Chapter 5 THERMO THERMO chemistry 5.4 Enthalpy of Reactions 5.5 Calorimetry 5.6 Hess s Law 5.7 Enthalpies of Formation
Chemical Equations 1 st WRITE the Chemical Equation 2 nd BALANCE the Chemical Equation 3 rd INTERPRET the Chemical Equation Now when Interpreting Chemical Equations must also include ENERGY
Involved in Chemical Processes Physical Process H 2 O(gas) H 2 O(liquid) + HEAT or Chemical Process CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O + HEAT
Chemical reactions can RELEASE when they occur CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O + HEAT or Chemical reactions can ABSORB when they occur 2 C (s) + 2 H 2 (g) + HEAT C 2 H 4 (g)
predict whether heat is absorbed endothermic or released exothermic by the system for : (1) An ice cube melts (2) butane is burned
Exo thermic and Endo thermic Processes thermic: transfers heat TO THE surroundings An exothermic process feels HOT thermic: absorbs heat FROM THE surroundings An endothermic process feels cold
Two (2) Ways to express heat 1. Heat can be expressed as q - or - 2. Expressed as H called Enthalpy - or - H Change in
Table 5.1 Sign Convention EXO thermic Heat is transferred FROM SYSTEM to the surroundings: q > 0 ; H < 0 ENDO thermic Heat is transferred FROM SURROUNDINGS to the system: q < 0 ; H > 0
System & Surroundings I. SYSTEM the portion of the universe that is singled out for study II. SURROUNDINGS everything outside the system III. CONSERVATION OF ENERGY
Open, Closed, & Isolated Systems
HEAT LOST = HEAT GAIN Something is gaining Heat While Something else looses Heat. IF YOU KNOW ONE OF THESE {Heat Lost - or - Heat Gained} THEN YOU KNOW
THERMO chemistry Two parts I. EXPERIMENTAL II. MATHEMATICAL
Energy Changes Involved With Part 1. PHYSICAL Changes a) Phase Changes b) WITHIN a Phase Part 2. CHEMICAL Changes
Part 1a Energy Changes Involved Physical Changes PHASE Changes Gas Liquid Solid
Energy Expressed In Terms of q for a Change Of State H 2 O (solid) H 2 O (liq) - Energy H 2 O(solid) H 2 O(liq) - 6.01 kj/mole The Sign Means Heat is Needed
Energy Can Also Be Expressed As A CHANGE In ENTHALPY ( H) H 2 O (solid) H 2 O (liq) - Energy H 2 O (solid) H 2 O (liq) H = + 6.01 kj/mole The Sign Means Heat is Needed!!!! NOTE THE SIGN CHANGE!!!!
Example 1: How much energy is needed to melt [ H 2 O (solid) H 2 O (liq) +/- Heat ] 18 grams of H 2 O at 0 o C? [The Heat of fusion of H 2 O = 6.008 KJ / mole] 18 grams of water = 1.0 mole of water therefore need KJ of heat
Example 2: What quantity of Heat is required to vaporize 18 grams of H 2 O at 100 o C? H 2 O (liq) H 2 O (gas) +/- Heat [The Heat of vaporization = 40.67 KJ / mole ] 18 grams of water = 1.0 mole of water therefore need KJ of heat
Part 1b [NO CHANGE IN PHASE] Energy Change WITHIN A State Gas Gas Liquid Liquid Solid Solid
Specific Heat Specific Heat- The heat required to raise the temperature of 1 gm of a substance by 1 o C EVERY SUBSTANCE HAS ITS OWN UNIQUE SPECIFIC HEAT (SH) For Water :S.H. = 4.18 JOULES (Grams) ( T)
UNIQUE SPECIFIC HEAT
Specific heat = quantity of heat transferred (grams of substance) x (temperature change)
How much heat energy is required to heat one pound of water from 25 o C (room temp) to its boiling point (100 o C)? LET THE UNITS SOLVE THE PROBLEM. Energy = (Specific Heat)x(grams)x(change in Temp) For Water :S.H. = 4.18 JOULES (Grams) ( T)
Energy required to heat one pound of water from 25 to 100 o C? Joules = Joules 4.18 C g ( ( ) o 454g (75 ) )( T ) Heat Energy = 142,329 Joules How many Significant Figures? Therefore answer is?????? Joules
Large beds of rocks are used in some solar-heated homes to store heat Calculate the quantity of heat absorbed by 50.0 kg of granite if the temperature increases by 12.0 C [The specific heat of granite is 0.79 J/g-K]
How much heat is absorbed by 50.0 kg of cement if the temperature increases 12.0 C [The specific heat of cement is 0.88 J/g-K] Let UNITS solve the problem Joules = (Specific Heat)x(grams)x(change in Temp) Joules = (0.88 J / g-k)(50.0x10 3 g)(12.0 C) Joules =
How much heat is absorbed by 50.0 kg of Rocks. 4.7 10 5 J Cement... 5.3 10 5 J Water 25.1 10 5 J What statement can be made about specific heat in terms of substances absorbing or releasing heat?
