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Thermochemistry

Thermochemistry:

- Energy
- Kinetic & Potential
- First Law of Thermo
- internal energy, heat & work
- endothermic & exothermic processes
- state functions
- Enthalpy
- Enthalpies of Reaction

Calorimetry

- heat capacity and specific heat
- constant-pressure calorimetry
- bomb calorimetry (constant-volume calorimetry)
- Hess’s Law
- Enthalpies of Formation
- for calculation of enthalpies of reaction
- Foods and Fuels

Energy

- work is a form of energy w = F x d
- energy is the capacity to do work or transfer heat
- Kinetic Energy
- energy of motion E = ½ mv2
- potential energy
- energy of position
- applies to electrostatic energy
- applies to chemical energy (energy of bonds)

energy units

- one joule = energy of a 2 kg mass moving at 1 m/s
- E = ½ mv2 (½)(2 kg) (m/s)2 = kg m2/s2 = 1 J
- 1 cal = 4.184 J 1 kcal = 1 food calorie (Cal)

Systems & Surroundings

- system -- chemicals in the reaction
- surroundings -- container & all outside environment

- closed system can exchange energy (but not matter) with its surroundings

energy(as heat or work)

2H2(g) + O2(g)

¯

2H2O(l)+ energy

(system)

no exchg of matterwith surroundings

First Law of Thermo.

- Energy is always conserved
- any energy lost by system, must be gained by surroundings
- Internal Energy -- total energy of system
- combination of all potential and kinetic energy of system
- incl. motions & interactions of of all components
- we measure the changes in energy E = Efinal - Einitial

+ D E = Efinal > Einitial system has gained E from surroundings

- - D E = Efinal < Einitial system has lost E to surroundings
- Relating D E to heat and work
- D E = q + w q is positive if heat goes from surroundings to system w is positive if work is done on system by surroundings

Endothermic

- system absorbs heat or heat flows into the system

Exothermic

- system gives off heat or heat flows out of the system

State Function

- a property of a system that is determined by specifying its condition or state (T, P, etc.)
- internal energy is a state function, \DE depends only on Efinal & Einitial

Enthalpy

- for most reactions, most of the energy exchanged is in the form of heat, that heat transfer is called enthalpy, H
- enthalpy is a state function
- like internal energy, we can only measure the change in enthalpy, DH
- DH = qp when the process occurs under constant pressureDH = Hfinal - Hinitial = qp
- - DH Þ exothermic process
- +DH Þ endothermic process

- D Hrxn = Hprod - Hreact
- enthalpy is an extensive property
- magnitude of D H depends directly on the amount of reactant
- C(s) + 2H2(g)® CH4(g)D H = -74.8 kJ/mol
- 2C(s) + 4H2(g)® 2CH4(g)D H = -149.6 kJ/2mol

enthalpy change for forward rxn is equal in magnitude but opposite in sign for the reverse rxn

- CH4(g) ® C(s) + 2H2(g)D H = +74.8 kJ/mol
- C(s) + 2H2(g)® CH4(g) D H = - 74.8 kJ/mol
- enthalpy change for a reaction depends on the state of the reactants and products
- C(g) + 2H2(g)® CH4(g)D H = -793.2 kJ/mol
- 2H2(g) + O2(g)® 2H2O(g)D H = -486.6 kJ/mol
- 2H2(g)+ O2(g)® 2H2O(l)D H = -571.7 kJ/mol

Practice Ex. 5.2:

- Hydrogen peroxide can decompose to water and oxygen . Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.
- 2H2O2(l)® 2H2O(l) + O2(g)D H = -196 kJ
- 5.00 g H2O2(l) x 1 mol = 0.147 mol H2O2(l) 34.0 g H2O2(l)
- 0.147 mol H2O2(l) x -196 kJ H2O2(l) = -14.4 kJ 2 mol

Calorimetry

- experimental determination of D H using heat flow

heat capacity

- measures the energy absorbed using temperature change
- the amount of heat required to raise its temp. by 1 K
- molar heat capacity -- heat capacity of 1 mol of substance

specific heat

- heat energy required to raise some mass of a substance to some different temp.
- specific heat = quantity of heat trans. (g substance) (temp. change)
- = q . m DT
- S.H. = joule g K
- q = (S.H.) (g substance) (D T)

remember: this is change in temp.

