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Enthalpy. Most chemical and physical changes occur under essentially constant pressure (reactors open to the Earth’s atmosphere) very small amounts of work are performed as system expands or contracts

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Enthalpy

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### Enthalpy

• Most chemical and physical changes occur under essentially constant pressure (reactors open to the Earth’s atmosphere)

• very small amounts of work are performed as system expands or contracts

• the change in internal energy occurs primarily, or exclusively, as heat that is gained or lost.

### Enthalpy

• If a process occurs at constant pressure and the only work done is PV work, the heat flow is described by the enthalpy of the system.

• Enthalpy (H):

• a state function defined by the equation:

H = E + PV

(Question: Are P and V state functions?)

### Enthalpy

• Although the enthalpy of a system cannot be measured, the change in enthalpy (D H) can.

• heat gained or lost by a system when a process occurs at constant pressure

D H = Hfinal - Hinitial = qP

where qP = heat gained/lost at constant pressure

### Enthalpies of Reaction

• The change in enthalpy can be found using:

D H = Hfinal - Hinitial

• For a chemical reaction:

• Hfinal = H products

• Hinitial = H reactants

• The enthalpy change for a chemical reaction is:

D H = Hproducts - Hreactants

### Enthalpies of Reaction

• The enthalpy change that accompanies a chemical reaction is called the enthalpy of reaction

• D Hrxn

• Also called heat of reaction

### Enthalpy

• For a chemical reaction,

D H = Hproducts - Hreactants

• D H < 0 (negative)

• heat is lost by system

• exothermic

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

### Enthalpy

• For a chemical reaction,

DH = Hproducts - Hreactants

• D H > 0 (positive)

• heat gained/absorbed

by the system

• endothermic

CO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g)

### Enthalpies of Reaction

P

• If D Hrxn = positive

• endothermic

• If D Hrxn = negative

• exothermic

• heat is given off

H

R

time

R

H

P

time

### Enthalpies of Reaction

• DHrxn is associated with a specific chemical reaction.

• extensive property

• Depends on the amount of material

• Thermochemical equationsare balanced chemical equations that show the associated enthalpy change

• balanced equation

• enthalpy change (DHrxn)

### Enthalpies of Reaction

• An example of a thermochemical equation:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

• The coefficients in the balanced equation show the # moles of reactants and products that produced the associated DH.

• If the number of moles of reactant used or product formed changes, then the DH will change as well.

### Enthalpies of Reaction

• For the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

-890. kJ-890. kJ

1 mol CH42 mol O2

-890. kJ-890. kJ

1 mol CO22 mol H2O

### Enthalpies of Reaction

• Guidelines for Using Thermochemical Equations:

• Enthalpy is an extensive property

• The magnitude of DH is directly proportional to the amount of reactant consumed or product produced

### Enthalpies of Reaction

• The thermochemical equation for burning 1 mole of CH4 (g):

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

• When 1 mole of CH4 is burned, 890. kJ of heat are released.

• When 2 moles of CH4 are burned, 1780. kJ of heat are released.

### Enthalpies of Reaction

Example: How much heat is gained or lost when 10.0 g of CH4 (g) are burned at constant pressure?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

Given:10.0 g CH4 (g)

-890. kJ

1 mol CH4

Find: heat

### Enthalpies of Reaction

Heat = 10.0 g CH4 x1 mole CH4 x-890. kJ

16.0 g CH41 mol CH4

= -556 kJ

Is this an endothermic or exothermic process?

### Enthalpies of Reaction

Example: How much heat is gained or lost when 10.0 g of water are formed at constant pressure in the following reaction?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

Given:10.0 g H2O (g)

-890. kJ

2 mol H2O

Find: heat

### Enthalpies of Reaction

Heat = 10.0 g H2O x1 mole H2O x-890. kJ

18.0 g H2O2 mol H2O

= -247 kJ

### Enthalpies of Reaction

• Guidelines for Using Thermochemical Equations (cont).

• The enthalpy change for a reaction is equal in magnitude but opposite in sign to the DH for the reverse reaction.

