Thermochemistry

Question 1

Thermochemistry

Answer 1

is the study of the energy changes that occur during changes in matter, including:
Physical changes
Chemical changes
Nuclear changes

Question 2

Thermal energy

Answer 2

is the total amount of energy in a substance. Thermal energy includes:
Potential energy (caused by position or composition)
Kinetic energy (caused by motion)
Temperature describes a substance’s average kinetic energy.

Question 3

Heat (q)

Answer 3

describes the transfer of thermal energy from warmer objects to cooler objects

Question 4

The Law of Conservation

of Energy

Answer 4

“Energy can be neither created nor destroyed, it can only be transferred or transformed.”

he transfer of energy in a chemical or physical change involves two parts:
System = the substances undergoing a physical or chemical change (ex. reactants and products)
Surroundings = all other nearby matter that interacts with the system

Systems may either be open or closed.
Open Systems: Energy and matter are exchanged with the surroundings (ex. a barbeque)
Closed Systems: Only energy can be exchanges with the surroundings (ex. glow sticks)

Question 5

The Law of Conservation of Energy continued

Answer 5

Physical and chemical changes either absorb or release thermal energy:
Endothermic processes absorb energy from the surroundings.
Exothermic processes release energy to the surroundings.

The amount of thermal energy transferred between the system and its surroundings can be represented as follows:
qsystem = − qsurroundings

Question 6

Specific Heat Capacity

Answer 6

Specific heat capacity (c) is the amount of thermal energy needed to raise the temperature of 1g of a substance by 1°C.
Units are in J/g•°C

Question 7

Measuring Energy Changes

Answer 7

Energy changes can be measured indirectly by measuring temperature changes that occur during a physical or chemical change.
This technique is called calorimetry and involves a device called a calorimeter.

Question 8

Measuring Energy Changes(continued)

Answer 8

The amount of thermal energy released or absorbed during a physical or chemical change can be calculated using the following equation:
q = mcΔT
Where:
q = total thermal energy/heat (in joules [J])
c = specific heat capacity (in J/g•°C)
m = mass (in grams [g])
ΔT = temperature change (in °C)

The magnitude of q tells us how much energy was transferred and the sign of q tells us the direction of energy transfer (into or out of the system).

Question 9

Measuring Energy Changes

Answer 9

When using calorimetry, four assumptions are made:

1. Any thermal energy transferred from the calorimeter to the outside environment is negligible.
2. Any thermal energy absorbed by the calorimeter itself is negligible.
3. All dilute, aqueous solutions have the same density (1.00g/mL) and specific heat capacity (4.18 J/(g•°C)) as water.
4. The chemical/physical change occurs at a constant pressure.

Question 10

Using a Simple Calorimeter

Answer 10

When you using a simple calorimeter, you must make the following assumptions:

1. The system is isolated
2. The thermal energy that is exchanged with the polystyrene cups, thermometer, stirring rod and lid is small enough to be ignored

Question 11

Enthalpy Change of Reaction

Answer 11

Since the reaction is at constant pressure, the heat exchanged by the system and the water is equal to the experimental enthalpy change of the system

ΔHsystem = - Qsolution

Question 12

Molar Enthalpy Change of Reaction

Answer 12

f you know the number of moles of reactant or product, you can determine the molar enthalpy change of the reaction (ΔHr)

ΔH = nΔHr

ΔHr = ΔH

     n

Question 13

MEASURING ENERGY CHANGES

Answer 13

Calorimetry provides accurate information for enthalpy changes in most situations, but it’s not always practical since some reactions:
occur too slowly
are too dangerous
have too small of a temperature change to measure

Hess’s Law and standard heats of formation are two other methods of calculating enthalpy changes. These methods can be used in any situation.

Question 14

HESS’S LAW

Answer 14

Hess’s Law states that the enthalpy change during a reaction is independent of the path taken and equal to the sum of the enthalpy changes in the individual steps in the process.

Question 15

HESS’S LAW(continud)

Answer 15

Using Hess’s Law involves manipulating chemical equations in several ways. Two rules must be followed when applying Hess’s Law:
If the chemical reaction is reversed (flipped), the sign (+/-) of ΔH is reversed.
If the coefficients in a balanced chemical equation are multiplied/divided by a factor, the magnitude of ΔH is multiplied/divided by the same factor.

Question 16

STANDARD HEATS OF FORMATION

Answer 16

A formation reaction is a chemical reaction that forms a substance from its elements.

