Thermochemistry Chapter 5 Energy The ability to do work or transfer heat. –Work: Energy used to cause an object that has mass to move. –Heat: Energy.

Thermochemistry Chapter 5

Energy The ability to do work or transfer heat. –Work: Energy used to cause an object that has mass to move. –Heat: Energy used to cause the temperature of an object to rise.

Potential Energy Energy an object possesses by virtue of its position or chemical composition.

Kinetic Energy Energy an object possesses by virtue of its motion. 1 2 KE =  mv 2

The energy something possesses due to its motion, depending on mass and velocity. Potential energy Energy in Energy out kinetic energy

Energy Kinetic Energy – energy of motion KE = ½ m v 2 Potential Energy – stored energy Batteries (chemical potential energy) Spring in a watch (mechanical potential energy) Water trapped above a dam (gravitational potential energy) massvelocity (speed) B A C

School Bus or Bullet? Which has more kinetic energy; a slow moving school bus or a fast moving bullet? Recall: KE = ½ m v 2 KE = ½ m v 2 BUSBULLET KE (bus) = ½ (10,000 lbs) (0.5 mph) 2 KE (bullet) = ½ (0.002 lbs) (240 mph) 2 Either may have more KE, it depends on the mass of the bus and the velocity of the bullet. Which is a more important factor: mass or velocity? Why?(Velocity) 2

Units of Energy The SI unit of energy is the joule (J). An older, non-SI unit is still in widespread use: The calorie (cal). 1 cal = 4.184 J 1 J = 1  kg m 2 s2s2

System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).

Work Energy used to move an object over some distance. w = F  d, where w is work, F is the force, and d is the distance over which the force is exerted.

Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects.

First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Use Fig. 5.5

Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. Use Fig. 5.5

Internal Energy By definition, the change in internal energy,  E, is the final energy of the system minus the initial energy of the system:  E = E final − E initial Use Fig. 5.5

Changes in Internal Energy If  E > 0, E final > E initial –Therefore, the system absorbed energy from the surroundings. –This energy change is called endergonic.

Changes in Internal Energy If  E < 0, E final < E initial –Therefore, the system released energy to the surroundings. –This energy change is called exergonic.

Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is,  E = q + w.

Relating  E to Heat and Work First Law of Thermodynamics  E = q + w

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.

Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system to the surroundings, the process is exothermic.

State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. –In the system below, the water could have reached room temperature from either direction.

State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so,  E depends only on E initial and E final.

State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its  E is the same. –But q and w are different in the two cases.

Work When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −P  V

Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

Enthalpy When the system changes at constant pressure, the change in enthalpy,  H, is  H =  (E + PV) This can be written  H =  E + P  V

Enthalpy Since  E = q + w and w = −P  V, we can substitute these into the enthalpy expression:  H =  E + P  V  H = (q+w) − w  H = q So, at constant pressure the change in enthalpy is the heat gained or lost.

Endothermicity and Exothermicity A process is endothermic, then, when  H is positive.

Endothermicity and Exothermicity A process is endothermic when  H is positive. A process is exothermic when  H is negative.

Enthalpies of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

Enthalpies of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ 2CH 4 (g) + 4O 2 (g)  2CO 2 (g) + 4H 2 O(g)  H = -1604 kJ Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: CO 2 (g) + 2H 2 O(g)  CH 4 (g) + 2O 2 (g)  H = +802 kJ Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: CO 2 (g) + 2H 2 O(g)  CH 4 (g) + 2O 2 (g)  H = +802 kJ CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g)  H = -802 kJ Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: 3.Change in enthalpy depends on state: Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: 3.Change in enthalpy depends on state: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)  H = -802 kJ Enthalpies of Reaction

For a reaction 1.Enthalpy is an extensive property (magnitude  H is directly proportional to amount): 2.When we reverse a reaction, we change the sign of  H: 3.Change in enthalpy depends on state: CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)  H = -802 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)  H = -890 kJ Enthalpies of Reaction

b) Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure. Enthalpies of Reaction Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ

Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ? Enthalpies of Reaction

Problem: 5.33 2 Mg(s) + O 2 (g)  2 MgO(s)  H = -1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

Problem: 5.33 2 MgO(s)  2 Mg(s) + O 2 (g)  H = 1204 kJ d) How many kilojoules of heat are absorbed when 7.50g of MgO is decomposed into Mg and O 2 at constant pressure? Enthalpies of Reaction

Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure  H through calorimetry, the measurement of heat flow.

