Thermodynamics

The study of the changes of heat in chemical reactions

Are temperature and heat the same thing? Temperature − A measure of the average kinetic energy of random motion of the particles in a substance Heat − A measure of the total amount of energy transferred from an object of high temperature to one of low temperature

The 1 st Law of Thermodynamics First law of thermodynamics: Because energy is conserved, the internal energy of a system changes as heat flows in or out of it.

Endothermic vs Exothermic Reactions Endothermic Reaction – absorbs heat Exothermic Reaction – releases heat

Endothermic vs. Exothermic When heat is transferred from surroundings to system, reaction is endothermic Heat enters the system. When heat is transferred from system to surroundings, reaction is exothermic Heat exits the system

HEAT MEASUREMENTS

Measuring Heat Two units for energy joule (J) SI unit for energy One joule is equal to calories. calorie (cal) The amount of energy required to raise the temperature of one gram of pure water by one degree Celsius. − A Calorie (capitalized) is 1000 calories or one kilocalorie. This is what is used in nutrition. One calorie is equal to joules

Imagine… A hot day in Arizona…in your back yard is a metal barbeque and a glass of water. Would you want to stick your hand in the water or touch the bbq? You would want to touch the water because different substances have different abilities to absorb heat!!

Specific Heat The specific heat of a substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius. The specific heat of water is 1 cal or J

Some Specific Heat Capacities Substance Specific heat capacity in J/g(oC) Water4.184 Copper0.385 Iron0.449

Calculating Heat Absorbed/Released q = cm∆T q = heat absorbed/released c = specific heat of the substance m = mass of the sample in grams ∆T = change in temperature (T final – T initial ) * If heat is being absorbed, q will always be positive. If heat is being released, q will always be negative

What quantity of heat is required to raise the temperature of g of water from 45.6  C to 52.8  C? The specific heat of water is J/g(  C). Practice Problem 1

Practice Problem 2 What is the specific heat of silver if the temperature of a 15.4 g sample of silver is increased from 20.0 o C to 31.2 o C when 40.5 J of heat is added?

Practice Problem 3 What is the final temp of silver if the temperature of a 5.8 g sample of silver starts out at 30.0 o C and 40.5 J of heat is added? The specific heat of silver is.235 J/g(  C).

CALORIMETRY

Calorimetry: A laboratory technique that determines the amount of heat transferred in a chemical or physical process

Calorimeter A calorimeter is an insulated device used for measuring the amount of heat absorbed or released in a chemical reaction or physical process.calorimeter

We will use q=cm∆T We will use this equation to solve calorimetry problems but we will be using it twice.

Calorimetry If a substance is placed in a calorimeter, heat flows from the substance to the water, or vice versa. The change in temperature of the water can be used to calculate the specific heat of the substance.

Calorimetry Example You put 125 grams of water into a calorimeter and find that its initial temperature is ⁰ C. Then you heat a 50.0 grams sample of unknown metal to ⁰ C and put the metal into the water. The temperature of the water raises to ⁰ C.

The heat gained by the water is equal to the heat lost by the metal. LAW OF CONSERVATION OF ENERGY This quantity of heat can be calculated using the equation q=cm∆T. Calorimetry Example

First calculate the heat gained by the water. q water = (4.184 J/g. ⁰ C )(125 g)(29.30 ⁰ C – ⁰ C ) q water = 1940 J The heat gained by the water is equal to the heat lost by the metal. q metal = q water q metal = J Calorimetry Example

Now solve the equation for the specific heat of the metal. q=cm∆T -1940J = c(50.0g)( ⁰ C ) -1940J = c(-4285g ⁰ C ) c= J/g. ⁰ C Calorimetry Example

You can use the table to identify the unknown metal.

Givens Titanium m = 20.8 g T i = 99.5 o C T f = 24.3 o C C= ? Water m = 75.0 g T i = 21.7 o C T f = 24.3 o C C = J/g o C

Water q = cm ∆ T q = (4.184)(75.0)( ) q = 816 J If water is gaining 816 J, then the titanium has lost 816 J so make it negative and use it for q titanium.

Titanium q = cm  T -816 = (C)(20.8)( ) -816 = -1560(C) C = J/g( o C)

Example problem 2: A 0.45 gram piece of copper (C=0.385 J/g( o C)) at a temperature of 87.0 o C is placed in a calorimeter. The calorimeter contained water (C=4.184 J/g( o C)) at a temperature of 23.0 o C. The solution came to equilibrium at 24.8 o C, what was the mass of water in the calorimeter?

Givens Copper m = 0.45 g Cu C Cu = J/g o C T initial = 87.0 o C T final = 24.8 o C Water m = ? C H 2 O = J/g o C T initial = 23.0 o C T final = 24.8 o C

Copper q = cm ∆ T q = (0.385)(0.45)( ) q = J That is the heat lost ( J) by the metal, so the water must have gained 10.8 J Therefore q water = 10.8 J

Water q = cm ∆ T 10.8= (4.184)m( ) m = 1.5 grams H 2 O Since we knew q for copper, we could use that number (with the opposite sign) to get q for water since heat lost by one is gained by the other.

ENTHALPY

Enthalpy Most reactions occur under constant pressure. Enthalpy (H) is the heat content of a system at constant pressure. You cannot measure the actual enthalpy of a substance, but you can measure the change in enthalpy.

Enthalpy of Reaction Enthalpy of reaction (∆H rxn ) is the heat released or absorbed in a chemical reaction carried out under constant pressure. ∆H rxn = H final – H initial Because the reactants are present at the beginning of the reaction and products are present at the end, ∆H rxn can also be defined by: ∆H rxn = H products – H reactants

Enthalpy Change H In endothermic reactions, the H products > H reactants Therefore, Endothermic reactions always have positive ∆ H In exothermic reactions, the H products < H reactants Therefore, Exothermic reactions always have negative ∆ H

Endothermic Reaction

Exothermic Reaction

Calculations with ∆ H Given an equation, you can use a ∆H of the reaction to find the ∆H of a specific amount of a substance in the reaction.

Steps: 1.Convert the grams to moles (using molar mass) 2.Convert moles to kJ using the H relationship established in the equation

Example: What is the ∆H for 1.0 g of H 2 O 2 in the following reaction? 2H 2 O 2  2H 2 O + O 2 ∆H = -190 kJ Answer: 1.0 g H 2 O 2 x 1 mol H 2 O 2 x -190 kJ 34.0 g H 2 O 2 2 molH 2 O 2 = kJ

Practice Problem: What is the change in enthalpy when 9.22 g of C 6 H 12 O 6 reacts? C 6 H 12 O 6 +6O 2  6CO 2 +6H 2 O ∆ H=-2803 kJ 9.22g C 6 H 12 O 6 x 1 mol x kJ g 1 mol = kJ