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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 • 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 • heat must be added • 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 CH4 2 mol O2 -890. kJ -890. kJ 1 mol CO2 2 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 CH4 1 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 H2O 2 mol H2O = -247 kJ The negative sign indicates that heat was released to the surroundings.
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 Capacity moles 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: • J cal • g.K g.oC Calorimetry Specific heat = 418 J = 4.18 J 50.0 g x 2.00 K g.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: • J cal • mol.K mol.oC Calorimetry Molar heat capacity = 9.00 J x 27.0 g Al = 24 J 5.00 g x 2.0 K mol Al mol.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 x 1000 g x 5.0 K g.K 1 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 g 1 kJ 0.45 J = 49 K DT = q m x C