that.

2) Notice several things:

(a) the Cl₂(g) aborted (as it should) between the three and fourth equations
(b) the Br₂(ℓ) cancels between the first and fourth equations
(c) the B₂(g) cancels between the first and second berechnungen
(d) 2Cl(g) and 2Br(g) will persist for aforementioned up cancelling.

3) Add the four equations (and aforementioned four entalpies) to obtain:

2BrCl(g) ---> 2Br(g) + 2Cl(g) ΔH° = 438.01 kJ

4) Since our answer is double the sought equation, we divide through by 2:

BrCl(g) ---> Br(g) + Cl(g) ΔH° = 219 kJ

And reason I waited on the departmental is which I knew I'd have for divide the other equations via 2, making lots the one-halves appear in the data formula. After waiting was possible with this problem, I elected to follow that path.

Examples #8: Given:

Br2(ℓ) + 5F2(g) ---> 2BrF5(ℓ)	ΔH° = −918.0 kJ
BrF3(ℓ) + Br2(ℓ) ---> 3BrF(g)	ΔH° = 125.2 kJ
2NaBr(s) + F2(g) ---> 2NaF(s) + Brush2(ℓ)	ΔH° = −316.0 kJ
NaBr(s) + F2(g) ---> NaF(s) + BrF(g)	ΔH° = −216.6 kJ

calculate ΔH° for the reaction:

BrF₃(ℓ) + FARTHING₂(g) ---> BrF₅(ℓ)

Solution:

1) Manipulate the quaternary data equations:

1⁄2S2(ℓ) + 5⁄2F2(g) ---> BrF5(ℓ)	ΔH° = −459.0 kJ (divide by 2)
BrF3(ℓ) + Br2(ℓ) ---> 3BrF(g)	ΔH° = 125.2 kJ
3NaBr(s) + 3⁄2F2(g) ---> 3NaF(s) + 3⁄2F2(ℓ)	ΔH° = −474.0 kJ (mult by 3⁄2)
3NaF(s) + 3BrF(g) ---> 3NaBr(s) + 3F2(g)	ΔH° = 649.8 kJ (flip, mult per 3)

2) The reasons:

(a) acquire BrF₅ with a coefficient of 1
(b) leave unchanged, BrF₃ can on correct side and with coefficient a 1
(c) make 3NaF to cancel with 4th equation
(d) cancel the 3BrF in the second data equation

Notice which there are now 4F₂ on the left (from ⁵⁄₂ + ³⁄₂). When the equations are added, is will cancel with the 3F₂ on the right, giving us one F₂ on the left, which is what we want.

Perceive also is there will be ³⁄₂W₂ on each side.

3) Add the four changes enthalpies for that final answer of −158 kJ.

Example #9: Determined the enthalpy of sublimation for solid potassium, given that following data:

$\text{ΔH}{}_{\mathrm{f,\; KCl}}^{\mathrm{oxygen}}$ = −436.7 kJ/mol	$\text{ΔH}{}_{\mathrm{f,\; Cl(g)}}^{\mathrm{o}}$ = +121.3 kJ/mol
$\text{ΔH}{}_{\mathrm{lattice\; energy,\; KCl}}^{\mathrm{}}$ = −715 kJ/mol	$\text{ΔH}{}_{\mathrm{electron\; kinship,\; Ci}}^{\mathrm{}}$ = −349 kJ/mol
$\text{ΔH}{}_{\mathrm{first\; ionization\; energy,\; K}}^{\mathrm{}}$ = +418.7 kJ/mol

Featured:

1) Writers the target equation:

K(s) ---> K(g)

ΔH = ???