Practice Ex. 5.3:

- Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temp. increases by 12.0 °C if the specific heat of the rocks is 0.82 J/gK.
- S.H. x g x DT = joules
- What unit should be in the solution?
- joules -- quantity of heat
- 0.82 J x 50.0 x 103 g x 12.0 K = 4.9 x 105 Jg K

Constant-Pressure Calorimetry

- D H = qp at constant pressure as in coffee cup calorimeter
- heat gained by solution = qsoln
- \qsoln = (S.H.soln)(gsoln)(DT)
- heat gained by solution must that which is given off by reaction
- \qrxn = - qsoln = - (S.H.soln)(gsoln)(DT)

must be opposite in sign

if DT is positive then qrxn is exothermic

Practice Ex. 5.4:

- When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from 22.30°C to 23.11°C. Calculate D H for this reaction, assuming that the combined solution has a mass of 100.0 g and a S.H. = 4.18 J/g °C.
- AgNO3(aq) + HCl(aq) ® AgCl(s) + HNO3(aq)
- qsoln = 4.18 J x 100.0 g soln x 0.81°C = 3.39 x 102 J g °C
- qrxn = - qsoln = - 3.39 x 102 J = - 68,000 J or 0.00500 mol - 68 kJ/mol

Bomb Calorimetry (Constant-Volume)

- bomb calorimeter has a pre-determined heat capacity
- sample is combusted in the calorimeter and D T is used to determine the heat change of the reaction
- qrxn = - Ccalorimeter x D T

heat capacity of calorimeter

because rxn is exothermic

Practice Ex. 5.5:

- A 0.5865 g sample of lactic acid, HC3H5O3, is burned in a calorimeter with C = 4.812 kJ/°C. Temp. increases from 23.10°C to 24.95°C. Calculate heat of combustion per gram and per mole. D T = +1.85°C
- qrxn = - (4.812 kJ/°C) (1.85°C) = - 8.90 kJ per 0.5865 g lactic acid-8.90 kJ = - 15.2 kJ/g 0.5865 g
- - 15.2 kJ x 90 .1 g = - 1370 kJ/mol 1 g 1 mol

Hess’s Law

- rxns in one step or multiple steps are additive because they are state functions
- eg.
- CH4(g) + 2O2(g)® CO2(g) + 2H2O(g)D H = - 802 kJ
- CH4(g) + 2O2(g)® CO2(g) + 2H2O(l)D H = - 890 kJ

2H2O(g)® 2H2O(l)D H = - 88 kJ

Practice Ex. 5.6:

- Calculate D H for the conversion of graphite to diamond:
- Cgraphite® Cdiamond
- Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ
- Cdiamond + O2(g)® CO2(g)D H = -395.4 kJ
- Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ
- CO2(g)® Cdiamond + O2(g)D H = 395.4 kJ

Cgraphite® CdiamondD H = + 1.9 kJ

Enthalpies of Formation

- enthalpies are tabulated for many processes
- vaporization, fusion, formation, etc.
- enthalpy of formation describes the change in heat when a compound is formed from its constituent elements, DHf
- standard enthalpy of formation, DHfo, are values for a rxn that forms 1 mol of the compound from its elements under standard conditions, 298 K, 1 atm

For elemental forms:

- eg. C(s) graphite, Ag(s) , H2(g) , O2(g) , etc.
- DHfo, for any element is = 0
- used for calculation of enthalpies of reaction, DHrxn
- DHrxn = S DHfo prod - S DHfo react

Practice Ex. 5.9:

- Given this standard enthalpy of reaction, use the standard enthalpies of formation to calculate the standard enthalpy of formation of CuO(s)(
- CuO(a) + H2(g)® Cu(s) + H2O(l)DHo = -130.6 kJ
- DHrxn = S DH f o prod - S DH f o react
- -130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)]
- DHfo CuO = -155.2 kJ/mol

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