Chemistry and Heat That’s hot stuff.

Energy is the capacity to do work. Energy is measured in Joules 1 Joule of energy can raise 1 N of weight exactly 1 meter 1 J=1Nm (PS: your diet offers ~5-10 million J/day

Energy is the capacity to do work. Forms include: Kinetic energy Gravitational potential energy Elastic potential energy Electrical energy Chemical potential energy Heat

Energy is the capacity to do work. Forms include: Kinetic energy Gravitational potential energy Elastic potential energy Electrical energy Chemical potential energy Heat …of the greatest interest to a chemist

Exothermic process System Surroundings An exothermic process releases energy Energy

Endothermic process System Surroundings Energy An endothermic process absorbs energy

If you add heat to matter, it may… a) b) c) d)

If you add heat to matter, it may… a) warm up. b) melt c) boil d) expand (tough to calculate, don’t bother)

Let’s try to warm up a cup of cold coffee. Step 1: Add heat.

Let’s try to warm up a cup of cold coffee. Step 1: Add heat. Well, that was easy.

Let’s try to warm up a cup of cold coffee. What if you add half as much heat?

Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) b) c)

Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) c)

Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) Use half as much coffee (and cup) c)

Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) Use half as much coffee (and cup) c) Use a different substance

The effect of heat, q! When something warms up: The heat, q, depends on: The mass of the sample (m) The change in temperature (  T) The nature of the sample (C)

The effect of heat (q) When something warms up: The heat, q, depends on: The mass of the sample (m) The change in temperature (  T) The nature of the sample (C) C is the specific heat capacity for a given substance. Its units are (J/g o C)

If you add heat to a sample, it may… a) warm up. q=mC  T b) melt c) boil d) expand (tough to calculate, don’t bother)

q=mC  T q – heat, in Joules m –mass, in grams C –specific heat capacity, in J/g o C  T—change in temperature (T final -T initial )

C water =4.184 J/g o C C water =4.2 J/g o C C ethanol =2.4 J/g o C C ice =2.1 J/g o C C Al =.90 J/g o C C Fe =.46 J/g o C C glass =.50 J/g o C C Ag =.24 J/g o C

How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T

How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T = 50.g x 4.18 J/g o C x (80. o C-15 o C)

How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T = 50.g x 4.18 J/g o C x (80. o C-15 o C) = 50.g x 4.18 J/g o C x (65 o C)

How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T = 50.g x 4.18 J/g o C x (80. o C-15 o C) = 50.g x 4.18 J/g o C x (65 o C) =14000 J (14 kJ)

What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC

What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C )

What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C ) = 31 o C

What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C ) = 31 o C If the temperature starts at 25 o C, it will heat up to …

What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C ) = 31 o C If the temperature starts at 25 o C, it will heat up to 56 o C

Calorimetry --the measurement of heat.

Calorimetry --the measurement of heat. If one thing gains heat…

Calorimetry --the measurement of heat. If one thing gains heat… …something else lost it.

If 75 g of a metal at 96 o C is placed in 58 g of water at 21 o C and the final temperature reaches 35 o C, what is the specific heat capacity of the metal?

Step 1 How much heat did the water gain?

Step 1 How much heat did the water gain? q=mC  T Mass of water, in grams Specific heat of water, 4.18 J/g o C Change in the temperature of water, in o C

Step 2 How much heat did the metal lose?

Step 2 How much heat did the metal lose? Heat lost = - heat gained q lost =-q gained

Step 3 What is the specific heat capacity of the metal?

Step 3 What is the specific heat capacity of the metal? C=q/m  T Mass of metal, in grams Specific heat of metal, in J/g o C Change in the temperature of metal, in o C Heat lost by metal

If 75 g of a metal at 96 o C is placed in 58 g of water at 21 o C and the final temperature reaches 35 o C, what is the specific heat capacity of the metal?.74 J/g o C

Thermochemistry 2H 2 (g)+O 2 (g)  2H 2 O(g)+ 443,000 J (read it.)

Thermochemistry 2H 2 (g)+O 2 (g)  2H 2 O(g)+ 443,000 J Two moles of hydrogen gas reacts with one mole of oxygen gas to form two moles of water vapor, releasing 443 kJ of heat.

Chemical Energy Chemical energy (enthalpy) is stored in bonds.

Chemical Energy Chemical energy (enthalpy) is stored in bonds. Forming bonds releases energy Breaking bonds requires energy

Chemical Energy Chemical energy (enthalpy) is stored in bonds. Forming bonds is exothermic Breaking bonds is endothermic

Chemical Energy Chemical energy (enthalpy) is stored in bonds. Exothermic reactions have a negative change in enthalpy Endothermic reactions have a positive change in enthalpy

Chemical Energy Chemical energy (enthalpy) is stored in bonds. Exothermic reactions have a negative change in enthalpy Endothermic reactions have a positive change in enthalpy The substances gain energy! The substances lose energy!