Heat Flow Heat spontaneously flows from a hot object to a cold one until the temperature of the two objects are the same
Every substance has its own unique specific heat Use that information to identify an unknown 1. Put into a Styrofoam cup ¾ full of water a piece of hot metal 2. Measure temperature change of water
A 2.61 gram block of metal was heated to 100.0 o C and put into an insulated cup containing 28.00 g of water. The water temperature rose from 25.00 to 26.48 o C
When the Hot Metal is dropped into the cup of water HEAT Flows from the metal to the water Heat Lost by Metal = Heat Gain by Water Heat spontaneously flows from a hot object to a cold one until the temperature of the two objects are the same
Metal: 2.61 grams ; initial temp = 100 o C Water: 28.00 grams; initial temp = 25.00 o C final temp = 26.48 0 C 1. What was the final temperature of the Metal? 2. How much energy did the water gain? 3. How much energy did the Metal lose? 4. What is the specific heat of Metal?
Joules Heat GAIN by Water Joules = 4.18 C g ( ( ) o 28.0g (26.48 25.00 ) )( T ) = 173 Joules {Heat Gained by Water equals Heat LOST by Metal Metal LOST 173 Joules FOR THE METAL: 173 Joules = S.H x (2.61 g) x (100 26.48)
SpecificHeat = ( 173Joules 2.61g )( 100 26.48) =???? WHAT WAS THE UNKNOWN METAL?
Review of Part 1 Energy Changes Involved With PHYSICAL Changes a) Phase Changes b) WITHIN a Phase
How much Energy required to heat 1.0 gram of ice at 10 o C to steam at 110 o C Ice Ice Liq Liq Gas Gas -10 o C 0 o C 0 o C 100 o C 100 o C 110 o C What information do you need to work this problem
Data Required for problem Heat of fusion = 6.008 kj / mole Heat of vaporization = 40.67 kj / mole Specific heat: Ice 2.092 J / g - K Liq 4.184 J / g K Steam 1.841 J / g - K
Energy required to heat 1.0 gram of ice at 10 o C to ice at 0 o C? Heat required to melt 1.0 gram of ice? Heat required to heat 1.0 gram of water at 0 o C to water at 100 o C? Heat required to vaporize 1.0 grams of H 2 O? Heat required to heat 1.0 gram of steam at 100 o C to steam at 110 o C?
How much Energy required to heat 1.0 gram of ice at 10 o C to steam at 110 o C Add all the numbers 21 J + 0.33 kj + 418 J + 0.59 kj + 18 J How many significant figures in answer?
