Thermochemistry – Heat and Chemical Change Thermochemistry is the study of heat transfer in chemical and physical processes.

Explain the relationship between energy and heat Objectives Explain the relationship between energy and heat Explain the difference between heat and temperature Construct equations that show the heat changes for chemical and physical processes Calculate heat changes in chemical and physical processes

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

Energy Potential energy (PE) is possessed by an object by virtue of its position. (Chemical potential energy is the energy stored in chemical bonds.) Kinetic energy (KE) is the energy of motion. This form of energy depends on the mass (m) and velocity (v) of the moving object. KE = ½ mv2

A. The Nature of Energy Potential energy (PE): Kinetic energy (KE): the energy of position the energy of motion

Law of conservation of energy Energy is neither created nor destroyed in a chemical or physical process. (Energy can be converted from one form to another.)

B. Temperature and Heat Temperature is a measure of the average kinetic energy in matter. Hot water (90oC) Cold water (10oC)

B. Heat (q) is…. the energy that transfers between two objects due to a temperature difference between them. Heat is transferred from a hot object to a colder object (never in the other direction). Hot water (90oC) Cold water (10oC) Water (50oC) Water (50oC)

Heat vs. Temperature - Question: Which represents more heat? A bathtub full of 100o C water or a teacup full of 100o C water? Answer: A bathtub full! The amount of heat depends partly on the mass of the substance.

In studying heat changes…. System – part of the universe on which we focus attention Surroundings – everything else in the universe Example: Burning a match

Exothermic and Endothermic Processes Exothermic – energy flows out of the system Endothermic – energy flows into the system

Units of energy: The joule (J) is the SI unit of energy. Another common unit of energy is the calorie (cal). 1 calorie = 4.18 joules

Question… How many joules are in 375 calories? Answer: 375 cal x 4.18 J/cal = 1567.5 joules

In food, a Calorie is the same as a kilocalorie. If an apple provides 120 Calories, that is the same as… 120 000 calories, or 501 600 joules of energy. (120 Cal)(1000 cal/1 Cal)(4.18 J/cal) = 501 600 J

Specific heat capacity (or specific heat) is… The amount of energy needed to change the temperature of 1.0 gram of a substance by 1.0 degree Celsius (or 1.0 kelvin).

The unit of specific heat is… Expressed in either joules or calories. The units are either J/(g . oC) or cal/(g . oC).

Table of Specific Heats:

The specific heat of water The specific heat of water is 4.18 J/g . K The specific heat of aluminum is 0.902 J/g . K. It takes over 4 times as much heat to raise the temperature of a gram of water by a certain amount than a gram of aluminum.

B. Calculating Heat Transfer To calculate the energy required for a reaction: Heat transferred = (specific heat) (mass) (change in temp) q = C m T

In solving heat problems: q = C m DT where the quantity of heat = q , and the change in temp (DT ) = (final temp – initial temp) or DT = (Tfinal - Tinitial).

Problem: How many joules must be transferred from a cup of coffee to your body if the temperature of the coffee drops from 60.0oC to 37.0oC (normal body temp)? Assume the cup holds 250. mL (with a density of 1.0 g/mL, the coffee has a mass of 250. g), and that the specific heat of coffee is the same as that of water.

Answer: q = C m DT = -24100 J = -24.1 kJ q = (4.18 J/g . K) (250. g) (37.0oC – 60.0oC) = -24100 J = -24.1 kJ Note: the negative value denotes the direction of heat transfer; it shows that energy is transferred from the coffee as the temperature declines.

Problem: In the previous problem, the cup of coffee lost 24.1 kJ when cooled from 60.0oC to 37.0oC. If this same amount of heat is used to warm a piece of aluminum weighing 250. g. what would be the final temperature of the aluminum if its initial temperature is 37.0oC? (The specific heat of aluminum is 0.902 J/g . K.)

(0.902 J/g . K) (250. g) (Tfinal - 37.0oC) Tfinal = 144oC Answer: q = C m DT 24100 J = (0.902 J/g . K) (250. g) (Tfinal - 37.0oC) Tfinal = 144oC

Calorimetry A. Thermochemistry Heat transfer is measured using a calorimeter.

Problem: Suppose you heat a 55.0 g piece of iron in the flame of a Bunsen burner to 425oC and then you plunge it into a beaker of water. The beaker holds 600. mL water (density = 1.00 g/mL), and its temperature before you drop in the hot iron is 25.0oC. What is the final temperature of the water and the piece of iron?