Thermochemistry 2H 2 +O 2  2H 2 O Breaking these bonds requires energy

Thermochemistry 2H 2 +O 2  2H 2 O Breaking these bonds requires energy

Thermochemistry 2H 2 +O 2  2H 2 O Breaking these bonds requires energy Forming these bonds releases a lot more energy

The mass to heat problem g kJ g 1 molkJ mol The heat of reaction,  H rxn

How about some stoichiometry? The oxidation of carbon releases 394 kJ/mol. How much heat is produced from the oxidation of 15 g C?

How about some stoichiometry? The oxidation of carbon releases 394 kJ/mol. How much heat is produced from the oxidation of 15 g C? 490 kJ

Special reactions: Formation Combustion Fusion Vaporization Dissolution

Special reactions: Formation —forming of 1 mole of a compound from its elements in their normal state Combustion —burning 1 mole of a substance in oxygen Fusion —freezing 1 mole of a substance at its melting point Vaporization —boiling 1 mole of a substance at its boiling point Dissolution —dissolving 1 mole of a substance in water

Special Heats The Heat of Formation  H formation of Mg(OH) 2 (s) =-925 kJ/mol says Mg(s)+O 2 (g)+H 2 (g)  Mg(OH) 2 (s)+ 925 kJ

Special Heats  H formation of CO 2 (g) = kJ/mol ? 2Na(s)+C(s)+1½O 2 (g)  Na 2 CO 3 (s)+1131 kJ ?

Special Heats  H formation of a pure element in its standard state is 0 kJ. (Ex. It takes no energy to make oxygen gas from oxygen gas.)

Special Heats The Heat of Combustion  H combustion of N 2 (g) =90.4 kJ/mol says N 2 (g)+O 2 (g) kJ  2NO(g)

Thermochemistry N 2 +O 2  2NO Breaking these bonds requires more energy… …than forming these bonds releases  H= kJ/mol

Special Heats  H combustion of H 2 (g) =-286 kJ/mol ? CH 3 OH+1½O 2 (g)  CO 2 (g)+H 2 O(l) +726 kJ ?

Special Heats The Heat of Fusion  H fusion of Fe =13.8 kJ/mol says Fe(s) kJ  Fe(l) at 1536 o C

Special Heats  H fusion of C 6 H 6 =9.87 kJ/mol ? CH 3 OH(s) J  CH 3 OH(l) at -98 o C ?

Special Heats The Heat of Vaporization  H vaporization of CS 2 =28 kJ/mol says CS 2 (l)+ 28 kJ  CS 2 (g) at 46.3 o C

Special Heats  H vaporization of CH 4 =8.2 kJ/mol ? O 2 (l) J  O 2 (g) at -183 o C ?

Special Heats The Heat of Solution  H solution of (NH 2 ) 2 CO =-14.0 kJ/mol says (NH 2 ) 2 CO(s)  (NH 2 ) 2 CO (aq) kJ

Special Heats  H solution of NH 3 (g) = -34 kJ/mol ? H 2 SO 4 (l)  H 2 SO 4 (aq) kJ ?

Special Heats Formation Combustion Fusion Vaporization Dissolution

Hess’s Law If: 2C + O 2  2CO+ 221 kJ and 2CO + O 2  2CO kJ

Hess’s Law If 2C + O 2  2CO+ 221 kJ and 2CO + O 2  2CO kJ

Hess’s Law If 2C + O 2  2CO+ 221 kJ and 2CO + O 2  2CO kJ …then 2C + 2O 2  2CO kJ

Hess’s Law If 2C + O 2  2CO+ 221 kJ and 2CO + O 2  2CO kJ …then 2C + 2O 2  2CO kJ Hess’s Law is usually used with heats of formation

Hess’s Law CH 4 + 2O 2  CO H 2 O Hess’s Law is usually used with heats of formation

Hess’s Law CH 4 + 2O 2  CO H 2 O CH kJ  C + 2 H 2 C + O 2  CO kJ 2H 2 + O 2  2 H 2 O kJ Hess’s Law is usually used with heats of formation

Hess’s Law CH 4 + 2O 2  CO H 2 O CH kJ  C + 2 H 2 C + O 2  CO kJ 2H 2 + O 2  2 H 2 O kJ Hess’s Law is usually used with heats of formation

Hess’s Law CH 4 + 2O 2  CO H 2 O CH kJ  C + 2 H 2 C + O 2  CO kJ 2H 2 + O 2  2 H 2 O kJ CH 4 + 2O 2  CO H 2 O kJ Hess’s Law is usually used with heats of formation

Hess’s Law CH 4 + 2O 2  CO H 2 O  H rxn =  H form,products -  H form,reactants Hess’s Law is usually used with heats of formation

Hess’s Law CH 4 + 2O 2  CO H 2 O  H rxn =  H form,products -  H form,reactants =(  H form,CO2 +2  H form,H2O )-  H form,CH4 =( kJ + 2x kJ)-( kJ) = kJ Hess’s Law is usually used with heats of formation

Hess’s Law Using table 11.6, What is  H rxn for… 1)2H 2 O 2  2H 2 O + O 2 2)CaCO 3  CaO + CO 2 3)2H 2 S + 3O 2  2H 2 O + 2SO 2 Hess’s Law is usually used with heats of formation

naphthalene Heating and cooling curve— naphthalene in a water bath

What is going on?

Notice the rate of cooling (from 75 o C)