thermochemistry

THERMOCHEMISTRY

Energy

• The ability to do work or transfer heat.– Work: Energy used to cause an

object that has mass to move.– Heat: Energy used to cause the

temperature of an object to rise.

Definitions #1

Energy: The capacity to do work or produce heat

Potential Energy: Energy due to position or composition

Kinetic Energy: Energy due to the motion of the object

2

2

1mvKE

Definitions #2

Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between formsThe First Law of Thermodynamics: The total energy content of the universe is constant

Definitions #3

Internal energy:

The internal energy of a system is identified with the random, disordered motion of molecules; the total (internal) energy in a system includes potential and kinetic energy.

The sample is at a height from the ground etc).  Symbol for Internal Energy Change: ΔU

• A) is the sum of the kinetic energy of all of its components

• B) is the sum of the rotational, vibrational, and translational energies of all of its components

• C) refers only to the energies of the nuclei of the atoms of the component molecules

• D) is the sum of the potential and kinetic energies of the components

• E) none of the above

1) The internal energy of a system __________.

Definitions #4

Enthalpy:

Enthalpy is the amount of heat content used or released in a system at constant pressure

E = q + wE = change in internal energy of a system

q = heat flowing into or out of the system

-q if energy is leaving to the surroundings+q if energy is entering from the surroundings

w = work done by, or on, the system

-w if work is done by the system on the surroundings

+w if work is done on the system by the surroundings

2) Which one of the following conditions would always result in an decrease in the internal energy of a system?

• A) The system loses heat and does work on the surroundings.

• B) The system gains heat and does work on the surroundings.

• C) The system loses heat and has work done on it by the surroundings.

• D) The system gains heat and has work done on it by the surroundings.

• E) None of the above is correct.

• 3) The change in the internal energy of a system that releases 2,500 J of heat and that does 7,655 J of work on the surroundings is __________ J.

• A) -10,155 B) -5,155 C) 1.9x107 D) 10,155 E) 5,155

Work problemsChapter 5

• 5.25• 5.27 A and B• 5.31 All

Calorimetry

The amount of heat absorbed or released during a physical or chemical change can be measured…

…usually by the change in temperature of a known quantity of water1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

The Joule

The unit of heat used in modern thermochemistry is the Joule

2

2111

s

mkgmeternewtonjoule

1 joule = 4.184 calories

A Bomb Calorimete

r

A Cheaper

Calorimeter

Specific HeatThe amount of heat required to raise the temperature of one gram of substance by one degree Celsius.Substance Specific Heat (J/g·K)

Water (liquid) 4.18

Ethanol (liquid) 2.44

Water (solid) 2.06

Water (vapor) 1.87

Aluminum (solid) 0.897

Carbon (graphite,solid) 0.709

Iron (solid) 0.449

Copper (solid) 0.385

Mercury (liquid) 0.140

Lead (solid) 0.129

Gold (solid) 0.129

Calculations Involving Specific Heat

Tmsq

s = Specific Heat Capacity

q = Heat lost or gainedT = Temperature

change

OR

Tm

qs

4) The temperature of a 15-g sample of lead metal increases from 22 °C to 37 °C upon the addition of 29.0 J of heat. The specific heat capacity of the lead is __________ J/g-K. A) 7.8B) 1.9 C) 29 D) 0.13

Problems

• 5.53 a and B

State Functions depend ONLY on the present state of the system

ENERGY IS A STATE FUNCTION

A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane!

WORK IS NOT A STATE FUNCTION

WHY NOT???

State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

State Functions• However, we do know that the

internal energy of a system is independent of the path by which the system achieved that state.– In the system below, the water could

have reached room temperature from either direction.

State Functions

• Therefore, internal energy is a state function.

• It depends only on the present state of the system, not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

State Functions

• However, q and w are not state functions.

• Whether the battery is shorted out or is discharged by running the fan, its E is the same.– But q and w are

different in the two cases.

Work

When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.

w = −PV

Work, Pressure, and Volume

VPw Expansion Compression+V (increase)

-V (decrease)

-w results +w results

Esystem decreases

Work has been done by the system on the surroundings

Esystem increases

Work has been done on the system by the surroundings

Energy Change in Chemical Processes

Endothermic:

Exothermic:

Reactions in which energy flows into the system as the reaction proceeds.

Reactions in which energy flows out of the system as the reaction proceeds.

+ qsystem - qsurroundings

- qsystem + qsurroundings

Endothermic Reactions

Exothermic Reactions

• 5) Which one of the following is an exothermic process?

• A) ice melting B) water evaporating C) boiling soup D) condensation of water vapor • E) Ammonium thiocyanate and

barium hydroxide are mixed at 25 °C: the temperature drops

Enthalpy

H = E + PV

At constant pressure and volume the change in enthalpy is the heat gained or lost

H = q

Enthalpies of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Enthalpies of Reaction

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Hess’s Law“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

Hess’s Law

Hess’s Law Example Problem

Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O    Reaction Ho  

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

CH4 C + 2H2 +74.80 kJ

Hess’s Law Example Problem

Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O    Reaction Ho  

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ

Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

C + O2 CO2 -393.50 kJ

Hess’s Law Example Problem

Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O    Reaction Ho  

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ

Step #3: Multiply reaction #3 by 2

2H2 + O2 2 H2O -571.66 kJ

Hess’s Law Example Problem

Calculate H for the combustion of methane, CH4:

CH4 + 2O2 CO2 + 2H2O    Reaction Ho  

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ

CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ

2H2 + O2 2 H2O -571.66 kJ

Step #4: Sum up reaction and H

CH4 + 2O2 CO2 + 2H2O -890.36 kJ

Calculation of Heat of Reaction Calculate H for the combustion of

methane, CH4: CH4 + 2O2 CO2 + 2H2OHrxn = Hf(products) - Hf(reactants)

    Substance Hf  

CH4 -74.80 kJ

O2 0 kJ

CO2 -393.50 kJ

H2O -285.83 kJ

Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]

Hrxn = -890.36 kJ

 13) Given the following reactions the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2SO2 (g) +O2 (g) 2SO3(g) is __________ kJ.

2SO2 2S + 2O2 ΔH = 594 kJ  2SO3 2S + 3 O2 ΔH = 790 kJ 

6.

2SO2 2S + 2O2 ΔH = 594 kJ keep

2SO3 2S + 3 O2 ΔH = 790 kJreverse

2S + 3 O2 2SO3 ΔH = -790 kJ

Ans: -196kJ

Problems

5.63