Calorimetry Use a calorimeter (measures heat flow). Two kinds: –Constant pressure calorimeter (called a coffee cup calorimeter) –Constant volume calorimeter is called a bomb calorimeter.

Specific Heat Capacity and Heat Transfer The quantity of heat transferred depends upon three things: 1.The quantity of material (extensive). 2.The difference in temperature. 3.The identity of the material gaining/losing heat.

Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1  C) is its heat capacity. Molar Heat Capacity – The amount of heat required to raise the temperature of one mole of a substance by 1 Kelvin.

Specific Heat (Specific Heat Capacity, c) – The amount of heat required to raise 1.0 gram of a substance by 1 Kelvin. Heat Capacity and Specific Heat

Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/mol-K), we can measure  H for the reaction with this equation: q = m  c   T

- Atmospheric pressure is constant  H = q P q system = -q surroundings -The surroundings are composed of the water in the calorimeter and the calorimeter. q system = - (q water + q calorimeter ) -For most calculations, the q calorimeter can be ignored. q system = - q water c system m system  T system = - c water m water  T water Calorimetry Constant-Pressure Calorimetry

Bomb Calorimetry Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water.

Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy,  E, not  H. For most reactions, the difference is very small.

-Special calorimetry for combustion reactions -Substance of interest is placed in a “bomb” and filled to a high pressure of oxygen -The sealed bomb is ignited and the heat from the reaction is transferred to the water -This calculation must take into account the heat capacity of the calorimeter (this is grouped together with the heat capacity of water). q rxn = -C calorimeter (  T) Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry)

Problem 5.50 NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  T water = 18.4 o C – 23.0 o C = -4.6 o C m water = 60.0g c water = 4.184J/g o C m sample = 3.88g q sample = -q water q sample = -c water m water  T water q sample = -(4.184J/g o C)(60.0g+3.88g)(-4.6 o C) q sample = 1229J - Now calculate  H in kJ/mol Calorimetry

Problem 5.50 NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq)  T water = 18.4 o C – 23.0 o C = -4.6 o C m water = 60.0g c water = 4.184J/g o C m sample = 3.88g Calorimetry q sample = 1229J moles NH 4 NO 3 = 3.88g/80.032g/mol = 0.04848mol  H = q sample /moles  H = 1229J/0.04848mol  H = 25.4 kJ/mol

A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane? Calorimetry 2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g

2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = -C cal (  T water ) Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = -C cal (  T water ) q rxn = -11.66kJ/ o C(7.42 o C) Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

2 C 8 H 18 + 25O 2  16 CO 2 + 18 H 2 O  T water = 28.78 o C – 21.36 o C = 7.42 o C C cal = 11.66kJ/ o C m sample = 1.80g q rxn = -C cal (  T water ) q rxn = -11.66kJ/ o C(7.42 o C) = -86.52kJ Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

 H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

 H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = -48.1 kJ/g Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

 H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = -48.1 kJ/g  H combustion (in kJ/mol)  H combustion = -86.52kJ/0.01575mol = Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

 H combustion (in kJ/g)  H combustion = -86.52kJ/1.80g = -48.1 kJ/g  H combustion (in kJ/mol)  H combustion = -86.52kJ/0.01575mol = -5492 kJ/mol Calorimetry A 1.800g sample of octane, C 8 H 18, was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/ o C. The temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of octane? Per mole of octane?

Hess’s Law  H is well known for many reactions, and it is inconvenient to measure  H for every reaction in which we are interested. However, we can estimate  H using  H values that are published and the properties of enthalpy.

