# Chapter 6 Thermochemistry.

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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

Exothermic Reactions Energy is released from a reaction
Product bonds are stronger than reactant bonds

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 grams of Aluminum metal from 22.0ºC to ºC? Specific heat of water is 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
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
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 Calories (that’s 2.50x105 calories). How many burgers would it take to supply the same amount of energy? 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 = 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!

Al(s) + Fe2O3(s)  Al2O3(s) + Fe(s)
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