Hess's Law

Brightstorm
8 Sept 201005:46

Summary

TLDRThis video explains Hess's Law and its application in determining the enthalpy change (Delta H) of a reaction when it's not directly available. The instructor demonstrates how to manipulate thermochemical equations to calculate Delta H by adding and adjusting other known reactions. The example provided involves calculating the Delta H for sulfur trioxide formation from sulfur and oxygen, showing how to reverse and scale reactions. The video concludes with a discussion of formation reactions, emphasizing the importance of having one mole of the product in such calculations.

Takeaways

  • 📉 Hess's Law helps calculate the enthalpy change (ΔH) of a reaction by adding up the ΔH values of multiple related reactions.
  • 📘 If ΔH for a reaction is unknown, you can either measure it using a calorimeter or use Hess’s Law with known reactions.
  • 🔁 Hess's Law states that you can reverse or multiply chemical reactions to match the desired overall reaction.
  • ⚖️ When multiplying a reaction, the ΔH value must also be multiplied by the same factor.
  • ↔️ Reversing a reaction changes the sign of its ΔH; an exothermic reaction becomes endothermic and vice versa.
  • 🧪 In an example with sulfur and oxygen forming sulfur trioxide, the reactions are manipulated to match the overall reaction.
  • ✖️ Intermediate products, like sulfur dioxide in this case, can cancel out when added across reactions.
  • 🔢 The total ΔH for the overall reaction is the sum of the ΔH values of the individual reactions.
  • 📊 In formation reactions, it’s common to have fractional coefficients, which is acceptable as long as there’s one mole of product.
  • 🌡️ Hess's Law is highly useful for calculating ΔH for reactions where direct experimental data is unavailable.

Q & A

  • What is the main method discussed in the video to calculate Delta H when it is not known?

    -The video discusses using Hess's Law to calculate Delta H by manipulating other known reactions that add up to the overall desired reaction.

  • What is Hess's Law?

    -Hess's Law states that it is possible to add two or more thermochemical equations to produce a final equation for a reaction, and the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction.

  • What example reaction is used in the video to explain Hess's Law?

    -The reaction used is sulfur (S) plus oxygen gas (O2) yielding sulfur trioxide (SO3), where the Delta H of the reaction is initially unknown.

  • How is the first reaction modified to match the desired reaction?

    -The first reaction, which has one mole of sulfur (S), is multiplied by 2 to have two moles of sulfur, matching the reactant side of the target reaction.

  • What happens to the Delta H when the reaction is multiplied by 2?

    -The Delta H is also multiplied by 2. For example, if the original Delta H was -297 kJ, it becomes -594 kJ after multiplying the reaction by 2.

  • Why is the second reaction flipped in the explanation?

    -The second reaction is flipped because sulfur dioxide (SO2) is on the product side, but it is needed on the reactant side to match the desired reaction. Reversing the reaction also changes the sign of the Delta H.

  • How do the sulfur dioxide (SO2) molecules get canceled out?

    -The two moles of sulfur dioxide (SO2) on the reactant side of the flipped reaction cancel out the two moles of sulfur dioxide on the product side of the first reaction.

  • What is the final Delta H for the overall reaction after using Hess's Law?

    -The final Delta H for the reaction is -792 kJ after adding the modified Delta H values from the two reactions.

  • Why is the equation divided by 2 at the end?

    -The equation is divided by 2 to convert it into a formation reaction, which involves forming one mole of the product, sulfur trioxide (SO3), from its elements. Formation reactions require one mole of the product.

  • What is the final Delta H per mole for the formation reaction?

    -After dividing by 2, the final Delta H for the formation of one mole of sulfur trioxide (SO3) is -396 kJ per mole.

Outlines

00:00

🧪 Understanding Hess's Law: Thermochemical Equations

The video begins by introducing the challenge of calculating the enthalpy change (Delta H) of a reaction, especially when it's not readily available. Various methods, such as using a calorimeter or leveraging other known reactions, are introduced. The focus then shifts to Hess's Law, which allows for the summation of enthalpy changes from individual reactions to calculate the enthalpy for a final reaction. This concept is simplified by using a practical example involving sulfur, oxygen gas, and sulfur trioxide to illustrate the application of Hess's Law.