Part 2 CHEMICAL REACTIONS Reactants Products +/-ENERGY
Chemical reactions can release or absorb heat Energy is stored in Chemical BONDS It TAKES Energy To BREAK Bonds -------------------------------------------------------------------------------------------------------------------------------------------------------------------- Energy Is RELEASED When A Bond Is Formed
Enthalpies of Reaction For a reaction: H = H final H initial = H products H reactants
HEAT OF REACTION CH 4 + 2 O 2 CO 2 + 2 H 2 O + HEAT The PLUS Sign Means Heat Is Given Off OR CH 4 + 2 O 2 CO 2 + 2 H 2 O H = - # The MINUS Sign Means Heat Is Given Off
DETERMINATION of HEATS of REACTIONS 1. THE DIRECT METHOD EXPERIMENTAL Use A Calorimeter 2. THE INDIRECT METHOD MATHEMATICAL Use HESS S Law
1. The Direct Method EXPERIMENTAL Go to Lab and Use A Calorimeter HEAT LOST = HEAT GAINED
Two (2) types of CALORIMETERS 1. OPEN { to the atmosphere and 2. CLOSED {to the atmosphere
OPEN CALORIMETER Also called a CONSTANT PRESSURE CALORIMETER Styrofoam cup
100 ml of 0.5 M HCl added to 100 ml of 0.5 M NaOH
Data Collected from experiment Volume of 0.500 M NaOH(aq) = 1.00 x 10 2 ml Volume of 0.500 M HCl(aq) = 1.00 x 10 2 ml Initial Temperature = 22.50 o C Final Temperature = 25.92 o C
Detemine the heat of the reaction PER MOLE (the Heat of neutralization) What GIVES OFF THE HEAT? The reaction! 1 HCl(aq) + 1 NaOH (aq) H 2 O + NaCl (aq) + HEAT What ABSORBS THE HEAT? The solution
How do you determine the Heat of the reaction? From the HEAT GAINED by the SOLUTION Joules = (Specific Heat) x (grams) x (Temp Change) of the solution from density from experiment
Number of GRAMS =? 100 ml HCl (aq) + 100 ml NaOH (aq) 1. 100 ml of each = ml total 2. Relationship between weight & volume? 3. DENSITY of solution 4. If Density = 1.00 g / ml 5. Have grams of solution
Let UNITS solve the problem Joules = (Specific Heat) x (grams) x (change in T) change in Temp =? grams = 200 grams of solution Specific Heat = Joules 4.18 gram x temp change Joules = (Specific Heat) x (grams) x (change in T) Joules = (4.184)(200)(3.42) =
Joules = (4.184)(200)(3.42) = 2.866 kj 1 HCl(aq) + 1 NaOH (aq) H 2 O + NaCl (aq) + Heat 2.8661 kj of heat given off when 1.00 x 10 2 ml of 0.500 M HCl (aq) is mixed with 1.00 x 10 2 ml of 0.500 M NaOH (aq) How much heat given off per mole of water formed?
heat of neutralization per mole? How many moles of water formed? 1.00 x 10 2 ml of both 0.500 M of HCl(aq) & NaOH Moles = Molarity x Volume = 0.0500 0.05 H + (aq) + 0.05 OH - (aq) 0.05 H 2 O + 2.866 kj 1.00 H + (aq) + 1.00 OH - (aq) 1.00 H 2 O +
Calorimeter [Used For Gas Reactions] CONSTANT VOLUME CALORIMETER It is called a BOMB CALORIMETER Because the Chemical Reaction Occurs in a CLOSED Container
See Text Fig 5.19 Page 186
Chemical Reaction takes place in Bomb HEAT IS GIVEN off by Reaction (SYSTEM) HEAT IS ABSORBED by SURROUNDINGS Surroundings are 1. WATER In Calorimeter 2. Everything else {thermometer, stirrer, metal bomb itself, etc
Example 1 2.431 grams Of Mg Was Burned In a Constant Volume Calorimeter Write & Balance the COMBUSTION Reaction Mg + O MgO + HEAT 2 Heat Lost by = Heat Gain by Chemical 1. Water + Reaction 2. Calorimeter
Data Collected from experiment The Calorimeter had a Heat Capacity = 1769 J/ 0 C Calorimeter Contained 3.00x10 2 ml of Water Initial Temperature = 22.5 o C Final Temperature = 42.4 o C
1. Heat Gain by Water = S. H. x grams x Temp q water = (4.184 x 300 x 19.9 ) 2. Heat Gain by Calorimeter = Heat Cap x Temp Change q Calorimeter = (1769 x 19.9 ) Total Heat Gained = Water + Calorimeter
= (1769 x 19.9 ) + (4.184 x 300 x 19.9 ) = (35,203.1) + (24,978.48) Joules = Joules for 2.431 grams of Mg HEAT LOST = HEAT GAIN How do you convert from grams to moles?
