THERMOCHEMISTRY CHAPTER 11
ENERGY AND HEAT nthermochemistry: The study of the energy changes that accompany chemical reactions and changes in the physical states of matter.
ENERGY AND HEAT nwork: Energy used moving objects against a force. nformula: Work = Force * Distance
ENERGY AND HEAT nheat: Energy transfer from one object to another. Represented by the symbol q.
ENERGY AND HEAT nenergy: The ability to do work or supply heat.
ENERGY AND HEAT n There are two main types of energy: A. Kinetic: Energy of motion
B. Potential: ENERGY AND HEAT Energy of position ninternal Energy The sum of the kinetic and potential energies of the particles making up a substance.
nsystem: ENERGY AND HEAT That which is under study, e.g.. a reaction. nsurroundings: Everything outside of the system.
ENERGY AND HEAT nuniverse: System + surroundings.
ENERGY AND HEAT nthere are two types of thermodynamic reactions: A. Endothermic B. Exothermic
ENERGY AND HEAT A. Endothermic: A chemical reaction or physical change in which heat is absorbed (q is positive).
ENERGY AND HEAT B. Exothermic: A chemical reaction or physical change in which heat is released (q is negative).
ENERGY AND HEAT n http://cwx.prenhall.com/petrucci/medialib/ media_portfolio/text_images/019_thermi TE.MOV
ENERGY AND HEAT nlaw of Conservation of Energy: Energy cannot be created or destroyed.
Measuring Heat ncalorie: A.The amount of heat energy needed to raise 1 g H 2 O at 1 C. B. 1 cal = 4.184 Joule
Measuring Heat njoule: A.The SI unit for energy B. One joule of heat raises the T of 1 g of water. C. 1 J = 0.2390 cal
Measuring Heat nspecific heat capacity (C) or (Cp) A. The amount of heat needed to raise the temperature of 1 gram of a substance by 1 C
Measuring Heat B. Equation: q = m x c x ΔT q = heat m= mass of material c = specific heat ΔT = change in T
Measuring Heat C. Units are J/g C.
nheat Capacity Measuring Heat A. An object's ability to absorb or release heat.
Measuring Heat B. Depends on the product of its specific heat capacity and the mass. ( c x m)
Measuring Heat C. Depends on heat divided by the change in temperature (q/ T).
D. Units are Measuring Heat usually J/ C.
Measuring Heat Examples: 1. The T of Cu with a mass of 95.4 g increases from 25 C to 48 C when the metal absorbs 849 J. What is the specific heat capacity of Cu?
Given: m = 95.4 g q = 849 J T 1 = 25 C T 2 = 48 C T = 23 C
Formula: q = m x c x T c = q m x T
c = 849 J 95.4 g x 23 C = 0.387 J / g C
2. How much heat is required to raise the temperature of 250.0 g of Hg to 52 C?
Given: C = 0.14 J/g C m = 250 g T = 52 C
Formula: q = m x c x T q = 250 g x 0. 14 J / g C x 52 C = 1820 J = 1.82 kj
3. Calculate the mass of water required to change the temperature to 56 C and produce 1324 J of energy?
Given: q = 1324 J C = 4.184 J/g C T = 56 C m =?
Formula: q = mc T m = q c T
m = 1324 J (4.184J /g C) (56 C) = 5.66 g
Calorimetry nwe can determine the heat flow ( H rxn ) associated with a chemical reaction by measuring the temperature change it produces.
Calorimetry ncalorimetry is The measurement of heat flow.
Calorimetry na calorimeter is An apparatus that measures heat flow.
Calorimetry nthere are two types: 1. Coffee cup 2. Bomb calorimeter
ncoffee cup Calorimetry A. Constant Pressure B. Used in heat changes involving reactions in aqueous solutions.
Calorimetry C. The reaction occurs in a known volume of water.
Calorimetry nbomb Calorimeter: A. Constant Volume B. Used to measure heat flows for gases and high temperature reactions.
Enthalpy A. The variable H B. The energy gained or lost C. q = H
nexample 1: 50 ml of 1.0 M HCl and 50 ml of NaOH are combined in a constant pressure calorimeter. The temperature of the solution is observed to rise from 21.0 C to 27.5 C. Calculate the enthalpy change for the reaction (assume density is 1.0 gram/ml, and that the specific heat of the solution is that of water).
T = 6.5 C c= 4.18 J/g C m = 100 ml x 1 g/ml = 100 g H solution =?
Formula: H solution = mc T (100 g)(4.18 J/g C) (6.5 C) = 2717 J
nthe heat absorbed by an aqueous solvent is equal to the heat given off by the reaction of the solutes: nq aq solution = -q rxn
Final answer is -2717 J
Why?
Because we were looking for the q rxn
Exothermic or Endothermic??
Exothermic Why? Negative sign
Example 2: A small pebble is heated and placed in a foam cup calorimeter containing 25 ml of water at 25 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble?
T = 1.4 C c = 4.18 J/g C m = 25 ml x 1 g 1 ml = 25 g
Formula: H = mc T = (25 g) (4.18 J/g C) (1.4 C) = 146 J
Exothermic or Endothermic?
Endothermic Because it is positive.