2 H2O2 (l) 2 H2O (l) + O2(g) DH = -196 kJ

2 H2O (l) + O2(g) 2 H2O2 (l) DH = +196 kJ

### Enthalpies of Reaction

• Guidelines for Using Thermochemical Equations (cont):

• The enthalpy change for a reaction depends on the physical state of the reactants and products.

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890. kJ

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) DH = -802 kJ

### Enthalpies of Reaction

• Why does the DHrxn depend on the physical state of the reactants and products?

• Energy is either absorbed or released when chemicals change from one physical state to another.

H2O (l) H2O (g) DH = + 44 kJ

### Calorimetry

• The enthalpy change associated with a chemical reaction or process can be determined experimentally.

• measure the heat gained or lost during a reaction (or process) at constant P

• Measure the change in temperature

### Calorimetry

• Calorimetry:

• the experimental measurement of heat gained or lost during a chemical reaction or process

• Calorimeter

• an instrument used to measure the heat gained or lost during a chemical reaction or process.

### Calorimetry

• Calorimetry is used to experimentally determine:

• Heat capacity or specific heat

• DHrxn

• Enthaply change for a reaction

• DHfusion

• Enthalpy change when a substance goes from the liquid to the solid state

• DHvaporization

• Enthalpy change when a substance goes from the liquid to the gas state

### Calorimetry

If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter?

Why is there a difference in temperature between the two objects?

### Calorimetry

• The temperature increase observed when an object absorbs a certain quantity of energy is determined by its heat capacity (C).

• Amount of heat required to raise the temperature of an object 1oC (or 1 K)

• As heat capacity increases, more heat must be added to produce a specific temperature increase.

### Calorimetry

• For pure substances, heat capacity is usually given for a specified amount of substance:

• Molar heat capacity:

• amount of heat required to raise the temperature of 1 mole of a substance by 1oC

• Specific heat:

• amount of heat required to raise the temperature of 1 g of a substance by 1oC

### Calorimetry

• Specific Heat = quantity of heat transferred

mass x temp change

= q

mass x DT

• Molar Heat = quantity of heat transferred

Capacitymoles x temp change

= q

mol x DT

Specific heat = q

m x DT

### Calorimetry

Example:If 418 J is required to increase the temperature of 50.0 g of water by 2.00 K, what is the specific heat of water?

Given:q = 418 J

m = 50.0 g

DT = 2.00K

Find: specific heat

• Common units for specific heat are:

• Jcal

• g.Kg.oC

### Calorimetry

Specific heat = 418 J =4.18 J

50.0 g x 2.00 Kg.K

Molar heat capacity = q

mol x DT

### Calorimetry

Example:What is the molar heat capacity of aluminum if it takes 9.00 J to raise the temperature of 5.00 g of aluminum from 298.0 K to 300.0 K?

Given:q = 9.00 J

m = 5.00 g

DT = 2.0 K

Find: molar heat

capacity

• Common units for molar heat capacity are:

• Jcal

• mol.Kmol.oC

### Calorimetry

Molar heat capacity =

9.00 J x 27.0 g Al =24 J

5.00 g x 2.0 Kmol Almol.K

### Calorimetry

Example:If the specific heat of Al (s) is 0.90 J/g.K, how much heat is required to raise the temperature of 10.0 kg of Al from 25.0oCto 30.0oC?

Given:C = 0.90 J/g.K

m = 10.0 kg

DT = 5.0 K

Find: heat

C = q

m x DT

### Calorimetry

q = C x m x DT

q = 0.90 J x10.0 kg x1000 g x 5.0 K

g.K1 kg

= 45,000 J = 45 kJ

C = q

m x DT

This equation is one that you will use OFTEN in calorimetry.

Specific heat = q

m x DT

### Calorimetry

Example:If the specific heat of Fe(s) is 0.45 J/g.K, what change in temperature would be observed when 1.0 kJ of heat is added to 45 g of Fe(s)?

Given:C = 0.45 J/g.K

m = 45 g

q = 1.0 kJ

Find:DT

### Calorimetry

Specific heat = q

m x DT

DT =1.0 kJ x 1000 J x g.K

45 g1 kJ 0.45 J

=49 K

DT = q

m x C