ch 5 thermochemistry power point

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Thermochemistry

Chapter 5

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Thermochemistry

• Thermochemistry:

Relationships between chemical reactions and energy changes

• Thermodynamics:

• Therm: heat Dynamics: power

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Energy: the capacity to do work

• Radiant energy

• Thermal energy

• Chemical energy

• Nuclear energy

• Potential energy- relative to position- stored in bondselectrostatic potential energy (C) = kQ1Q2

d

• Kinetic energy Ek = ½ mv2

• Units of Energy: J = 1kg-m2/s2 1 cal= 4.184 J

Energy

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Energy Changes in Chemical Reactions

900C 900C

Which one has more thermal energy?

Coffee cup? Bathtub?

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900C400C

Which one has more thermal energy?

What additional information is needed?

How does it relate to heat?

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Heat:

the transfer of _______ between two bodies that are at ________temperatures.


Temperature is a measure of the ________.

Measure of the ______________

Temperature = Thermal Energy

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The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

Energy Changes

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Energy Changes

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Exothermic process: is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat is supplied to the system from the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

Enthalpy

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Enthalpy

• Enthalpy: a thermodynamic quantity used to describe heat changes(absorbed/released) at constant pressure (most reactions occur at constant pressure).

H = E + PV

Equation:

ΔHreaction = ΔHproducts - ΔHreactants

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Enthalpy: Enthalpy Diagrams

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0

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Enthalpy: Enthalpy Diagrams

ΔHrxn= ΔHproducts – ΔHreactants

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Enthalpy: Examples

Indicate the sign of enthalpy change in the

following processes carried under atmospheric

pressure, and indicate whether the process is

exothermic or endothermic:

1. An ice cube melts

2. 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O

3. A bowling ball is dropped from a height of 8 ft into a bucket of sand.

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Thermochemical Equations

H2O (s) → H2O (l) H = 6.01 kJ

Is H negative or positive?

System absorbs heat

Endothermic

H > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

6.3

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Thermochemical Equations

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) H = -890.4 kJ

Is H negative or positive?

System gives off heat

Exothermic

H < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

6.3

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H2O (s) H2O (l) H = 6.01 kJ

Stoichiometric coefficients: equal the number of moles of a substance

Thermochemical Equations: Rules

If you reverse a reaction, the sign of H changes

H2O (l) H2O (s) H = - 6.01 kJ

If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l) H = 12.0 kJ

2 x 6.01kJ/mol =12.0 KJ/2 mol

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When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes.

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The physical states of all reactants and products must be specified in thermochemical equations.

H2O(s) → H2O(l) ΔH = 6.00 kJ

H2O(l) → H2O(g) ΔH = 44.0 kJ

• What will be the ΔH when 1 mole of ice at 0ºC is changed into one mole of steam at the boiling point?

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How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x-3013 kJ1 mol P4

x [ -6470 kJ]

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Enthalpy: Examples

4. The complete combustion of liquid octane, C8H18, to produce carbon dioxide and liquid water at 25 ºC and at constant pressure gives off 47.9 kJ per gram of octane. Write a chemical equation to represent this information.

[2C8H18(l) + 25 O2(g) → 16 CO2(g) + 9H2O(l)

ΔH = -1.09 x 104 kJ]

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Enthalpy: Examples

5. Given the thermochemical equation:SO2(g) + ½ O2(g) →SO3(g) ΔH = -99.1 kCal

Calculate the heat evolved when 74.6 g of SO2

(MM = 64.07 g/mol) is converted to SO3. [ -115 kJ]

6. Hydrogen peroxide can decompose to water and oxygen by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH = -196 kJ

Calculate the value of q (heat evolved or absorbed) when 5.00 g of H2O(l) decomposes at constant pressure.

[ -14.4 kJ]

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Enthalpy: Examples

7. Given the equation

H2(g) + I2(s) → 2HI(g) ΔH = +52.96 KJ

Calculate ΔH for the reaction

HI(g) → ½ H2(g) + ½ I2(s)[ -26.48 kJ]

8. Given the equation:

3O2(g) → 2O3(g) ΔH = +285.4 kJCalculate ΔH for the following reaction:

3/2 O2(g) → O3(g)[ 142.7 kJ]

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Calorimetry and Heat Exchanges in a Chemical Reaction

• Read Section 6.4 (Pages 213-2200)• Examples: pp. 216 – 220: 6.1 – 6.3• Exercises: pp. 240: 6.19, 6.23, 6.26, 6.28

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Calorimetry

Calorimetry is performed in devices called calorimeters

Calorimetry is based on the Law of Conservation of Energy

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Calorimetry

9. The specific heat of aluminum is 0.895 J/g ºC. Calculate the heat necessary to raise the temperature of 40.0 g of aluminum from -20.0 ºC to 32.3 ºC. Specific heat of aluminum is 0.215-cal/g ºC.

[1.87 x 103 Joules]

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Calorimetry: Heat Capacity

• Calorimetry:Experimental measurement of heat produced in chemical and physical processes

• Heat Capacity, C, of a system:

the quantity of heat needed to raise the temperature of matter 1 K.