05:04

🔄 Manipulating Reactions Using Hess's Law

In this section, a specific reaction involving sulfur and oxygen is used to demonstrate how Hess's Law works. The example begins by multiplying a reaction to balance the sulfur atoms on both sides, ensuring the thermochemical equations add up correctly. The process of adjusting the enthalpy values according to the manipulation of the reactions is explained step-by-step, highlighting how reversing reactions changes the sign of Delta H. This process ultimately leads to determining the overall Delta H for the desired reaction.

➗ Simplifying to Formation Reactions

The video further refines the example by introducing the concept of formation reactions, where sulfur trioxide is formed from its elemental components. To simplify the reaction into a formation reaction, everything is divided by two, including the enthalpy, yielding a more standard format with one mole of product. It also explains that in formation reactions, fractional coefficients are acceptable as long as there's one mole of the product. The section concludes by emphasizing the usefulness of Hess's Law in finding Delta H for complex reactions.

Mindmap

Keywords

💡Thermochemical equations

Thermochemical equations describe chemical reactions along with the associated enthalpy changes (ΔH), indicating whether energy is absorbed or released. In the video, the speaker refers to how these equations can be used to determine the heat change of reactions, which is crucial when calculating the ΔH of a reaction.

💡Delta H (ΔH)

Delta H represents the change in enthalpy, or heat content, during a reaction. It's a key concept in the video, as the speaker is trying to find the ΔH for a reaction using Hess's law. ΔH can indicate whether a reaction is exothermic (negative ΔH) or endothermic (positive ΔH).

💡Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the path taken, as long as the initial and final conditions are the same. This law allows chemists to calculate the enthalpy change of a reaction by adding the enthalpy changes of known reactions, which is demonstrated when the speaker manipulates and adds reactions to find the unknown ΔH.

💡Calorimeter

A calorimeter is an instrument used to measure the amount of heat released or absorbed during a chemical reaction. The speaker mentions it as one method to determine ΔH for a reaction, but emphasizes that using Hess's law is often more practical when a calorimeter is unavailable.

💡Sulfur trioxide (SO3)

Sulfur trioxide is a chemical compound formed from sulfur and oxygen. In the video, the speaker is trying to find the ΔH for the formation of sulfur trioxide using Hess’s Law. It serves as the final product of the reactions being manipulated.

💡Formation reaction

A formation reaction involves creating one mole of a compound from its elements in their standard states. The speaker concludes by discussing how sulfur trioxide is formed from sulfur and oxygen, and explains that in formation reactions, fractional coefficients are allowed as long as one mole of the product is produced.

💡Reverse reaction

A reverse reaction is when the products of a reaction are turned back into the reactants. The speaker flips one of the reactions in the video to balance sulfur dioxide and sulfur trioxide, explaining that reversing a reaction changes the sign of its ΔH.

💡Exothermic reaction

An exothermic reaction releases energy in the form of heat, resulting in a negative ΔH. In the video, the speaker reverses a reaction, turning it into an exothermic one, and explains how this affects the overall enthalpy.

💡Endothermic reaction

An endothermic reaction absorbs energy, which results in a positive ΔH. The speaker initially discusses the forward reaction, which is endothermic, but when reversed, it becomes exothermic.

💡Fractional coefficients

Fractional coefficients are used in balanced chemical equations when partial moles of a substance are involved, particularly in formation reactions. The speaker demonstrates this when adjusting the equation to ensure that only one mole of sulfur trioxide is formed, leading to fractional oxygen in the equation.

Highlights

Hess's law allows you to calculate the Delta H of a reaction by adding up the enthalpy changes of multiple reactions.

If you don't know the Delta H of a reaction, you can measure it with a calorimeter or use Hess's law to calculate it using other reactions.

Hess's law states that the sum of the enthalpy changes of individual reactions equals the enthalpy change of the final reaction.