Hess’s Law Hess’s law states that “If a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Hess’s Law Because  H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Hess’s Law –rxns in one step or multiple steps are additive because they are state functions eg. CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g)  H = - 802 kJ CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l)  H = - 890 kJ 2H 2 O (g)  2H 2 O (l)  H = - 88 kJ

Practice –Calculate  H for the conversion of graphite to diamond: C graphite  C diamond C graphite + O 2(g)  CO 2(g)  H = -393.5 kJ C diamond + O 2(g)  CO 2(g)  H = -395.4 kJ C graphite + O 2(g)  CO 2(g)  H = -393.5 kJ CO 2(g)  C diamond + O 2(g)  H = 395.4 kJ C graphite  C diamond  H = + 1.9 kJ

Enthalpies of Formation -There are many type of  H, depending on what you want to know  H vapor – enthalpy of vaporization (liquid  gas)  H fusion – enthalpy of fusion (solid  liquid)  H combustion – enthalpy of combustion (energy from burning a substance)

Enthalpies of Formation -A fundamental  H is the Standard Enthalpy of Formation ( )

Enthalpies of Formation -A fundamental  H is the Standard Enthalpy of Formation ( ) Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure.

Enthalpies of Formation -A fundamental  H is the Standard Enthalpy of Formation ( ) Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure. “The standard enthalpy of formation of the most stable form on any element is zero”

Enthalpies of Formation

Calculation of  H Imagine this as occurring in 3 steps: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l)

C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g) 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g) 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H The sum of these equations is:

Calculation of  H We can use Hess’s law in this way:  H =  n  H f(products) -  m  H f(reactants) where n and m are the stoichiometric coefficients. 

C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H  H= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)] = [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)] = (-2323.7 kJ) - (-103.85 kJ) = -2219.9 kJ

Enthalpies of Formation Problem 5.72 a) N 2 O 4 (g) + 4 H 2 (g)  N 2 (g) + 4 H 2 O(g) N 2 O 4 (g)9.66 kJ/mol H 2 (g)0 kJ/mol N 2 (g)0 kJ/mol H 2 O(g) -241.82 kJ/mol  H = [1mol(  H(N 2 )) + 4mol(  H(H 2 O))] – [1mol(  H(N 2 O 4 )) + 4mol(  H(H 2 ))]  H = [1mol(0kJ/mol) + 4mol(-241.82kJ/mol)] – [1mol(9.66kJ/mol) + 4mol(0kJ/mol)] = -976 kJ

Enthalpies of Formation Problem 5.72 b) 2 KOH(s) + CO 2 (g)  K 2 CO 3 (s) + H 2 O(g) KOH(s) -424.7 kJ/mol CO 2 (g) -393.5 kJ/mol K 2 CO 3 (s) -1150.18 kJ/mol H 2 O(g) -241.82 kJ/mol  H = [1mol(  H(K 2 CO 3 )) + 1mol(  H(H 2 O))] – [2mol(  H(KOH)) + 1mol(  H(CO 2 ))]  H = [1mol(-1150.18kJ/mol) + 1mol(-241.82kJ/mol)] – [2mol(-424.7kJ/mol) + 1mol(-393.5kJ/mol)] = -149.1kJ

Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.

Foods 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. Energy in our bodies comes from carbohydrates and fats (mostly). Intestines: carbohydrates converted into glucose: C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O,  H = -2816 kJ Fats break down as follows: 2C 57 H 110 O 6 + 163O 2  114CO 2 + 110H 2 O,  H = -75,520 kJ Fats contain more energy; are not water soluble, so are good for energy storage. Foods and Fuels

Fuels The vast majority of the energy consumed in this country comes from fossil fuels.

Fuels Fuel value = energy released when 1 g of substance is burned. Most from petroleum and natural gas. Remainder from coal, nuclear, and hydroelectric. Fossil fuels are not renewable. In 2000 the United States consumed 1.03  10 17 kJ of fuel. Hydrogen has great potential as a fuel with a fuel value of 142 kJ/g. Foods and Fuels

Thermochemistry Chapter 5 Energy The ability to do work or transfer heat. –Work: Energy used to cause an object that has mass to move. –Heat: Energy.

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Thermochemistry Chapter 5 Energy The ability to do work or transfer heat. –Work: Energy used to cause an object that has mass to move. –Heat: Energy.

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Presentation on theme: "Thermochemistry Chapter 5 Energy The ability to do work or transfer heat. –Work: Energy used to cause an object that has mass to move. –Heat: Energy."— Presentation transcript:

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