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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 HEATn There are two main types of energy:
A. Kinetic: Energy of motion
ENERGY AND HEATB. Potential: Energy of position
nInternal Energy The sum of the kinetic and potential energies of the particles making up a substance.
ENERGY AND HEATnSystem: 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 HEATB. 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_THERMITE.MOV
ENERGY AND HEATnLaw of Conservation of Energy: Energy cannot be created or destroyed.
Measuring Heat
nCalorie: A.The amount of heat energy needed to raise 1 g H2O 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 HeatB. 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.
Measuring HeatnHeat Capacity
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).
Measuring HeatD. Units are
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 JT1 = 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°Cm = 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 = ?
q = mcTFormula:
m = q
cT
m = 1324 J
(4.184J /g°C) (56 °C)
= 5.66 g
Calorimetry
nWe can determine the heat flow (Hrxn) 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
CalorimetrynCoffee cup 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.
EnthalpyA. 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
Hsolution = ?
Formula:
Hsolution = mcT
(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:
nqaq solution = -qrxn
Final answer is
-2717 J
Why?
Because we were looking for the qrxn
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 g1 mL
= 25 g
H = mcTFormula:
= (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) N2(g) H = -941 kJ
nO2(g) 2O(g) H = +502 kJ
Enthalpies of Phase Changes
nEnthalpy of fusion (Hfus):
Heat to melt 1 mole of solid to liquid.
Always positive.
Enthalpies of Phase ChangesnEnthalpy of Vaporization (Hvap):
Heat to evaporate 1 mol of liquid
Always positive.
Enthalpies of Phase Changes
nEnthalpy of Condensation (Hcond):
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 Hfus for water
1 mol = 6.01 kJ
What are you looking for?
Mass of ice melted
2.25 kJ 1 mol6.01 kJ
18 g H2O1 mol
Hfus GFM H2O
= 6. 74 g
2. How much heat in kJ is absorbed when 63.7 g H2O (l) at 100°C is converted to steam at 100°C?
What kind of phase change is it?
Vaporization
Find the Hvap for water
1 mol = 40.7 kJ
What are you looking for?
Amount of heat
63.7 g H2O 1 mol
18 g H2O40.7 kJ1 mol
= 144 kJ
GFM H2O Hvap
Enthalpy of formation (Hf)
nThe enthalpy associated with the reaction that forms a compound from its elements in their most thermodynamically stable states.
Enthalpy of formation (Hf)nEqual 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 B5H9 reacts with excess oxygen according to the following reaction:
2 B5H9 (g) + 12 O2 (g) 5 B2O3 (g) + 9 H2O (g)
Compound Hf
B5H9 73.2 kJ/mol
B2O3 -1272.77 kJ/mol
O2 0 kJ/mol
H2O -241.82 kJ/mol
Sum of Reactants2 mol B5H9 73.2 kJ
1 mol= 146.4 kJ
12 mol O2 0 kJ
1 mol = 0 kJ
Sum of Reactants:
146.4 kJ + 0 kJ =
146.4 kJ
Sum of Products
5 mol B2O3 -1272.77 kJ1 mol
=
- 6363.85 kJ
Sum of Products
9 mol H2O -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
∑Hproducts ∑ Hreactants
= -8686.63 kJ
Hess’s LawnHeat 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 LawnIf 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 Lawn 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 LawnExample: Nitrogen and oxygen gas combine to form nitrogen dioxide according to the following reaction:
N2 (g) + 2 O2 2 NO2 (g)
Hess’s LawCalculate the change in enthalpy for the above overall reaction, given:
N2 (g) + O2 (g) 2NO (g)2 NO (g) + O2 (g) 2 NO2 (g)
N2 (g) + 2 O2 (g) 2 NO2 (g)
ΔH = 181 kJ + - 131 kJ= 50 kJ
From the following enthalpy changes:
OF2 (g) + H2O (l) O2 (g) + 2 HF (g) ∆H° = -277 kJ
SF4 (g) + 2 H2O (l) SO2 (g) + 4 HF (g) ∆H° = -828 kJ
S (g) + O2 (g) SO2 (g) ∆H° =-297 kJ
Calculate the value of ∆H for the reaction:
2 S (g) + 2 OF2 (g) SO2 (g) + SF4 (g)
TARGET
OF2 (g) + H2O (l) O2 (g) + 2 HF (g)
How do you need to change this equation to look like the target equation?
2 S (g) + 2 OF2 (g) SO2 (g) + SF4 (g)
Multiply it by 2
SF4 (g) + 2 H2O (l) SO2 (g) + 4 HF (g)How do you need to change this equation to look like the target equation?
2S(g) + 2 OF2 (g) SO2 (g) + SF4 (g)Flip the equation
S (g) + O2 (g) SO2 (g)
How do you need to change this equation to look like the target equation?2S(g) + 2 OF2 (g) SO2 (g) + SF4 (g)
Multiply by 2
2 OF2(g) + 2 H2O (l) 2 O2 (g) + 4 HF (g)
SO2 (g) + 4 HF (g)SF4 (g) + 2 H2O (l) 2 S(g) + 2 O2 (g) 2 SO2 (g)
2 S(g) + 2 OF2 (g) SO2 (g) + SF4 (g)
ΔH = -554 kJ + 828 kJ +- 594 kJ
= -320 kJ