• What do you visualize when thinking about chemical reactions? You probably imagine…o Heat, Light, Sound

• These are energy, and chemical reactions involve energy because chemical bonds are being broken and made!

• Does the reaction in the picture release energy or absorb energy?

Chemical Reactions That Involve Heat

• Energy: the capacity to do work.• What kinds of energy are there?o Radiant (solar)o Thermal (motion of atoms & molecules)o Chemical (stored within chemical bonds)o Potential (available because of an object’s

position)o Kinetic (energy of motion)

• Heat is the transfer of thermal energy between two objects at different temperature.

• Thermochemistry is the study of heat changes that occur in chemical reactions.o ‘Thermes’ is Greek for “heat.”

• Thermodynamics is the broader study of energy and work changes during such processes.

Learning Target:Be able to calculate the amount of energy absorbed or released during a chemical or physical change.

Learning Outcomes:Complete the endothermic exothermic

minilab and calculations for ∆H.

CHAP. 12 OUTLINE DUE TODAY!!!

• Remember the “Law of Thermodynamics”o “Energy cannot be created or destroyed, but may be

converted from one form to another.”

• To study thermochemistry we must be careful about the ‘system’ involved.o Open System: allows exchange of both energy and mass.o Closed System: allows the exchange of energy, but not

mass.o Isolated System: allows the exchange of neither energy nor

mass.

OPEN

CLOSED

ISOLATEDEnergy Lossor Gain

Mass Lossor Gain

Energy Lossor Gain

Lab ActivityBreak out into lab groups to collect data

on the reaction that is given to you at your lab station.

When you have collected your data we will compile it to investigate endothermic versus exothermic reactions.

• Exothermic reactions release heat energy.o ‘Exo’ means outside.

CH4(g) + 2O2(g) →CO2(g) + 2H2O(g) + Heat

• Endothermic reactions absorb heat energy.o ‘Endo’ means inside.

2HgO(s) + Heat →2Hg(l) + O2 (g)

• Most chemical processes we use occur at normal, constant atmospheric pressure, and heat energy is either gained or lost during the chemical reaction.

• Chemists use a special term, enthalpy, to describe heat changes for chemical reactions at constant pressure.

• Enthalpy is the energy of a substance plus a small adjustment for pressure-volume work. o Enthalpy (H) = E + PV = Internal Energy + Pressure-Volume (Work) = Heat absorbed or released by a reaction

• U.S. Army “MREs” (“Meals ready to eat”)o Magnesium reacts with water to release heat and

hydrogen gas. Write the reaction:o Mg + 2H2O →Mg(OH)2 + H2 + Heato MREs heat the food to 60° C in 15 minutes!o EXOTHERMIC Reaction

• Ammonium Nitrate (A common fertilizer)o NH4NO3 dissolves in water, causing a drop in

temperature. Write the reaction:o NH4NO3 + H2O + Heat →NH4

1+ (aq) + NO31- (aq)

o NH4NO3 is used in cold packs for sports injuries.o ENDOTHERMIC Reactiono Note: NH4NO3 can also undergo an EXOTHERMIC

reaction when it explodes, hence its use as a blasting agent in mining and (unfortunately) in terrorists’ weapons, such as that used in the Oklahoma City attack.

Learning Target:Be able to calculate the amount of energy absorbed or released during a chemical reaction.

Learning Outcomes:Complete the Heat Calculation Problems.

• Enthalpy change, ΔH, for a chemical reaction that occurs at constant pressure is the heat released or absorbed in the reaction.

ΔH is just the difference between the enthalpy of the products minus the enthalpy of the reactants:

ΔH = ∑Hproducts – ∑Hreactants

Think of enthalpy in terms of a savings account balance. You are just keeping track of money going into the account and the money spent vs. the net balance.

• We use diagrams to show enthalpy changes.• For exothermic reactions, Hproducts is lower than

Hreactants so ΔH is negative.o Heat is released (lost by the system).

• For endothermic reactions, Hproducts is higher than Hreactants so ΔH is positive.o Heat is absorbed (gained by the system).

• See diagrams on the next slide.

