Energy and the First Law of Thermodynamics

Energy and the First Law of Thermodynamics

Section 1

As you know from physics, energy can be kinetic or potential. In our world potential energy is often stored in molecules that are processed later. For example, the energy in gasoline is stored until your car converts it into kinetic energy, and the energy of the food you eat is stored until your body uses it.

Where does this energy come from? How is it stored?

Where does this energy come from? How is it stored?

Chemical energy is the energy associated with chemical bonds. Changes in energy that occur during a chemical reaction are due to the making and breaking of chemical bonds.

Thermodynamics (heat + moving) studies this process of the changes in energy.

Thermodynamics might study the following:

2 C(g) + 6 H(g) + O(g) CH3OCH3(g)(dimethyl ether)

When this occurs 3151 kJ of energy is released.

When dimethyl ether is burned:CH3OCH3(g) + 3O2(g) 2CO2(g) + 2H2O(g)

1327 kJ of energy is released.

2 C(g) + 6 H(g) + O(g) CH3OCH3(g)

When this occurs 3151 kJ of energy is released.

In terms of bonding, what is happening in the first reaction? Where does the energy come from? Turn this into a general rule.

CH3OCH3(g) + 3O2(g) 2CO2(g) + 2H2O(g)1327 kJ of energy is released.

In terms of bonding, what is happening in the second equation? Why is less energy released than when dimethyl ether was made? Turn this into a general rule.

When energy is released, it must go somewhere, and when energy is absorbed it must come from somewhere. This somewhere is considered the surroundings and the movement of energy into and out of the system to the surroundings is called heat, abbreviated q.

Note heat and temperature are not the same thing. Temperature is not a direct measurement of heat, but just a comparison.Does 20°C have more heat than 10°C? Does it have twice as much heat?

When objects gain or lose heat, we expect them to change in temperature, but as temperature is not a direct measure of heat we cannot say that a 15° change means a 15 kJ change in heat. There are two other factors also involved.

1. The amount of matter to gain or lose heat will have an effect. Which would warm up the most on a hot summer day, a bucket of water or a swimming pool full of water?

Typically, the more matter is in the system, the more heat it will gain or lose before changing in temperature. The amount of matter is, of course, measured in grams.

2. All matter also has a characteristic amount of heat that can be gained or lost before changing temperature, regardless of the mass. This is called specific heat, and is abbreviated Cp.

To calculate the temperature change associated with a change in heat:

ΔT = or q = m Cp ΔT

q (heat) is measured in J or kJT (temperature) is measured in °Cm (mass) is measured in g Cp (specific heat) is measured in J g-1 °C-1

or kJ g-1 °C-1 that is Jg °C

qm Cp

Also, when heat is gained or lost it must flow into another part of the system or surroundings, so the heat lost by one part must equal the heat gained by another part.

-qpart 1 = qpart 2

-(mpart 1 Cp part 1 ΔT part 1) = mpart 2 Cp part 2 ΔT part 2

And ΔT = Tfinal - Tintial for both sides.

What is the negative showing?What would be true of Tfinal for two things

touching and trading heat?

It is also possible to use moles instead of mass, in which case use the Molar Specific Heat

q = n c ΔT

q (heat) is measured in J or kJT (temperature) is measured in °Cn (moles) is measured in molc (molar specific heat) is measured in

J mol-1 °C-1 or kJ mol-1 °C-1

Energy and the First Law of Thermodynamics

Section 2

A state function is one that depends only on the present state of the system. It does not depend on how the past state was changed to arrive at the present state, just that it went from one state to another.

State FunctionsA story of two beakers.

Once upon a time there were two beakers with exactly the same amount of pure water.

Beaker 1 was the boring sort. The water in beaker 1 warmed from 15°C to 75°C.

Beaker 2 was the exciting sort. The water in beaker 2 warmed from 15°C to 23°C, then cooled to 4°C, then warmed to 97°C, and then cooled to 75°C.

What is the change in temperature of each?

Is temperature a state function? Explain.

Energy is also a state function, it only depends on the old and new state (initial and final), not how it changed in-between.

ΔE = Efinal - Einitial

Regardless of the type of change as the state of energy changes, energy must be conserved.

This is the First Law of Thermodynamics:The total energy before and after a change must be the same. Does this sound familiar?

If energy is lost from the system, where must it go?

If energy is gained in the system, where did it come from?

