7.02a Work

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Chemistry

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University of Kentucky

Chemistry

588 classificações

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.

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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

Conheça os instrutores

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

In this problem we are going to be reacting C2H4 and C2H4 is a gas.

It doesn't say up there but I'm letting you know it's a gas.

It's reacting with hydrogen.

Now we know that that's a gas.

To produce C2H6.

This too is a gas.

Now in your past, you may have learned about Avogadro's law.

You should have learned about Avogadro's law.

And a constant temperature and

pressure, we know that we can think about the balanced equation,

not in terms of just moles, but also in terms of volume units.

So, if you have 100 millilitres of this substance reacting with 100

millilitres of this substance, it's equal amounts, and you're going to, according to

the 1 to 1 to 1 ratio, you would produce 100 millilitres of this substance.

And that's because it's at constant temperature and pressure.

Now we're going to calculate the amount of work done in this reaction.

When a gas is involved, work can be calculated by this equation.

Work is equal minus p times delta v,

where p is the external pressure of 1.5 atmostpheres.

Delta V is a change of volume of final minus initial.

Now, I have got a conversion factor between the work unit or

the energy unit of liter atmospheres to joules.

I know how to go between those two.

So I'm not going to put millilitres into this equation.

I'm going to put it in terms of liters.

So final that's product.

I have 100 millimeters of gas.

That would be 0.1 liters.

And I'm, initially started with 200 millilitres, or 0.2 litres.

Now, when you multiply this out,

you're going to get 0.15 liter atmospheres.

So that is a work unit that is an energy unit.

And it's going to be, since it's an a, it's 0.1 minus 0.2.

So, this is going to be negative and I changed the signs.

So, we have positive work.

Positive work is always what happens when a gas compresses in, gets smaller.

So the work is being done by the surroundings on the system, and

that is a positive work.

And it works out that way with our equation as well.

But we don't want to leave it as liter atmospheres because it asks for

it in joules.

So I've got a relationship between liter atmospheres and joules.

There are 101.3 joules in a liter atmosphere.

Now the two significant figures, that is 15 joules.

So that's how much work is done as this reaction takes place.

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