CH4(g) + 2O2(g) →CO2(g) + 2H2O(g) + Heat

ENTHALPY

REACTION PROGRESS (TIME)

CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)

ΔH = -803kJ(EXOTHERMIC; HEATRELEASED FROM THE‘SYSTEM’ TO THESURROUNDINGS)

NOTE: Hproducts ISLOWER THANHreactants

2HgO(s) + Heat →2Hg(l) + O2 (g)

ENTHALPY

REACTION PROGRESS (TIME)

2Hg(l) + O2(g)

2HgO(s)

ΔH = +181kJ(ENDOTHERMIC; HEATABSORBED BY THE‘SYSTEM’ FROM THESURROUNDINGS)

NOTE: Hproducts ISHIGHER THANHreactants

What is the enthalpy change for the following reaction?

The ΔH for Mg(s) and H2(g) is 0 kJ/mol

a. Calculate the ΔH for the above reaction.b. Is the reaction exothermic or endothermic?c. Create an enthalpy diagram.

Mg(s) + 2H2O(l) → Mg2+(aq) + 2OH-

(aq) + H2(g)

Learning Target:Be able to use thermochemical equations to determine enthalpy change (ΔH) associated with a given amount of reactant.Learning Outcomes:1.Know how to calculate ΔH, and produce an enthalpy diagram. 2.Know how to calculate enthalpy for a given amount of reactant.

Learning Target:Be able to calculate the amount of energy absorbed or released during a chemical or physical change.

Learning Outcomes:Complete the Heat Calculation Problems.

Warm UpCalculate the enthalpy change (∆H) for the

following reaction:b. C2H5OH(l) + 3O2 (g) → 2CO2(g) + 3H2O(g)

How much heat will be released when 8.76g of oxygen gas reacts with excess sulfur by the reaction:

2S(s) + 3O2 (g) →2SO3 (g) ΔH° = -791.4kJ

Solution:

.

Calculate the enthalpy (∆H°) for the following reaction:

2NH3(g) + 4H2O(l) → 2NO2(g) + 7H2(g)

How much energy would be released if 50.0g of water were reacted?

How much energy is released if 10.0g of ethanol, C2H5OH , were burned?

C2H5OH(l) + 3O2 (g) → 2CO2(g) + 3H2O(g)

∆H° = -1234.8 kJ

Warm UpUsing the standard enthalpies of formation

table, calculate standard enthalpy change (ΔH°) for the following reactions (Balance first):

CH3OH(l) + O2(g) → CO2(g) + H2O(g)

Using the above balanced equation and the ΔH° you calculated, how much heat will be released when 24.0 g of methanol (CH3OH) reacts with excess oxygen?

Learning Target:Be able to determine the amount of standard enthalpy change (ΔH°) for a series of reactions by applying Hess’ Law.Learning Outcomes:

Understand how to calculate ΔH° for a series of reactions.

• To be exact when comparing enthalpy changes chemists must define the conditions, such as pressure, temperature and phase.

• The “Standard State” of a substance is its pure form at the standard pressure of 1 atmosphere and 25° C.o Standard state of oxygen is as the diatomic gas.o Standard state of carbon is graphite.o Standard state of phosphorus is P4 (white phosphorus).

• Enthalpy changes measured under “standard state” conditions are called “Standard Enthalpy Changes” (ΔH°).

• Hess’s Law: If a series of chemical reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.

Hess’ Law ExampleExample

N2 (g) + O2 (g) 2NO(g) ΔH1 = 181 kJ2NO (g) + O2 (g) 2NO2(g) ΔH2 = -113 kJ

Net equation: N2 (g) + 2 O2 (g) + 2NO (g) 2NO(g) + 2NO2(g)

To write the net equation, the sum of the reactants is placed on the left hand side, and the sum of the products is placed on the right side. Substances on both sides cancel out – like algebra. In example above the next reaction is:N2 (g) + 2 O2 (g) 2NO2(g)

ΔH = ΔH1 + ΔH2 = 181kJ + -113 kJ = 68kJ (endo)

Applying Hess’s LawIf the coefficients of an equation are multiplied

by a factor, the enthalpy change the reaction is multiplied by the same factor.

If the equation is reversed, the sign of the ΔH changes also.For example

C (s) + H2O(g) CO(g) + H2(s) ΔH = 113kJ

Then CO(g) + H2(s) C (s) + H2O(g)

ΔH = -113kJ

Hess’s Law Example• Calculate the ∆H for the reaction:

2C(s) + O2(g) → 2CO(g) from

C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ

2CO(g) + O2(g) → 2CO2(g) ΔH = -566.0 kJ

• The trick is to simply combine equations so that only the relevant parts are present (everything else cancels out). Remember, whatever you do to the equations, you must do to the value of ΔH for the reaction.