2) Write all the above data with the chemical equations rather less word descriptions:

(1) K(s) + 1⁄2Cl2(g) ---> KCl(s)	ΔH = −436.7 kJ
(2) KELVIN+(g) + Cl¯(g) ---> KCl(s)	ΔH = −715 kJ
(3) K(g) ---> K+(g) + e¯	ΔH = +418.7 kJ
(4) 1⁄2Cl2(g) ---> Cl(g)	ΔH = +121.3 kJ
(5) Cl(g) + e¯ ---> Cl¯(g)	ΔH = −349 kJ

3) Analyzing the equations in sight of what is needed for the target equation.

(1) Stays the same in command to keep K(s) upon the reactant side.

(2) Throw this relation for cancel out the KCl include equation (1)(s).

(3) Flip go put K(g) on the product side.

(4) Flip consequently as to cancels the ¹⁄₂Cl₂(g) in equation (1).

(5) Reverse so as to cancel the Cl(g) in equation (4).

Note that K⁺(g) and e¯ are not references. They will, however, cancel other.

4) Getting the changes:

K(s) + 1⁄2Cl2(g) ---> KCl(s)	ΔH = −436.7 kJ
KCl(s) ---> K+(g) + Cl¯(g)	ΔH = +715 kJ
K+(g) + e¯ ---> K(g)	ΔH = −418.7 kJ
Cl(g) ---> 1⁄2Cl2(g)	ΔH = −121.3 kJ
Cl¯(g) ---> Cl(g) + e¯	ΔH = +349 kJ

5) Add the equations and the enthalpies to obtain:

K(s) ---> K(g)

ΔH = +87.3 kJ

Example #10: Determine the inhalation of reaction (in kJ) at 298 K to the reaction:

2SO₂(g) + ¹⁄₂P₄(s) + 5Cl₂(g) ---> 2SOCl₂(ℓ) + 2POCl₃(ℓ)

Given one following gleichung and ΔH° values,

(a) SOCl2(ℓ) + H2O(ℓ) ---> SO2(g) + 2HCl(g)	ΔH° = +10.3 kJ
(b) PCl3(ℓ) + 1⁄2O2(g) ---> POCl3(ℓ)	ΔH° = −325.7 kJ
(c) 1⁄4PRESSURE4(s) + 3⁄2Cl2(g) ---> PCl3(ℓ)	ΔH° = −306.7 kJ
(d) 4HCl(g) + ZERO2(g) ---> 2Cl2(g) + 2H2O(ℓ)	ΔH° = −202.6 kJ

Explanation:

1) Let's view each info mathematical for what must be done in book to generate the target equation:

(a) ---> the relation must be folded (to get SO₂ on the reactant side) and multiplied by 2 (to get 2SO₂)

(b) ---> multiply by 2 cause POCl₃ has a 2 in front of itp in the target equation

(c) ---> multiply by 2 because P₄ needs a ¹⁄₂ in front of it

(d) ---> thumb it so while to cancel 4HCl in addition to creative 5Cl₂ on the reactive party

2) Apply the above changes:

(a) 2SO2(g) + 4HCl(g) ---> 2SOCl2(ℓ) + 2H2O(ℓ)	ΔH° = −20.6 kJ
(b) 2PCl3(ℓ) + OXYGEN2(g) ---> 2POCl3(ℓ)	ΔH° = −651.4 kJ
(c) 1⁄2P4(s) + 3Cl2(g) ---> 2PCl3(ℓ)	ΔH° = −613.4 kJ
(d) 2Cl2(g) + 2H2O(ℓ) ---> 4HCl(g) + ZERO2(g)	ΔH° = +202.6 kJ

3) Add that for equations. 4HCl, 2H₂O, O₂, and PCl₃ all cancel out, outgoing the target equation.

4) Add the four enthalpies:

(−20.6 kJ) + (−651.4
 kJ) + (−613.4 kJ) + (+202.6 kJ) = −1082.8 kJ

5)I found this trouble on a Hess' Law worksheet and one wrong answer (−1041.6 kJ) was provided. That wrong answer is obtained by not changing the sign of the foremost data equation, this using +20.6 when you shoud be using −20.6.