Ch 6 thermochemistry
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CH 6: Thermochemistry. Renee Y. Becker Valencia Community College CHM 1045. Energy. Energy : is the capacity to do work, or supply heat. Energy = Work + Heat Kinetic Energy : is the energy of motion. E K = 1 / 2 mv 2 (1 Joule = 1 kg m 2 /s 2 ) (1 calorie = 4.184 J)

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CH 6: Thermochemistry

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Ch 6 thermochemistry

CH 6: Thermochemistry

Renee Y. Becker

Valencia Community College

CHM 1045


Energy

Energy

  • Energy: is the capacity to do work, or supply heat.

    Energy = Work + Heat

  • Kinetic Energy: is the energy of motion.

    EK = 1/2mv2(1 Joule = 1 kgm2/s2)

    (1 calorie = 4.184 J)

  • Potential Energy: is stored energy.


E k e p

Ek & Ep


Example 1 ke

Example 1: KE

Which of the following has the greatest kinetic energy?

  • A 12 kg toy car moving at 5 mph?

  • A 12 kg toy car standing at the top of a large hill?


Energy1

Energy

  • Thermal Energy is the kinetic energy of molecular motion

  • Thermal energy is proportional to the temperature in degrees Kelvin. Ethermal T(K)

  • Heat is the amount of thermal energy transferred between two objects at different temperatures.


Ch 6 thermochemistry

  • In an experiment:Reactants and products are the system; everything else is the surroundings.

    • Energy flow from the system to the surroundings has a negative sign (loss of energy). (-E or - H)

    • Energy flow from the surroundings to the system has a positive sign (gain of energy). (+E or +H)


Ch 6 thermochemistry

  • The law of the conservation of energy: Energy cannot be created or destroyed.

  • The energy of an isolated system must be constant.

  • The energy change in a system equals the work done on the system + the heat added.

    DE = Efinal – Einitial = E2 – E1 = q + w

    q = heat, w = work


Ch 6 thermochemistry

  • Pressure is the force per unit area.

    (1 N/m2 = 1 Pa)

    (1 atm = 101,325 Pa)

  • Work is a force (F) that produces an object’s movement, times the distance moved (d):

    Work = Force x Distance


Ch 6 thermochemistry

The expansion in volume that occurs during a reaction forces the piston outward against atmospheric pressure, P. Work = -atmospheric pressure * area of piston * distance piston moves


Example 2 work

Example 2: Work

How much work is done (in kilojoules), and in which direction, as a result of the following reaction?


Ch 6 thermochemistry

  • The amount of heat exchanged between the system and the surroundings is given the symbolq.

    q = DE + PDV

    At constant volume (DV = 0): qv = DE

    At constant pressure: qp = DE + PDV = DH

    Enthalpy change: DH = Hproducts – Hreactants


Example 3 work

Example 3: Work

The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 489 L at room temperature. How much PV (in kilojoules) work is done during the explosion? Assume P = 1 atm, T = 25°C.

2 C7H5N3O6(s)  12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)


Ch 6 thermochemistry

  • Enthalpies of Physical Change:

Enthalpy is a state function, the enthalpy change from solid to vapor does not depend on the path taken between the two states.

Hsubl = Hfusion + Hvap


Ch 6 thermochemistry

  • Enthalpies of Chemical Change:Often called heats of reaction (DHreaction).

    Endothermic:Heat flows into the system from the surroundings and DH has a positive sign.

    Exothermic:Heat flows out of the system into the surroundings and DH has a negative sign.


Ch 6 thermochemistry

  • Reversing a reaction changes the sign of DH for a reaction.

    C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ

    3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ

  • Multiplying a reaction increases DH by the same factor.

    3 [C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l)] DH = 3(-2219) kJ

    DH = -6657 kJ


Example 4 heat

Example 4: Heat

  • How much heat (in kilojoules) is evolved or absorbed in each of the following reactions?

    a)Burning of 15.5 g of propane:C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)

    DH = –2219 kJ/mole

    b)Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride:

    Ba(OH)2·8 H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)

    DH = +80.3 kJ/mole


Ch 6 thermochemistry

  • Thermodynamic Standard State:Most stable form of a substance at 1 atm pressure and 25°C; 1 M concentration for all substances in solution.

  • These are indicated by a superscript ° to the symbol of the quantity reported.

  • Standard enthalpy changeis indicated by the symbol DH°.


