7.07b Strontium carbonate

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来自 University of Kentucky 的课程

Chemistry

614 个评分

University of Kentucky

Chemistry

614 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

So we're given several thermochemical reactions, and the,

we have the reaction and the delta H for them, and we're trying to

calculate the standard heat of formation of this strontium carbonate.

Now, what does that mean?

The heat of formation of a reaction is the energy changed to produce 1 mol,

or the enthalpy change to produce 1 mol of this substance from the elements in

their most stable form.

So Sr is a solid that's a metal, it's a solid, that's its most stable form.

Carbon's most stable,

stable form is graphite, and oxygen's most stable form is O2.

And we need to balance it for 1 mol, so we have to keep a one on this side,

because that's the definition of heat of formation.

So we are going to put, let's see a, three halves here to give me 3 oxygens.

So we have 1 carbon, 1 strontium already.

So what we're trying to do is, determine the delta H of this reaction.

The delta H of this reaction would then be, by definition, the heat of formation.

So let's try to construct this reaction from these puzzle pieces.

We need strontium on the left-hand side of the equation, 1 mol of it.

We look up here and the only equation that has strontium is this first one, but

it's 2 mols on the right-hand side.

So we have to reverse the reaction and divide by 2.

So we will reverse the reaction and

divide by 2, okay.

If we do that, we can determine the delta H of this reaction.

By changing the sign, because we reversed the reaction, and

dividing it by 2, because we cut it in half.

All right so that gets our strontium in place.

Now we're ready to get a carbon in place.

So here's the graphite equation, this third equation here,

it's got 1 mol of graphite on the left-hand side, that's what we needed, so

we use it exactly as we see it.

Carbon graphite plus O2

gas producing carbon dioxide gas.

Since we used it just as we saw it,

take the delta H just as it comes off the table there 394 kilojoules.

So then I got the graphite in place.

The O2, so far we've got one and a half mols of O2, okay?

So it looks like it's good.

Now let's make sure we get the SrCO3 in place, okay?

So we look up here and SrCO3 come from the second equation.

It has 1 mole on the right-hand side, so we're going to use it just as we see it.

SrO solid plus CO2 gas producing CrCO3.

Since it's just Sr,

sorry CO3 and that's not a gas.

It's a solid, okay.

So the delta H of this reaction, I read it right off of here is a negative 200, boy,

I'm having a hard time writing, 234 kilojoules.

Now, let's verify that all of these do add up.

To give me the reaction I'm looking for.

I'm trying to obtain this reaction.

So we've got the strontium.

We've got one and half mols of oxygen.

We have Sr0, cancelling with Sr0.

We've got carbon dioxide, cancelling with carbon dioxide.

And that leaves me with exactly what we see up here, one Strontium, one carbon,

and a total of one and a half oxygens, three half oxygens.

And the SrCO3.

So yes, that does construct this reaction.

So these have to add up, to give me the delta H of the reaction.

When you add all of these up you will get a negative 1220 kilojoules.

So, that is the delta H of this reaction, and by definition that would

be the calculating standard heat of formation of the strontium carbonate.

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