Record Data Record results in Tabular Form
Heats of Combustion (- H) per mole in kj at 25 o C HC 2 H 3 O 2 (aq) { Acetic Acid (aq) CH 4 (g) { Methane C 2 H 6 (g) C 6 H 6 (liq) C 6 H 12 (liq) Sucrose (s) 874.5 890.4 1541.4 3277.7 3928.8 5690.2
Enthalpy is an EXTENSIVE property ( H is directly proportional to amount): CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(liq) H = - 890 kj 2 CH 4 (g) + 4 O 2 (g) 2 CO 2 (g) + 4 H 2 O(liq) H = -1780 kj Phase (solid, liquid, gas) is Important CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(gas) H = - 802 kj 4 CH 4 (g) + 8 O 2 (g) 4 CO 2 (g) + 8 H 2 O(g) H = -3208 kj Note the state of water
Next, DETERMINATION OF HEATS OF REACTIONS Using 2. THE INDIRECT METHOD MATHEMATICAL Use Of HESS S Law
I. COMBUSTION of hydrocarbons CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) II. FORMATION Reactions e.g., CH 4 (Methane) C(s) + 2 H 2 (g) CH 4 (g)
Standard State The Standard State of a substance is the state the pure substance is in at atmospheric pressure ( 1 atm) and 25 o C The standard state of carbon is graphite and not diamond The standard state of hydrogen is H 2 not H
STANDARD ENTHALPIES OF FORMATION For example H 2 O (liq) is formed from its elements as they exist in nature H 2 (gas) + ½ O 2 (gas) H 2 O (liq) H =-285.8 kj
STANDARD ENTHALPIES OF FORMATION THE STANDARD ENTHALPY OF FORMATION ( H f ) OF ANY ELEMENT IN ITS MOST STABLE FORM IS ZERO (BY DEFINATION)
HESS S LAW ENTHALPY CHANGES ARE ADDITIVE
Example 1 Calculate [using Hess Law] the heat of reaction for CO(g) + ½ O 2 (g) CO 2 (g) What DATA Do You Need From Table?
H f From Table WRITE AND BALANCE REACTIONS Formation of CO (g) is : 1. C (s) + ½ O 2 (g) CO(g) H = - 110.5 kj Formation of CO 2 (g) is : 2. C (s) + O 2 (g) CO 2 (g) H = - 393.5 kj
Want CO (g) + ½ O 2 (g) CO 2 (g) 1. C (s) + ½ O 2 (g) CO(g) H = - 110.5 kj REWRITE Eq 1 1b. CO (g) C (s) + ½ O 2 (g) H = + 110.5 kj also 2. C (s) + O 2 (g) CO 2 (g) H = - 393.5 kj
CO (g) C (s) + ½ O 2 (g) H = + 110.5 kj C (s) + O 2 (g) CO 2 (g) H = - 393.5 kj Add Equations To Get Reaction Wanted: CO (g) + ½ O 2 (g) CO 2 (g) Add H s To Get : H = + 110.5 kj - 393.5 kj =
Example 2 Find the Enthalpy of Formation of C 3 H 8 (g) Given Enthalpy of combustion of C 3 H 8 (g) = -2043 kj Enthalpy of formation of CO 2 (g) = -393.5 kj Enthalpy of formation of H 2 O(g) = -241.8 kj
Write and balance the following reactions Combustion of one mole of C 3 H 8 (g) Formation of one mole of CO 2 (g) Formation of one mole of H 2 O(g) Formation of one mole of C 3 H 8 (g) 3 C (s) +4 H 2 (g) C 3 H 8 (g) H =???