When applying Hess's law, you can manipulate reactions by multiplying them to match the desired reaction.

If you flip a reaction, you also need to flip the sign of the Delta H.

To match the desired equation, sulfur needed to be multiplied by two, and the corresponding Delta H was doubled.

Reversing the reaction to swap the sides of sulfur dioxide and sulfur trioxide requires reversing the sign of Delta H.

The sulfur dioxide molecules cancel out when combining two reactions to form sulfur trioxide, leaving the final reaction balanced.

Adding the Delta H values of both reactions gives the final Delta H for the reaction, which is -792 kilojoules.

For formation reactions, it’s important to reduce the reaction to produce one mole of the product, which involves dividing the entire equation by two.

Formation reactions often involve fractional coefficients to ensure that the product has one mole.

When dividing the reaction for a formation process, the enthalpy change also needs to be halved.

Hess's law is useful when the Delta H of a reaction is unknown and allows you to calculate it easily by using related reactions.

Formation reactions focus on creating one mole of a product from its elements, and fractional coefficients are acceptable in these equations.

Using Hess's law simplifies finding the enthalpy changes without performing direct experimental measurements.

Transcripts

play00:00

[Music]

play00:02

all right sometimes when you're dealing

play00:04

with thermo chemical equations you might

play00:06

not know the Delta H of your reaction so

play00:08

how there's different ways you can go

play00:10

about getting the Delta H of the

play00:11

reaction you can actually do the

play00:12

reaction itself and use a calorimeter

play00:13

and figure out the Delta H or that you

play00:16

can actually use other reactions they

play00:18

might add up to your particular action

play00:20

what does that mean okay there's this

play00:22

law called Hess's law

play00:23

Hess's law states that it is possible to

play00:25

add two or more thermo chemical

play00:27

equations to produce a final equation

play00:28

for a reaction and the sum of the

play00:30

enthalpy changes for the individual

play00:31

reactions is is the enthalpy change for

play00:34

the final reaction now this is

play00:35

definitely a mouthful it's much easier

play00:37

to explain if you actually do it so

play00:39

let's actually go over here and let's

play00:41

say I had this reaction sulphur pulse

play00:43

oxygen gas yields me sulphur trioxide

play00:45

and I have no idea what the Delta H is

play00:48

okay well there are many many many

play00:50

reference books that tell you what the

play00:51

Delta H is for other reactions but this

play00:53

one for some reason I couldn't find or

play00:55

my reference book didn't have it or I

play00:57

didn't want to do the particular

play00:58

reaction in lab or something along those

play00:59

lines so how can I find this well Hess's

play01:02

law says that I can manipulate these

play01:04

reactions in order to get this overall

play01:06

reaction and that way I can figure out

play01:07

my Delta H if I add up the two H's so

play01:10

first I have to make sure that I have

play01:13

these reactions actually do add up to

play01:16

the final reaction so let's look at this

play01:18

first thing I need to sulfurs on my

play01:20

reactant on my reactant side well here's

play01:23

sulfur but I only have one so I'm going

play01:24

to multiply this whole reaction by 2 so

play01:26

I can get two sulfur's on my reactant

play01:28

side so I'm going to multiply this by 2

play01:31

so that makes two sulfur plus two oxygen

play01:36

gases yields me two sulfur dioxides and

play01:42

because I multiplied the reaction by two

play01:44

this also must be multiplied by two

play01:46

because this this negative 297

play01:48

kilojoules is telling me how much

play01:50

energies can be released for one mole of

play01:51

sulfur but I'm now doing two moles of

play01:53

sulfur so I'm going to multiply this by

play01:54

two so my Delta H is now going to be

play01:57

what is it negative 594 kilojoules okay

play02:04

great

play02:04

so now I have my two moles of sulfur on

play02:06

my reactant side that's awesome but now

play02:08

I have sulfur dioxide on my product side

play02:10

and I don't want it i want sulphur

play02:11

trioxide so I'm gonna look at this