Standard enthalpy of formation (ΔHf°) is the enthalpy change that occurs when one mole of a substance is formed from its elements under standard conditions.
Standard conditions = SATP (25°C (298 K) and 100 kPa)

Question 17

STANDARD HEATS OF FORMATION

continued

Answer 17

Data tables of standard enthalpies of formation and the following equation can be used to calculate the enthalpy change of a chemical reaction:

ΔH = ΣnΔHf(products) – ΣnΔHf(reactants)

IMPORTANT: The standard enthalpy for any element in its standard state (most stable state) is always zero.

Question 18

SAMPLE PROBLEM(Hess’s Law)

Answer 18

Use standard enthalpies of formation to calculate the enthalpy change that occurs during the complete combustion of methane gas:

CH4(g) + O2(g) 🡪 CO2(g) + 2 H2O(l)

ΔH = ΣnΔHf(products) – ΣnΔHf(reactants)
= [(1 x CO2) + (2 x H2O)] – [(1 x CH4) + (1 x O2)]
= [(1 x – 393.5) + (2 x – 285.8)] – [(1 x – 74.6) + (2 x 0)]
= – 965.1 – (–74.6)
= – 890.5 kJ

∴ The enthalpy change for this reaction is − 890.5 kJ

Question 19

Efficiency

Answer 19

the ratio of useful energy produced (output) to energy used in its production (input), expressed as a percentage.

Question 20

efficiency eqaution

Answer 20

efficiency = energy output x 100%

energy input

Question 21

Sample Problem(Effciency)

Answer 21

Propane, C3H8(g), is a commonly used barbecue fuel. Determine the efficiency of the barbecue as a heating device if 5.10g of propane is required to change the temperature of 250.0g of water contained in a 500.0g stainless steel pot (c = 0.503 J/g.oC) from 25.0oC to 75.0oC.
What information do you have?
What information do you need?
Which energy is the input, and which is the output?

Question 22

sample problem (Energy input)

Answer 22

Useful’ energy plus ‘wasted’ energy - calculate using Hess’ law and ΔH = nΔHx
Write a balanced chemical equation for the complete combustion of propane, and calculate ΔHocomb by using enthalpies of formation

Calculate the amount of moles of propane combusted from the mass and determine the theoretical energy content of the fuel

calculations
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
ΔHor = ∑ (nΔHof products) - ∑ (nΔHof reactants)
= [(3mol)(ΔHof CO2(g)) + (4mol)(ΔHof H2O(g)] - [(1mol)(ΔHof C3H8(g)) + (5mol)(O2(g))]
= [(3mol)(-393.5kJ.mol)+(4mol)(-241.8kJ/mol)]-[(1mol)(-103.8kJ/mol)+(5mol)(0kJ/mol)]
= (-2147.7kJ) - (-103.8kJ)
= -2043.9 kJ

Therefore, the ΔHocomb of propane is -2043.9kJ/mol

) Amount of moles of propane combusted
n = m/MM
= 5.10g / 44.11g/mol
= 0.1156mol

Therefore, the energy content of propane is
ΔH = nΔHocomb
= (0.1156mol)(-2043.9kJ/mol)
= -236kJ

Question 23

sample problem efficiency(energy output)

Answer 23

Determine how much energy was absorbed by the pot and the water using Q=mcΔt
Qtotal = Qwater + Qsteel
= (mcΔt)water + (mcΔt)steel
= (250.0g)(4.19J/g.oC)(75.0oC-25.0oC) + (500.0g)(0.503 J/g.oC)(75.0oC-25.0oC)
= 52,375J + 12,575J
= 64950J
= 6.50 x 104J or 65.0 kJ

efficiency = energy output x 100%
energy input

                                                    = (65.0kJ / 236kJ) x 100%
                                                    = 27.5% Therefore, as a heating device, the barbecue was 27.5% efficient.

Question 24

Factors Affecting Reaction Rate

Answer 24

Nature of the reactants
Concentration 
Temperature
Pressure (gases)
Surface area (solid)
Presence of a catalyst

Question 25

1. Nature of the Reactants(factors affecting rate)

Answer 25

Small molecules react faster than large molecules Gases react faster than liquids; liquids react faster than solids Ions react faster than molecules The same is true for acids and bases, due to the attraction of oppositely charged ions

Question 26

. Concentration(factors affecting rate)

Answer 26

Increasing the concentration of reactants leads to a greater number of collisions per unit time, because there are more particles in the same volume

Question 27

3. Temperature(factors affecting rate)

Answer 27

When the temperature increases, particles have more kinetic energy. he frequency of collisions increases, and the number of effective collisions increases per unit time. This is because, as the temperature of the sample increases, the fraction of collisions with energy equal to or greater than Ea increases.