C m DTmetal = - C m DTwater Answer (set up): Heat lost by metal = - (Heat gained by water) qmetal = - qwater C m DTmetal = - C m DTwater (55.0 g) (0.451 J/g . K) (Tfinal - 425oC) = - (600. g) (4.18 J/g . K) (Tfinal - 25oC)

Problem solved… - (600. g) (4.18 J/g . K) (Tfinal - 25oC) 24.805 (Tfinal - 425) = - 2508 (Tfinal - 25) 24.805 Tfinal – 10542 = -2508 Tfinal + 62700 2532.8 Tfinal = 73242. Tfinal = 29oC

Remember: always subtract the initial temp from the final temp; this results in a calculation that indicates an increase (+) or a decrease (–) in the heat transferred. If q is +, heat is transferred into the object (endothermic), if q is –, heat is transferred out of the object (exothermic).

Problem: You want to cool down a cup of coffee, and you do so by dropping in a cold block of aluminum. The aluminum block has a mass of 250. g and you want to cool 250. g of coffee (with a specific heat of 4.18 J/g . K) from 60.oC to 45oC. To accomplish this, what must the initial temperature of the aluminum block be?

Answer: Heat transferred from coffee = (250. g) (4.18 J/g . K) (45oC – 60.oC) = - 15675 J Heat transferred to aluminum = 15675 J = (250. g) (0.902 J/g . K) (45oC – Tinitial) Tinitial = -25oC

Molar Heat Capacity… Another way to express the capacity of a substance to absorb heat is to give its molar heat capacity. This measure uses 1 mole of a substance rather than 1 gram. Its unit is J/mol . K.)

Enthalpy For systems at constant pressure, the heat content is also called the enthalpy (H) of the system. All of the heat changes we have discussed occur at constant pressure, so for these processes, q = DH.

Thermochemical equations CaO + H2O  Ca(OH)2 + 65.2 kJ You can treat heat change in a chemical reaction like any other reactant or product in a chemical equation. An equation that includes the heat change is called a thermochemical equation. This equation includes a heat of reaction.

Thermochemical Equations When carbon disulfide is formed from its elements, heat is absorbed. Calculate the amount of heat (in kJ) absorbed when 5.66 g of carbon disulfide is formed. C(s) + 2S(s)  CS2(l) ∆H= +89.3 kJ C(s) + 2S(s) + 89.3 kJ  CS2(l)

Thermochemical equations The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction. How many kilojoules of heat are produced when 3.40 mol Fe2O3 reacts with an excess of CO? Fe2O3 + 3CO  2Fe + 3CO2 + 26.3 kJ Fe2O3 + 3CO  2Fe + 3CO2 ∆H= - 26.3 kJ

In thermochemical equations… If heat is a reactant, energy is absorbed and the reaction is endothermic. If heat is a product, energy is released and the reaction is exothermic.

Heat of fusion The heat absorbed by one mole of a substance in melting is the molar heat of fusion (DHfus). The heat lost when one mole of a substance solidifies is the molar heat of solidification (DHsolid).

DHfus = - DHsolid The amount of heat absorbed by melting a solid is exactly the same as the amount of heat lost when the liquid solidifies (but opposite in direction of heat flow). DHfus = - DHsolid

Heat of vaporization The amount of heat needed to vaporize one mole of a substance is called its molar heat of vaporization (DHvap). The amount of heat released when one mole of vapor condenses is called the molar heat of condensasion (DHcond).

DHvap = - DHcond The amount of heat absorbed by melting a solid is exactly the same as the amount of heat lost when the liquid solidifies (but opposite in direction of heat flow). DHvap = - DHcond

During a phase change… As a substance changes phase, no temperature change occurs so the heat involved is described as the latent heat of the phase change. Heat is added, but no temperature change is observed as a substance boils, melts, etc.

To solve for the amount of heat involved in a change of phase… q = m (DH)

Heating curve

Problem: How much heat is absorbed when 75.0 g H2O(l) at 100.oC is converted to steam at 100.oC? (The DHvap for water is 40.7 kJ/mol.)

Answer: Convert grams to moles: (75 g H2O)(1 mol/18.02 g) = 4.16 mol q = m (DH) q = (4.16 mol) (40.7 kJ/mol) = 169 kJ

B. Energy and Our World Fossil fuel – carbon based molecules from decomposing plants and animals Energy source for United States

B. Energy and Our World Petroleum – thick liquids composed of mainly hydrocarbons Hydrocarbon – compound composed of C and H

B. Energy and Our World Natural gas – gas composed of hydrocarbons

B. Energy and Our World Coal – formed from the remains of plants under high pressure and heat over time

B. Energy and Our World Effects of carbon dioxide on climate Greenhouse effect

B. Energy and Our World Effects of carbon dioxide on climate Atmospheric CO2 Controlled by water cycle Could increase temperature by 10oC

B. Energy and Our World New energy sources Solar Nuclear Biomass Wind Synthetic fuels