To show if energy is being lost or gained, we use a +/- sign in front of the energy. For example:

energy is gained by the system, and lost by the surroundings

ΔEsystem = -ΔEsurroundings

energy is lost by the system, and gained by the surroundings

-ΔEsystem = ΔEsurroundings

The energy we’ve talked about so far is the heat energy of the system, the energy the molecules or atoms or whatever innately have at the state the system is in.

But we know gases like to move around and spread out, and all this moving around and spreading out takes energy as well.

From physics we know that w=FΔd, but in chemistry, work is done by gases as they move around and spread out.

So the force of a gas is called pressure (P) and the distance the gas moves is called volume (V).

So w=PΔV

w=PΔVA system with gas in a piston could have the gas push on the piston and do work on the surroundings, -w (work out or lost by the system). Using the equation, why is w -?

Or a system with gas in a piston could have the gas be pushed smaller and have work done on the system by the surrounding, +w (work in or gained by the system). Using the equation, why is w +?

w=PΔVWhen would be a case where no work is done? (In physics is work done if nothing moves?)

When no work is done by or on the system, the energy of a system is just the heat energy of the system, therefore:

E = qv

If a system has both heat energy and work energy, we want to find the total energy. We call this total energy enthalpy (H). Enthalpy is also a state function.

H = E + wor

H = E + PΔV

H = E + w

If solid water turns into liquid water, how would H and E compare? Explain.

If liquid water turns into water vapor, how would H and E compare? Explain.

If this seems like a lot to keep track of, there is good news. Most chemistry happens at constant pressure - reactions happen in open containers on a lab bench. Why is this constant pressure?

In this situation, the system will adjust it’s volume according to the heat energy gained or lost without any extra work, so

H = qp

If enthalpy is a state function, and if a chemical reaction occurs open to the atmosphere (so no work), then to find how the enthalpy changes during the reaction we just need to take the final enthalpy minus the initial enthalpy.

ΔH = Hfinal - Hinitial

With a chemical reaction, initial is reactants and final is products, so we’ll write it this way:

ΔH = Hproducts - Hreactants

ΔH Energy Movement

Esystem Tsystem

+

-

Δw Work Movement

Esystem (also H)

Tsystem

+

-

Energy and the First Law of Thermodynamics

Section 3

As we mentioned earlier, the energy in a reaction comes from the bonds between atoms. Specifically, breaking bonds takes energy and forming bonds releases energy.

For example:CO(g) + H2O(g) CO2(g) + H2(g)

As specifically as possible, describe what has to happen to turn the reactants into products.

CO(g) + H2O(g) CO2(g) + H2(g)

It might be easier to view this as lewis diagrams (you’ll need to draw these):

:C ≡ O: + H - O - H O = C = O + H - H

Now specifically state what happens to turn reactants into products.

..

.... .... ..

It is probably too simple for real life, but let’s pretend this reaction happens in two steps (you will learn about reaction steps in detail later this year):

CO(g) + H2O(g) C(g) + 2O(g) + 2H(g)and

C(g) + 2O(g) + 2H(g) CO2(g) + H2(g)

What is happening in the first reaction?What is happening in the second reaction?

As enthalpy is a state function, to figure out the enthalpy change for the entire reaction, we only need to know the enthalpy of each separate reaction and add them together.

CO(g) + H2O(g) C(g) + 2O(g) + 2H(g) ΔH = 2002.7 kJ mol-1 Why is this positive?

andC(g) + 2O(g) + 2H(g) CO2(g) + H2(g) ΔH = -2043.8 kJ mol-1 Why is this negative?

CO(g) + H2O(g) C(g) + 2O(g) + 2H(g) ΔH = 2002.7 kJ mol-1

andC(g) + 2O(g) + 2H(g) CO2(g) + H2(g) ΔH = -2043.8 kJ mol-1

Thus ΔH = -41.1 kJ mol-1 for the overall reaction. When a reaction gives off more enthalpy in forming bonds than it uses to break bonds, it is called exothermic.What is the relationship between the sign of ΔH and being exothermic or endothermic?

Very few reaction ever just break bonds or form bonds, almost all reactions are combination of the two, thus you must be careful to use the terms exothermic or endothermic only for the specific reaction in discussion at that moment.

Many students find it useful to view change in enthalpy as a diagram: (activation energy is the energy needed to start a reaction -

more about that will come later in the year)

We now know that the equation below is exothermic:

CO(g) + H2O(g) CO2(g) + H2(g)ΔH = -41.1 kJ mol-1

What would happen if we reversed the equation? What would you expect the change in enthalpy to be? Explain.