Question 28

4. Pressure(factors affecting rate)

Answer 28

Increasing the pressure increases the number of collisions per unit time, increasing the reaction rate. According to Boyle’s law, the pressure can be increased by adding more reactant gas particles to a fixed volume or by decreasing the volume of the reaction container.

Question 29

5. Surface Area(factors affecting rate)

Answer 29

Smaller pieces of reactants have a greater amount of exposed surface area compared to larger pieces that have the same total mass. Therefore the chances of effective collisions increases and the rate of reaction increases

Question 30

6. Catalysts(factors affecting rate)

Answer 30

When the Ea is lowered, a larger fraction of reactants have enough kinetic energy equal to or greater than Ea and the reaction proceeds more quickly. The catalyst may take part in the chemical reaction, but it is not consumed during the reaction. It is returned to its original condition by the end of the reaction. A catalyzed reaction is a reaction in which a catalyst has been used.

Question 31

collision theory

Answer 31

The rate of a reaction is proportional to the rate of reactant collisions: The orientation of the reactants (the collision geometry) must be favourable The collision must occur with sufficient energy (activation energy)

Question 32

Orientation of Reactants(collision theory)

Answer 32

In order for a reaction to occur, reacting particles must collide with the proper orientation relative to one another. This is known as collision geometry.

Question 33

Activation Energy (Ea)

Answer 33

Reactant particles must collide with one another with sufficient energy to occur

Question 34

Maxwell-Boltzman Distribution

Answer 34

he Maxwell-Boltzman Distrubution is a graph comparing the total number of particles in a chemical system and their kinetic energy. The area under the curve represents the distribution of the kinetic energy of collisions at a given temperature. The shaded part of the graph indicates the fraction of the total collisions that have energy equal to or greater than the activation energy.

Question 35

Activated Complex(PE diagrams)

Answer 35

During the transition state, an-in between product called the activated complex forms

Question 36

Activation Energy and Enthalpy

Answer 36

Generally, endothermic and exothermic reactions with low activation energy tend to proceed more quickly at room temperatures, and those with high activation energies tend to proceed more slowly The enthalpy change of a reaction is independent of the activation energy

Question 37

Reversible Reactions

Answer 37

Many chemical reactions can proceed in two directions: forward and reverse A potential energy diagram can represent the reaction in both directions

Question 38

Reaction Mechanism

Answer 38

he process, or pathway, by which a reaction occurs An overall reaction may consist of two or more elementary steps (where rate law is directly derived from stoichiometric coefficients!) The complete sequence of elementary steps that make up a overall reaction is called a reaction mechanism

Question 39

Reaction Mechanism(example)

Answer 39

The experimentally derived reaction rate for 2 NO2Cl → 2 NO2 + Cl2 is rate = k[NO2Cl] a) What SHOULD the rate be if it was an elementary (single step) mechanism? answer: What would be the expected rate law for this one-step reaction: NO2 + CO → CO2 + NO

Question 40

reaction mechanisms (example)

Answer 40

Since the reaction IS NOT second order, the actual mechanism MUST NOT BE ELEMENTARY. In fact, experimentally-derived, this is a two-step process: NO2Cl → NO2 + Cl NO2Cl + Cl → NO2 + Cl2 Which step is slow? answer: The actual rate law for this reaction is rate = k[NO2 ]2

Question 41

Concentration and Reaction Rates

Answer 41

The rate of a chemical reaction is directly related to the concentration of the reactants

Question 42

Initial Reaction Rates

Answer 42

To determine the relationship between concentration and rate chemists run a chemical reaction several times using different starting concentrations each time and keeping the temperature and other conditions constant. They then determine the initial rate of reaction (the instantaneous rate of reaction at time zero)

Question 43

Rate Law

Answer 43

Any expression that relates reaction rate to the concentration of reactants is called a rate law In chemistry, slope is represented by the letter k, so the general equation for this type of reaction is rate = k[A]

Question 44

Rate law

Answer 44

k is also referred to as the rate constant, and depends on the nature of the reactants and on temperature The general rate law can be expressed as: rate = k[A]m[B]n The exponents “m” and “n” are called orders

Question 45

Order of Reaction

Answer 45

order of reaction describes how much a change in the amount of each substance affects the overall rate The values of “m” and “n” can only be determined experimentally example If rate = k[B], the reaction is first order for reactant “B” If rate = k[A]2, the reaction is second order for reactant “A”

Question 46

order of reaction

Answer 46

The overall order of the reaction = “m” + “n” For example: rate = k[A]2[B] Overall this is a third order reaction (i.e. 2+1)