Chapter 5

Energy (section 5.1)

Energy is- the capacity to do work or transfer heat.

So what are work and heat?Work-energy used to cause an object with

mass to moveHeat-energy used to cause the temperature of

an object to increase.

EnergyKinetic (Ek)-energy of motion

EK = ½mv2

Magnitude depends on the object’s mass (m) and its velocity (v).

Potential-depends on the object’s position relative to other objects. In chemistry, it is expressed as electrostatic potential energy

Potential Energy

Eel= (KQ1Q2) /d, where K is a constant, and Q is the electrical charge on the two objects, and d is the distance between them.

K = 8.99x109 J-m/C2 (C is in coulombs)Q will usually be about the size of electron

charge (1.6x10-19 C)When both Q’s have the same sign, the

charges will repel one another, etc.When Eel is positive, what is the situation?

UnitsSI unit for energy is the Joule (J) 1 kg-m2/s2

Joules are tiny, so kJ will be appropriate.calorie=4.184J1000cal=1 Cal=1kcalWhat is the kinetic energy in joules of a 45g

golf ball moving at 61 m/s?What is this energy in calories?What happens to this energy when the ball

lands in a sand trap?

Work and HeatTwo players: the system (that which we are

focused upon), and the surroundings (everything else).

Work (w)= F*d A force (F) moves an object a distance (d).

When we (surroundings) cause an object (system) to move, we are performing work on the system, or transferring energy to it.

Heat, q, can enter or leave the system.

First Law of Thermodynamics section 5.2

Energy(E) is neither created nor destroyed.The internal energy of a system is the sum of

the kinetic & potential energies of all the components of the system. (We can’t know these values, but we can measure their change)

∆E=Efinal – Einitial

Positive ∆E means the system gained energy.Negative ∆E means the system lost energy.

First Law, cont.Positive ∆E denotes an endothermic process.Negative ∆E denotes an exothermic process.Endothermic-heat flows into the system from

the surroundings.Exothermic-heat flows out of the system to

the surroundings.

State FunctionsA property of a system that is determined by

specifying the system’s condition, or state (in temp., pressure, location, etc.)

The value of a state function depends only on the present state of the system, not the path it took to get to that state.

∆E is a state function, since it is derived from an initial and final state, not how it got from initial to final. It describes the amount of change only.

q (heat) and w (work) are NOT state functions.However, ∆E = q + w

Enthalpy H section 5.3

In chem., much of our work is in open systems (like a beaker), so work, w, is not readily obvious.

In a closed system, with a piston, pressure is constant, and the piston can be moved via a rxn. That generates a gas. This work would be visible.

This would be called pressure-volume work.H accounts for heat flow in processes where

pressure is constant, and the only work done is pressure-volume work.

H2 gas is generated and moves the piston. Pressure-volume work.

Enthalpy cont.So, H = E + PV (internal energy plus pressure-volume work

done)

Therefore, when a change occurs at constant pressure, ∆H = ∆E + P∆V

If ∆E= q + w, and w = -P∆V, then:∆H = (qP + w) – w, or ∆H = qP

This simply means that in most cases, ∆H = qPositive ∆H means the system has gained heat from

the surroundings (endothermic).Negative ∆H means the system has lost heat to the

surroundings (exothermic).

Enthalpies of Reaction section 5.4

∆H = Hproducts – Hreactants

This change is known as the heat of reaction. It is the heat change that acoompanies a reaction.

Example:CH4 + 2O2 → CO2 + 2H2O ∆H = -890 kJ

The production of water from its elements is an exothermic reaction the gives off 483.6 kJ of heat.

This is a thermochemical equation.

Enthalpies of reaction cont.Enthalpy is an extensive property. If you double the

amount reactants consumed, ∆H doubles also. Enthalpy is stoichiometric.

The enthalpy change for a reaction is equal in magnitude, but opposite in sign, for its reverse reaction.

The enthalpy change for a reaction depends on the state (phase) of the reactants and products.

∆H of reaction is usually in the units of kJ/mol.

The value of ∆H for the reaction below is -72 kJ.

__________ kJ of heat are released when 1.0 mol of HBr is formed in this reaction.

A) 144 B) 72 C) 0.44 D) 36 E) -72

d

2 2H (g) Br (g) 2HBr(g)

The value of ∆H for the reaction below is -126 kJ.

The amount of heat that is released by the reaction of 25.0 g of Na2O2 with water is __________ kJ.

A) 20.2 B) 40.4 C) 67.5 D) 80.8 E) -126

a

2 2 2 22Na O (s) 2H O(l) 4NaOH(s) O (g)

The value of ∆H for the reaction below is -1107 kJ:

How many kJ of heat are released when 15.75 g of Ba(s) reacts completely with oxygen to form BaO(s)?

A) 20.8 B) 63.5 C) 114 D) 70.3 E) 35.1

b

22Ba(s) O (g) 2BaO(s)

Calorimetry section 5.5

∆H can be measured experimentally, and this measurement of heat flow is called calorimetry.