CO2(g) + H2(g) CO(g) + H2O(g)ΔH = ????? kJ mol-1

We now know that the equation below is exothermic:

CO(g) + H2O(g) CO2(g) + H2(g)ΔH = -41.1 kJ mol-1

What would happen if we reversed the equation? What would you expect the change in enthalpy to be? Explain.

CO2(g) + H2(g) CO(g) + H2O(g)ΔH = 41.1 kJ mol-1

In the previous example, you were given the change in enthalpy for simple reactions. What if that information is not readily available? It is certainly possible to determine the change in enthalpy if we account for all the bonds broken and all the bonds formed during a chemical reaction. We call these bond energies, and in general the stronger the bond (also the shorter the bond) the more energy it takes to break or the more is given off when it forms (see Coulomb’s Law).

To determine the ΔH using bond energies:

ΔH = B.E.broken + B.E.formed

NOTE: This is the only Δ equation with a “+”!

Bond energies will always be given to you, typically in a chart like the one at the end of your notes.

Example: Propane is burned in air to produce carbon dioxide and water. How much heat energy will one mole of propane be able to provide for cooking hamburgers?

What should always be your first step?

C3H8 + 5O2 3CO2 + 4H2O

There is a special case of determining the change in enthalpy using bond energies - dissolving a salt in water.

What happens to solid NaCl when placed in water? Compare and contrast this to determining bond energies as we’ve done before.

Not only does the bond not break evenly, but now that the ions are in water, water - a polar molecule - can form intermolecular bonds with the ions. Is forming bonds + or - enthalpy? Is this endothermic or exothermic? What would you feel touching the container? Let’s draw this:

The energy from bonding with water is called enthalpy of solution (or heat of solution). If the change in enthalpy of an ionic compound dissolving (solvating) in water is desired, the heat of solution would also be added (or subtracted) in the equation. Overall bond breaking and solvation could be exothermic or endothermic as you’ve seen in the web activity.We will learn later how the heat of solution might make a nonspontaneous reaction become spontaneous.

As you’ve no doubt noticed, it makes for a long calculation to determine the change in enthalpy from bond energy data, plus we are limited to the information given in the bond energy table. Fortunately, as enthalpy is a state function, there is a faster and more easily measured way to get the enthalpy of a reaction. Instead of calculating bond by bond, we will calculate compound by compound.

Scientists have measured the enthalpy change when many compounds are formed from their constituent atoms (usually at STP and in their standard states). This information is called the enthalpy (or heat) of formation (Hf), and is measured per mole of compound formed.

Why would it be called enthalpy of formation?

Example:What is the change in enthalpy of the reaction below, and is it endothermic or exothermic?

2 B5H9(g) + 12 O2(g) 5 B2O3(s) + 9 H2O(g)Use the data in your notes.

One other great thing about using enthalpies of formation is that you can add entire equations that you have enthalpy data for to calculate the enthalpy of an equation that you don’t have data for. This is called Hess’s Law.

For example:What is the change in enthalpy for

H2O(l) H2O(g)if H2(g) + ½ O2(g) H2O(l) ΔH = -285.83 kJ mol-1

and H2(g) + ½ O2(g) H2O(g) ΔH = -241.82 kJ mol-1

For example:What is the change in enthalpy for

H2O(l) H2O(g)if H2(g) + ½ O2(g) H2O(l) ΔH = -285.83 kJ mol-1

and H2(g) + ½ O2(g) H2O(g) ΔH = -241.82 kJ mol-1

First, get the equations to add together to match the overall equation. If you flip the equation, flip the sign of ΔH.

Second, add everything up, cancelling like terms on both sides and add up the enthalpies.

For example:What is the change in enthalpy for

H2O(l) H2O(g)if H2(g) + ½ O2(g) H2O(l) ΔH = -285.83 kJ mol-1

and H2(g) + ½ O2(g) H2O(g) ΔH = -241.82 kJ mol-1

For example:What is the change in enthalpy for

C(s) + H2O(g) CO(g) + H2(g)C(s) + ½ O2(g) CO(g) ΔH = -110.53 kJ mol-1

C(s) + O2(g) CO2(g) ΔH = -393.51 kJ mol-1

CO(s) + ½ O2(g) CO2(g) ΔH = -282.98 kJ mol-1

H2(g) + ½ O2(g) H2O(g) ΔH = -241.82 kJ mol-1