Chapter 6

Thermochemistry

Thermochemistry

• Deals with the energy changes

• Chemical reactions

– Determines whether a reaction is exothermic or endothermic

• Physical Changes

– Energy changes

Vocabulary

• Energy - The ability to do work (E) in J

• Potential Energy – Energy of position (PE)

• Kinetic Energy – Energy of motion (KE)

• Temperature – Random motion of particles

• Heat – Transfer of energy (q) in J

• Work – Force x Distance (W) in J

• Enthalpy – Energy Change for reactions – (ΔH) in J

Vocabulary

• Pathway – How energy is transferred

• State Function – Properties that depend on current state. (Independent of pathway)

• Energy is a state function– Temperature, Volume, Pressure

• Heat and Work are not state functions

Example of Energy Change

Endothermic Reactions• Energy is put into a reaction

• Product bonds are weaker than reactant bonds

http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig10.gif

Exothermic Reactions• Energy is released from a reaction

• Product bonds are stronger than reactant bonds

http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig03.gif

Signs for Exo vs. Endo.

Calorimetry

• Measuring heat associated with a reaction

• Uses a calorimeter – Coffee Cup Calorimeter

• At constant pressure q = -ΔH

q=m*s* ΔT

• q = heat, m = mass, ΔT = Change in Temp

• s=specific heat capacity

Specific Heat Capacity

• Energy required to heat one gram of a substance 1 degree Celcius

• Units J/g*ºC

• Molar Heat Capacity = Just per mole

• Metals have low heat capacities

• Example

How much energy is required to heat 55.3 grams of Aluminum metal from 22.0ºC to 100.0 ºC? Specific heat of water is 4.18 J/g*ºC the specific heat of aluminum is 0.89 J/g*ºC

• Example

#46 p. 282

Homework

• P. 281 #’s 32, 37,40, 43, 48

Hess’s Law

• In going from a particular set of reactants to a particular set of products, the change in enthalpy (H) is the same whether the reaction takes place in one step or in a series of steps

• Energy is a state function

Comparison

• One step:

N2(g) + 2O2(g) 2NO2(g) H1 = 68kJ

• Two step

N2(g) + O2(g) 2NO(g)H2 = 180kJ

2NO(g) + O2(g) 2NO2(g) H3 = -112kJ

N2(g) + 2O2(g) 2NO2 (g) H2 + H3 = 68kJ

Hess’s Law Rules

• Start with the end reaction in mind and work backward rearrange your equations

• Work to cancel like terms

• When reversing the equation the sign of ΔH must be changed

• When multiplying an equation by a coefficient ΔH must be multiplied by the same coefficient

• Example #54 p. 283

NH3 ½ N2 + 3/2 H2 H = 46 kJ

2H2 + O2 2H2O H = -486 kJ

2N2 + 6H2O 3O2 + 4NH3 H = ?

Based on enthalpy is this a useful synthesis?

• Example #52 p. 283

C4H4 + 5O2 4CO2 + 2H2O H = -2341kJ

C4H8 + 6O2 4CO2 + 4H2O H = -2755kJ

H2 + ½ O2 H2O H = -286kJ

C4H4 + 2H2 C4H8 H = ?

Homework

• P. 283 #’s 51,53,55,57

• Calculate how much energy (kJ) is required to heat 50.0 L of tap water from 10.0 ºC to 45.0 ºC. Assume the density of water is 1.00 g/mL and specific heat of water is 4.18 J/g* ºC

– 7320 kJ

• A McDonalds hamburger contains 250. Calories (that’s 2.50x105 calories). How many burgers would it take to supply the same amount of energy? 1 calorie = 4.18 J

– 7.00 burgers

Enthalpy of Formation

• The change in enthalpy that accompanies the formation of one mole of a compound from its elements with all elements in their standard state

• Symbol = ΔHfº

– f is formation– Degree symbol means standard states

Standard States

• Standard States

– Temperature is 25 ºC

– Pressure = 1.00 atm

– Concentration = 1.00 M

– The standard state is how something exists at these conditions

– O2(g), Na(s), Br2(l)

Reactions

• 4C(s) + 2H2(g) CH4(g) ΔHfº = -75 kJ/mol

• H2(g) + ½ O2(g) H2O(l) ΔHfº = -286 kJ/mol

• Write the enthalpy of formation reaction for gaseous carbon dioxide. Don’t worry about the value

• C(s) + O2(g) CO2(g)

• Values are in Appendix 4 (A21)

• ΔHfº = -393.5 kJ/mol

Enthalpy of Formation Values

• Most values are negative– Compounds are more stable than element

• All elements are zero• Even diatomic elements!• States matter

– Liquid water is -286– Gaseous water is -242

• Use this information to calculate enthalpy of the reaction!

Standard Enthalpy of Reaction

• Standard Enthalpy of Reaction is the energy change at standard conditions

• Symbol = ΔHº

• ΔHº = Σ ΔHfº(Products) - Σ ΔHfº(Reactants)

• Remember to include coefficients

• Remember elements are zero!

• Calculate the standard enthalpy of reaction for the following

Al(s) + Fe2O3(s) Al2O3(s) + Fe(s)

Al = 0, Fe = 0, Fe2O3 = -826, Al2O3 = -1676

Homework

• P. 284 #63