The temp. change a substance undergoes when it absorbs heat is its heat capacity. Every pure substance has its own.

Heat capacity, C, is the heat required to raise the temperature 1°C. Specific heat capacity, s, is the heat required to raise 1g of that substance 1°C. Molar heat capacity, Cmolar……1 mole of a substance 1°C.

Constant-pressure calorimeterReally just a styrofoam cup.

Styrofoam is an excellent insulator, so it holds heat very well.

Constant pressure because it’s open to the atmosphere.

Constant-volume calorimeter

Called constant-volume because it is a closed system, so pressure varies, but the volume of the system does not.

Used to measure the energy stored in samples of fuels and foods.

Calorimetry cont.Heat, q, = m*s*∆THeat lost by the system is gained by the

surroundings, and vice versa.As stated earlier, ∆H and q are the same thing at

constant pressure, so when we measure heat, we are measuring ∆H directly.

When mixing solutions,

qSOLN = (spec. heat of solution)*(g of solution)*∆T

For solutions, the specific heat will be that of water: 4.184 J/g-°C

CalorimetryWhen 50.0mL of 0.100M AgNO3 and 50.0mL of 0.100M HCl

are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30°C to 23.11°C. The temperature increase is caused by the following reaction:

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

Calculate ∆H for the reaction, assuming 100.0g of solution, and a specific heat of 4.184 J/g-°C.

-68000 J/mol or -68 kJ/mol (q= 338.9 J)

Would the ∆H value be different if the molarity of your reactants was different?

Write a net ionic equation for the reaction, just for fun.

Hess’s Law section 5.6

Allows you to calculate ∆H for a reaction without measuring anything, using known ∆H values for other reactions.

Stated: if a reaction is carried out in a series of steps, ∆H for the overall reaction will equal the sum of the enthalpy changes for the individual steps.

Some examples:

Enthalpy of Formation section 5.7

Using Hess’s Law, the enthalpy change can be calculated for many different kinds of reactions:

∆Hvap-heat of vaporization-converting liquid to gas

∆Hfus-heat of fusion-melting a solid

∆Hcomb-heat of combustion-combusting a substance in oxygen

∆Hf-heat of formation-heat associated with the formation of a compound from its elements. This one is kind of important.

Since the amount of enthalpy change depends on temp., pressure, and state (phase), it helps to compare reactions at what is called standard state, which is defined as 1 atm pressure and 298K or 25°C. From this, we have developed tables of enthalpy data called standard enthalpy changes.

Standard enthalpy changeSymbolized as ∆H°, standard enthalpy change is the enthalpy

change that occurs for a reaction when all of the reactants and products are at their standard state. That is, whatever state they are in at 1 atm and 298K.

So, a reaction with Fe(l) or H2O(s) could not be occurring at standard state conditions.

Even more specific, ∆H°f, or standard enthalpy of formation, is the enthalpy change that occurs when 1 mole of a compound is formed from its elements at standard state conditions.

These are listed in Appendix C in the back of the textbook.

∆H°f Std. enthalpy of formationLet’s try writing a few equations, with their ∆H°f component

alongside (from Appendix C):HBr

AgNO3

Hg2Cl2

C2H5OH

NH3

SO2

RbClO3

NH4NO3

∆H°f Std. enthalpy of formation

If an element exists in more than one form at standard conditions, the most stable form of the element is listed. For example Cgraphite would be listed instead of Cdiamond.

∆H°f of elements is zero, because no energy is needed, the element already would exist in its standard state at standard conditions.

∆H°f equation or not (and if not, why not?):

2Na(s) + ½O2(g) → Na2O(s)

2K(l) + Cl2(g) → 2KCl(s)

C6H12O6(s) → 6C(diamond) + 6H2(g) + 3O2(g)

Using ∆H°f to find ∆HrxnC3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆Hcomb = ?

We can “add” these three equations to find ∆Hcomb.

C3H8(g) → 3C(s) + 4H2(g) ∆H1 = -∆H°f[C3H8(g)]

3C(s) + 3O2(g) → 3CO2(g) ∆H2 = 3∆H°f[CO2(g)]

4H2 + 2O2(g) → 4H2O(l) ∆H3 = 4∆H°f[H2O(l)]

_______________________________________________ C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆Hrxn = ∆H1 + ∆H2 + ∆H3

You can use Hess’s Law and ∆H°f equations from a table to calculate ∆Hrxn for any reaction, so long as you choose the necessary reactions to work with.

The short version: To find the heat of reaction, ∆Hrxn,total all of the ∆H°f‘s for the products, total all of the ∆H°f’s for the reactants, and subtract the reactant total from the product total.

Or:

∆H°rxn= Σn∆H°f(products) – Σm∆H°f(reactants)

Now try # 69 and 71 in the textbook.