play02:13

reaction

play02:13

here wellhere's call for dioxide on the

play02:16

product side and sulfur trioxide on the

play02:18

reactant side so I actually want to flip

play02:20

those around so I'm going to do the

play02:23

reverse reaction so I'm going to say

play02:24

oxygen gas I'm going to make sure I note

play02:29

that oxygen gas plus two moles of sulfur

play02:35

dioxide gas is going to give me two

play02:40

moles of sulfur trioxide gas okay and

play02:43

because I flipped it this the only this

play02:46

forward direction is going to be an

play02:47

endothermic reaction but if I reverse it

play02:49

I mean it's going to becoming an

play02:50

exothermic reaction so this is going to

play02:51

be negative so my Delta H is going to be

play02:54

negative 198 kilojoules okay so let's

play02:59

make sure this works so if I were to add

play03:00

these two reactions together I note that

play03:04

I have two moles of sulfur dioxide here

play03:07

and two moles of sulfur dioxide here I

play03:09

can cross those out because this is a

play03:12

product and this is the reactant so I

play03:13

can just cross those out okay

play03:16

and now if I add everything up I can't

play03:18

nothing else crosses out so I'm going to

play03:19

add everything up so I have two moles of

play03:21

sulfur solid plus two moles of oxygen

play03:25

which should be oxygen to sorry

play03:27

Oh to gas plus another bulb

play03:30

Oh two guest yields two moles of so3 gas

play03:35

okay so what if I add these together two

play03:39

plus one is three second say three moles

play03:41

of oxygen gas so two sulfur atoms are

play03:45

bold two moles of sulfur plus 3 moles of

play03:47

oxygen yields two moles of sulfur

play03:49

trioxide which is exactly my original

play03:51

equation is so yes I did this properly I

play03:53

can check it off and so I want to find

play03:55

my Delta H I'm just going to add these

play03:57

guys up

play03:57

so negative 594 plus negative 198 is

play04:00

going to give me negative seven ninety

play04:02

two so negative 792 kilojoules so that

play04:10

actually now I could know this is

play04:12

negative 792 kilojoules because houses

play04:14

law says that I can do this okay so this

play04:19

is basically houses all on a nutshell

play04:20

you might have several different

play04:21

reactions you're going to manipulate but

play04:23

actually once you get the hang of it

play04:24

it's actually kind of fun and pretty

play04:25

easy

play04:26

but most going to do one left thing this

play04:29

is a reaction it is a formation reaction

play04:32

meaning we're taking or forming sulfur

play04:33

trioxide from its elements so in order

play04:36

to do formation reactions the best way

play04:38

is we want to get this to be one mole so

play04:40

I'm going to divide the whole thing by

play04:41

two because I don't like information

play04:44

reactions if you want to learn more

play04:46

about them and there's a video on that

play04:47

as well but a formation reactions have

play04:50

one mole of the product so we're

play04:52

dividing everything by two which ends up

play04:54

was sulfur solid plus three-halves

play04:58

oxygen gas yields sulfur trioxide gas

play05:03

and then my Delta H is also going to be

play05:05

divided by 2 oops

play05:07

which is going to be negative 396

play05:11

kilojoules and we can actually say

play05:13

kilojoules per mole because we have one

play05:15

mole of this so this is a formation

play05:17

reaction and notice this is the fraction

play05:20

in it but in formation reactions is

play05:22

totally okay we are allowed to have

play05:24

fractional coefficients and our

play05:25

formation reactions because our main our

play05:28

main important thing is that we have one

play05:30

mole of product so this is has its law

play05:32

in action and then we put a little bit

play05:33

of formation Delta H of formation in

play05:35

there as well so Hess's law is really

play05:37

really useful when dealing with

play05:38

reactions where you do not know our

play05:40

Delta HS and we need to find it pretty

play05:42

easily

play05:44

[Music]

Rate This

5.0 / 5 (0 votes)

الوسوم ذات الصلة
ThermochemistryHess's LawDelta HChemical ReactionsFormation ReactionsSulfur TrioxideCalorimetryEnthalpy ChangeChemistry TipsReaction Manipulation
هل تحتاج إلى تلخيص باللغة الإنجليزية؟