Example 3 Find the Enthalpy of Formation of C 3 H 8 (g) From Enthalpy of combustion of C 3 H 8 (g) = -2043 kj Enthalpy of formation of CO 2 (g) = -393.5 kj Enthalpy of formation of H 2 O(g) = -241.8 kj answer 3 C (s) +4 H 2 (g) C 3 H 8 (g) H = - 104.7 kj
Example 4 Find the Heat of Vaporization of Water From the following HEATS of COMBUSTION C 3 H 8 (g) + 5O 2 (g) 3CO 2 + 4H 2 O(g) C 3 H 8 (g) + 5O 2 (g) 3CO 2 + 4H 2 O(l) H = - 2043 kj H = - 2219kJ NOTE: These reactions have Nothing to do with the vaporization of water
Want H for H 2 O (liquid) H 2 O (gas) C 3 H 8 (g) + 5O 2 (g) 3CO 2 + 4H 2 O(g) C 3 H 8 (g) + 5O 2 (g) 3CO 2 + 4H 2 O(l) H = - 2043 kj H = - 2219kJ Rewrite last equation to get 3CO 2 + 4H 2 O(l) C 3 H 8 (g) + 5O 2 (g) H = + 2219 kj WHY?
Want H for H 2 O (liquid) H 2 O (gas) C 3 H 8 (g) + 5O 2 (g) 3CO 2 + 4H 2 O(g) H = - 2043 kj 3CO 2 + 4H 2 O(l) C 3 H 8 (g) + 5O 2 (g) H = + 2219 kj Add Equations getting 4 H 2 O (liquid) 4 H 2 O (gas) H = + 176 kj H 2 O (liquid) H 2 O (gas) H =
Given the data N 2 (g) + O 2 (g) 2 NO(g) H = +180.7 kj 2 NO(g) + O 2 (g) 2 NO 2 (g) H = - 113.1 kj 2 N 2 O(g) 2N 2 (g) + O 2 (g) H = - 163.2 kj Use Hess s law to calculate H For the reaction N 2 O(g) + NO 2 (g) 3 NO(g)
Example 4 Find H for N 2 O(g) + NO 2 (g) 3NO(g) Given N 2 (g) + O 2 (g) 2 NO(g) H = +180.7 kj 2 NO(g) + O 2 (g) 2 NO 2 (g) H = - 113.1 kj 2 N 2 O(g) 2N 2 (g) + O 2 (g) H = - 163.2 kj ---------------------------------- Add The Following Equations N 2 (g) + O 2 (g) 2 NO(g) H = +180.7 kj NO 2 (g) NO(g) + ½ O 2 (g) H = ½ [+ 113.1 kj] N 2 O(g) N 2 (g) + ½ O 2 (g) H = ½ [- 163.2 kj]
Energy From Foods
Energy in our bodies comes from fats and carbohydrates (mostly) Carbohydrates converted into glucose, then: C 6 H 12 O 6 + 6 O 2 6 CO 2 + 6 H 2 O H = - 2816 kj Fats: contain more energy; are not water soluble, so are good for energy storage. 2 C 57 H 110 O 6 + 163 O 2 114 CO 2 + 110 H 2 O H = - 75,520 kj
Which releases the greatest amount of energy per gram upon metabolism (a) carbohydrates (b) proteins (c) fats
Energy From Fuels
Which releases the greatest amount of energy per gram upon combustion (a) Methane (b) gasoline (c) hydrogen