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

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.

EK = 1/2mv2(1 Joule = 1 kgm2/s2)

(1 calorie = 4.184 J)

• Potential Energy: is stored energy.
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?
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.
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)
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

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

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

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

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

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)

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

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.

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
• 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

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:

Is an endothermic reaction a favorable process thermodynamically speaking?

• Yes
• No
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

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
• 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

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
• 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)]

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

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

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:

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

• -H
• + H
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 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.
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 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

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

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
• 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?