Thermochemical Equations nan equation that includes the heat change ( H) for the reaction.
Thermochemical Equations nexamples: n2n(g) N 2 (g) H = -941 kj no 2 (g) 2O(g) H = +502 kj
Enthalpies of Phase Changes nenthalpy of fusion ( H fus ): Heat to melt 1 mole of solid to liquid. Always positive.
Enthalpies of Phase Changes nenthalpy of Vaporization ( H vap ): Heat to evaporate 1 mol of liquid Always positive.
Enthalpies of Phase Changes nenthalpy of Condensation ( H cond ): Heat released when 1 mol of vapor condenses. Always negative.
Examples: 1. How many grams of ice at 0 C & 101325 Pa could be melted by the addition of 2.25 kj of heat?
What kind of phase change is it? Fusion
Find the H fus for water 1 mol = 6.01 kj What are you looking for? Mass of ice melted
2.25 kj 1 mol 6.01 kj 18 g H 2 O 1 mol H fus GFM H 2 O = 6. 74 g
2. How much heat in kj is absorbed when 63.7 g H 2 O (l) at 100 C is converted to steam at 100 C?
What kind of phase change is it? Vaporization
Find the H vap for water 1 mol = 40.7 kj What are you looking for? Amount of heat
63.7 g H 2 O 1 mol 18 g H 2 O GFM H 2 O 40.7 kj 1 mol H vap = 144 kj
Enthalpy of formation ( H f ) nthe enthalpy associated with the reaction that forms a compound from its elements in their most thermodynamically stable states.
Enthalpy of formation ( H f ) n Equal to In the above reaction, n and m are the coefficients of the products and the reactants in the balanced equation.
Example: Calculate the heat given off when one mole of B 5 H 9 reacts with excess oxygen according to the following reaction: 2 B 5 H 9 (g) + 12 O 2 (g) 5 B 2 O 3 (g) + 9 H 2 O (g)
Compound B 5 H 9 H f 73.2 kj/mol B 2 O 3-1272.77 kj/mol O 2 0 kj/mol H 2 O -241.82 kj/mol
Sum of Reactants 2 mol B 5 H 9 73.2 kj = 146.4 kj 1 mol 12 mol O 2 0 kj 1 mol = 0 kj
Sum of Reactants: 146.4 kj + 0 kj = 146.4 kj
Sum of Products 5 mol B 2 O 3-1272.77 kj = 1 mol - 6363.85 kj
Sum of Products 9 mol H 2 O -241.82 kj = 1 mol -2176.38 kj
Sum of Products -6363.85 kj + -2176.38 kj = -8540.23 kj
H = -8540.23 kj - 146.4 kj H products H reactants = -8686.63 kj
Hess s Law nheat transferred, or change in enthalpy (H), in a reaction is the same regardless whether the reaction occurs in a single step or in several steps.
Hess s Law nif a series of reactions are added together, the net change in the heat of the reaction is the sum of the enthalpy changes for each step.
Hess s Law n Rules for Using Hess s Law: 1. If the reaction is multiplied (or divided) by number, H must also be multiplied (or divided) by that number.
Hess s Law 2. If the reaction is reversed (flipped), the sign of H must also be reversed.
Hess s Law nexample: Nitrogen and oxygen gas combine to form nitrogen dioxide according to the following reaction: N 2 (g) + 2 O 2 2 NO 2 (g)
Hess s Law Calculate the change in enthalpy for the above overall reaction, given:
N 2 (g) + O 2 (g) 2NO (g) 2 NO (g) + O 2 (g) 2 NO 2 (g) N 2 (g) + 2 O 2 (g) 2 NO 2 (g) ΔH = 181 kj + - 131 kj = 50 kj
From the following enthalpy changes: OF 2 (g) + H 2 O (l) O 2 (g) + 2 HF (g) H = -277 kj SF 4 (g) + 2 H 2 O (l) SO 2 (g) + 4 HF (g) H = -828 kj S (g) + O 2 (g) SO 2 (g) H =-297 kj
Calculate the value of H for the reaction: 2 S (g) + 2 OF 2 (g) SO 2 (g) + SF 4 (g) TARGET
OF 2 (g) + H 2 O (l) O 2 (g) + 2 HF (g) How do you need to change this equation to look like the target equation? 2 S (g) + 2 OF 2 (g) SO 2 (g) + SF 4 (g) Multiply it by 2
SF 4 (g) + 2 H 2 O (l) SO 2 (g) + 4 HF (g) How do you need to change this equation to look like the target equation? 2S(g) + 2 OF 2 (g) SO 2 (g) + SF 4 (g) Flip the equation
S (g) + O 2 (g) SO 2 (g) How do you need to change this equation to look like the target equation? 2S(g) + 2 OF 2 (g) SO 2 (g) + SF 4 (g) Multiply by 2
2 OF 2 (g) + 2 H 2 O (l) 2 O 2 (g) + 4 HF (g) SO 2 (g) + 4 HF (g) SF 4 (g) + 2 H 2 O (l) 2 S(g) + 2 O 2 (g) 2 SO 2 (g) 2 S(g) + 2 OF 2 (g) SO 2 (g) + SF 4 (g)
ΔH = -554 kj + 828 kj + - 594 kj = -320 kj