Example 5

Example 5:

Is an endothermic reaction a favorable process thermodynamically speaking?

  • Yes

  • No


Hess s law

Hess’s Law

  • Hess’s Law:The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.(not a physical change, chemical change)

    3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ


Ch 6 thermochemistry

  • Reactants and products in individual steps can be added and subtracted to determine the overall equation.

    (1) 2 H2(g) + N2(g)  N2H4(g) DH°1 = ?

    (2) N2H4(g) + H2(g)  2 NH3(g) DH°2 = –187.6 kJ

    (3) 3 H2(g) + N2(g)  2 NH3(g) DH°3 = –92.2 kJ

    DH°1 + DH°2 = DH°reaction

    Then DH°1 = DH°reaction - DH°2

    DH°1 = DH°3 – DH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ


Example 6 hess s law

Example 6: Hess’s Law

  • The industrial degreasing solvent methylene chloride (CH2Cl2, dichloromethane) is prepared from methane by reaction with chlorine:

    CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g)

    Use the following data to calculate DH° (in kilojoules) for the above reaction:

    CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)

    DH° = –98.3 kJ

    CH3Cl(g) + Cl2(g)  CH2Cl2(g) + HCl(g)

    DH° = –104 kJ


Ch 6 thermochemistry

  • Standard Heats of Formation (DH°f): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states.

  • The standard heat of formation for any element in its standard state is defined as being ZERO.

    DH°f = 0 for an element in its standard state


Standard heats of formation

Standard Heats of Formation

  • Calculating DH° for a reaction:

    DH° = DH°f (Products) – DH°f (Reactants)

  • For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient.

    aA + bB  cC + dD

    DH° = [cDH°f(C) + dDH°f(D)] – [aDH°f(A) + bDH°f(B)]


Standard heats of formation1

CO(g)

-111

C2H2(g)

227

Ag+(aq)

106

CO2(g)

-394

C2H4(g)

52

Na+(aq)

-240

H2O(l)

-286

C2H6(g)

-85

NO3-(aq)

-207

NH3(g)

-46

CH3OH(g)

-201

Cl-(aq)

-167

N2H4(g)

95.4

C2H5OH(g)

-235

AgCl(s)

-127

HCl(g)

-92

C6H6(l)

49

Na2CO3(s)

-1131

Standard Heats of Formation

Some Heats of Formation, Hf° (kJ/mol)


Example 7 standard heat of formation

Example 7: Standard heat of formation

Calculate DH° (in kilojoules) for the reaction of ammonia with O2 to yield nitric oxide (NO) and H2O(g), a step in the Ostwald process for the commercial production of nitric acid.


Example 8 standard heat of formation

Example 8: Standard heat of formation

Calculate DH° (in kilojoules) for the photosynthesis of glucose and O2 from CO2 and liquid water, a reaction carried out by all green plants.


Example 9

Example 9:

Which of the following would indicate an endothermic reaction? Why?

  • -H

  • + H


Heat of phase transitions from h f

Heat of Phase Transitions from Hf

Calculate the heat of vaporization, Hvap of water, using standard enthalpies of formation

Hf

H2O(g)-241.8 kJ/mol

H2O(l)-285.8 kJ/mol


Calorimetry and heat capacity

Calorimetry and Heat Capacity

  • Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters:

    • Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q = DE.

    • Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q = DH.


Ch 6 thermochemistry

Constant Pressure

Bomb


Calorimetry and heat capacity1

Calorimetry and Heat Capacity

  • Heat capacity (C)is the amount of heat required to raise the temperature of an object or substance a given amount.

    Specific Heat:The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C.

    q = s x m x t

    q = heat required (energy)

    s = specific heat

    m = mass in grams

    t = Tf - Ti


Calorimetry and heat capacity2

Calorimetry and Heat Capacity

  • Molar Heat:The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C.

    q = MH x n x t

    q = heat required (energy)

    MH = molar heat

    n = moles

    t = Tf - Ti


Example 10 specific heat

Example 10: Specific Heat

What is the specific heat of lead if it takes 96 J to raise the temperature of a 75 g block by 10.0°C?


Example 11 specific heat

Example 11: Specific Heat

How much energy (in J) does it take to increase the temperature of 12.8 g of Gold from 56C to 85C?


Example 12 molar heat

Example 12: Molar Heat

  • How much energy (in J) does it take to increase the temperature of 1.45 x104 moles of water from 69C to 94C?


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