7.06a Rocket Fuel

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来自 University of Kentucky 的课程

Chemistry

481 个评分

University of Kentucky

Chemistry

481 个评分

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

从本节课中

Week 7

In this unit, we will learn about thermochemistry, which is the study of the thermal energy transfer (heat) in chemical reactions. We will learn how these measurements of heat are made via calorimetry. Then, learn to calculate heat transfer using various methods.

7.01 Thermochemistry: Foundational Definitions 13:56

7.02 Introduction to Thermodynamics 19:26

7.03 Enthalpy 12:48

7.04 Thermochemical Equations 19:11

7.04a Heat Produced from Na 2:12

7.04b Delta H for aluminum and chlorine 3:46

7.05 Enthalpy Change and the First Law of Thermodynamics 12:10

7.05a Delta E for acetylene 5:02

7.06 Calorimetry 25:01

7.06a Rocket Fuel 5:05

7.06b Delta H for CaO 5:02

7.07 Calculating Enthalpy Change 41:15

7.07a Heat capacity of bomb 11:24

7.07b Strontium carbonate4:27

与讲师见面

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

Okay, we are given information from a bomb calorimeter where hydrazine is combusted,

okay, so that means it's reacting with oxygen.

And it's giving us information about how much water was placed around the bomb and

the temperature change of this system.

And it wants us to calculate the heat of combustion for one mole of hydrazine.

Now hydrazine, as it says, is rocket fuel.

And when it reacts with oxygen,

it will produce nitrogen gas which I don't expect you to know,

I'm just telling you this, and water as the products.

And if we balance it, we'll, we can balance it by putting a 2 right here.

Now this reaction will be balanced for one mole of hydrazine.

So if we knew the delta H for this reaction,

that would be the heat of combustion, because it's called a combustion reaction.

So that's our goal here, is to determine the delta H for this balanced reaction.

Now they did not place a mole of hydrazine in this bomb.

They only put two grams which is far less than a mole into the bomb calorimeter.

So what we're going to do, is figure out first of all, how much heat is

produced when two grams react of this hydrazine reacts and then we'll be able to

translate that into how much would be produced when one mol reacts.

Now any time you're dealing with calorimetry,

what you want to do as a starting point is to know that,

the heat of the system equals the negative the heat of the surroundings.

That's the starting point for any type of transfer or thermal energy.

Now the system is my reaction, so

that's what I'm going to determine, the heat of the reaction.

The surroundings is made up of two things.

You've got your bomb calorimeter.

That says bamb, but it's supposed to be bomb.

And you've got your water that surrounds the bomb calorimeter.

The bomb calorimeter can be further broken apart into the heat capacity

of the bomb times the change in temperature, and that's given to me.

The heat capacity of the bomb is 927 joules per degree Celsius.

[SOUND].

The water can be further broken apart into the mass of the water,

the specific heat of the water, and the change in temperature.

So we can place in there the information that we know.

The C is equal to 927 joules per degrees Celsius.

The change in temperature, let's go ahead and calculate the change in temperature.

It would be final temperature, 32.19, minus the initial temperature,

25.15, and that will give us a change in temperature of 7.04 degrees Celsius.

And then play that, place that right here.

So this is times 7.04 degrees Celsius.

The mass of the water is 1,000 grams.

The specific heat of water is 4.184 joules per gram degree Celsius.

And the change in temperature of the water will be the same as that of

the bomb, 7.04.

So if we multiply all that information out,

what we would obtain is that this is equal to a negative 35,981

joules or to three significant figures which is all we can really know this.

To three significant figures, that would be a negative 300 or

sorry negative 36.0 kilojoules.

Does not look like a seven, I'm going to fix that right there.

Okay, so there is the amount of heat that would be generated when two grams reacts,

so lets write that ratio.

This is negative 36.0 kilojoules

for 2 grams of N2H4 reacting.

Now remember, we don't want it per gram.

We want it for mole.

And we want it for one mole, because that's what the balanced equation is.

So, when we put in the molar mass of hydrazine, which is 32.06 grams per mole.

We will, be as we see here, left with the units of kilojoules for mole hydrazine.

I suppose I ought to be good, and put N2H4 here, and N2H4 here,

so we have things perfectly cancelling, and that will give me the value

of a negative 577 kilojoules of heat released for one mole of hydrazine.

And so now we can say with certainty what the value of the delta H for

this reaction is as balanced with one mole.

This would equal a negative 577 kilojoules.

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