• Calculate the ∆H for the reaction: 2S(s) + 2OF2(g) → SO2(g) + SF4(g)

OF2(g) + H2O(l) → O2(g) + 2HF(g) ∆H = -277 kJ

SF4(g) + 2H2O(l) → SO2(g) + 4HF(g) ∆H = -828kJ

S(s) + O2(g) → SO2(g) ∆H = -297kJ

Exit QuestionFrom the following enthalpy changes, H2S(g) + 3/2O2(g) → H2O(l) + SO2(g) ∆H° = -563kJCS2(l) + 3O2(g) → CO2(g) + 2 SO2(g) ∆H° = -1075kJ

Calculate the value of ∆H° for CS2(l) + 2H2O(l) → CO2(g) + 2H2S(g)

• What happens during chemical reactions?o During an exothermic reaction, the surroundings gain heat from

the ‘system.’o During an endothermic reaction, the surroundings lose heat to

the ‘system.’

• Calorimetry is the study of heat flow and heat measurement during chemical reactions.

• Calorimetry experiments use accurate measurements of temperature changes to determine the enthalpy changes (heat flow).

12-4 Calorimetry

Calorimeter Experiment

Learning Target:Be able to use the equation “q = m . C . ΔT” to solve for heat absorbed or released, mass, specific heat or temperature change of a substance during a chemical or physical change.Learning Outcomes:

Understand how to calculate for the variables in the heat equation.

Finish calculations for Hess’s Law Lab

• Heat Capacity: the amount of heat needed to raise the temperature of an object by 1 degree Celsius (1° C).

• Specific Heat: the heat capacity of one gram of a substance.o Specific Heat of water = 4.184 J/g.C°o To raise the temperature of one gram of liquid water by 1

degree C requires 4.184 J of heat energy, or 1 calorie (cal.)

• Note that heat and temperature are related, but different, concepts.

• We detect heat changes by measuring temperature changes.o A small temperature change does not mean a small

amount of heat transfers! (Consider melting a lake!)o Use specific heat and mass to measure heat changes.o Heat equation:

q = m X C X (Tf – Ti)

HEAT vs. TEMPERATURE

Hess’ Law LabOn a separate sheet of paper answer the

following:1.Write 2 specific experimental errors that

occurred in your lab? (Don’t just give me I made a mistake!)

2.How did they impact your results?

A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25°C to 175°C. Calculate the specific heat capacity of iron.  

How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C?

• Calorimeter: a well insulated container filled witha known mass of water, a way to conduct a chemical reaction, a stirrer and a thermometer.

• Measure the initial temperature of the water (Ti).• Conduct the chemical reaction.• Measure the temperature periodically.• Determine the final temperature (Tf).• Note that all heat released by a reaction is gained by the water in the calorimeter, and vice versa!• Mathematically we may say: (qrxn = -qsur)• Calculate the quantity of heat absorbed by the water (qsur) from the

mass of water, the Specific Heat of water and the temperature change (Tf – Ti). (And vice versa.)

qsur = m X C X (Tf – Ti)

Calorimeter Experiment

Calorimeter Experiment

• Our body is like a chemical factory, taking in raw materials of various kinds to make new products that keep us alive and help us grow.

• These reactions involve bond-breaking and bond-making, and they meet the energy needs of our body.

• Carbohydrates and fats are major food sources, and they release lots of energy (exothermic reactions).

• Glucose releases -2803kJ/mol C6H12O6

• See Table on page 399.• “Calories Count” Project• Recall 1000 cal = 1 Kcal = 1 Cal (the food kind!)

• “Caloric Theory”o Heat was thought of as an invisible, weightless fluid capable

of flowing from a hot object to a colder one.• Benjamin Thompson (Count Rumford) showed that

friction heat between objects was continuous as long as two objects were rubbed together.o This was not consistent with ‘Caloric Theory.’

• James Joule used a paddlewheel experiment to convert mechanical energy into heat, thus showing there is a ‘mechanical equivalent of heat.’

• Modern “Kinetic Theory”o Heat is defined as the transfer of kinetic energy from a hotter

object to a colder one.