• Units: J/ºC

T

qC

• Molar Heat Capacity:

The heat required to raise 1 mole of a substance, 1 K.

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Calorimetry: Specific Heat

• Specific heat of a substance

(also represented by cp):

the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius ( or 1 K).

Tq

eatSpecific

massSpecific

p

p

mc

Tm

q

m

1 (

T

q

m

C c : h

T

q C but

m

C

capacity heat heat

)

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Calorimetry: Definitions

Heat (q) absorbed or released:

q = mcpt

q = Ct

t = tfinal - tinitial

When q>0 endothermic reaction

When q<0 exothermic reaction

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Calorimetry: Specific Heat

10. How much heat, in joules and kilojoules, does it take to raise the temperature of 225 g of water from 25.0 ºC to 100.0 ºC

[ 7.05 x 104 J; 70.5 kJ]

11. Calculate the heat capacity, specific heat and the molar heat capacity of an aluminum block that has mass of 5.7 g that must absorb 629 J of heat from its surroundings for its temperature to rise from 22 ºC to 145 ºC.

[C = 5.11 J/ºC; cp = 0.90 J g-1 ºC-1, 24 J mol-1 ºC-1 ]

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Calorimetry: Problems

12. What will be the final temperature if a 5.00-g silver ring at 37.0 ºC gives off 25.0 J of heat to the surroundings? Specific heat of silver = 0.235 J g-1 ºC-1

[15.7 ºC]

13. A 15.5 g sample of a metallic alloy is heated to 98.9 ºC and then dropped into 25.0 g of water in a calorimeter. The temperature of the water rises from 22.5 ºC to 25.7 ºC. Calculate the specific heat of the alloy.

[0.29 J g-1 ºC-1]

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Calorimetry: Problems

14. Suppose a piece of gold with a mass of 21.5 g at temperature of 95.00 ºC is dropped into an insulated calorimeter containing 125.0 g of water at 22.00 ºC. What will be the final temperature of the water?

[22.4 ºC]

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Measuring Enthalpy Changes for Chemical Reactions: Constant-

Volume Calorimetry

Bomb Calorimeter

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Constant Volume Calorimeter: Isolated System

Isolated system:

No exchange of matter and energy

Uses Cp of water

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Constant-Volume Calorimetry: Isolated system

qsys = qwater + qbomb + qrxn = 0

There is no heat exchange with the surroundings

qrxn = - (qwater + qbomb)

qbomb = Cbombt qwater = mwcpΔt

qrxn = -(qwater + CbombΔt) = -(mwcpΔt + CbombΔt)

qrxn = - (mwcp + Cbomb)ΔtMeasured heat is not enthalpy: constant volume

Cbomb= mbombx cbomb

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Constant Volume Calorimetry: Problems

15. In a preliminary experiment, the heat capacity of a bomb calorimeter assembly is found to be 5.15 kJ/°C. In a second experiment. A 0.480 g sample of graphite (carbon) is placed in the bomb with an excess of oxygen. The water, bomb, and other contents of the calorimeter are in thermal equilibrium at 25.00 °C. The graphite is ignited and burned, and the water temperature rises to 28.05 °C. Calculate ΔH for the reaction.

C(graphite) + O2(g) → CO2(g)[ΔH = -393 kJ]

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Measuring Enthalpy Changes for Chemical Reactions

Coffee-Cup Calorimeter: Constant pressure

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Constant-Pressure Calorimetry

No heat enters or leaves!Insulated system! Pressure is constant.

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcalor)

qwater = mcpt

qcal = Ccalt

Reaction at Constant P

ΔH = qrxn = - (qwater + qcalor)

Measured heat is the enthalpy

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Calorimetry

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Constant Pressure Calorimetry: Problems

16. (a) A 50.0 mL sample of 0.250 M HCl at 19.50 °C is added to 50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a calorimeter. After mixing, the solution temperature rises to 21.21 °C. Calculate the heat of this reaction. Assume te volumes of the solutions are additive. Ignore the heat absorbed by the calorimeter.

[-715 J]

(b) Determine the value of ΔH that should be written in the equation for the neutralization reaction:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH = ?

[-57.2 kJ]

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Calorimetry: Problem

17. A quantity of 1.00 x 102 mL of 0.500M HCl is mixed with 1.00 x 102 mL of 0.500 M NaOH in a constant pressure calorimeter having a heat capacity of 335 J/ ºC. The initial temperature of the HCl and NaOH solutions is the same, 22.50 ºC, and the final temperature of the mixed solution is 24.90 ºC. Calculate the heat change for the neutralization reaction and the molar heat of the neutralization reaction. Assume s = 4.18 J/gºC for the solution.

[-2.81 kJ; -56.2 kJ/mol]

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Chemistry in Action:

Fuel Values of Foods and Other SubstancesC6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) H = -2801 kJ/mol

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J

One food Calorie:

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Standard Enthalpy of Formation

Absolute value of the enthalpy of a substance cannot be measured. We need a frame of reference.

Standard states for elements and compounds has to be established.

Standard state for a solid or liquid substance is the pure element or compound at 1 atm at a given temperature.

Standard state for gas: pure gas behaving as an ideal gas at 1 atm and temperature of interest.

Standard enthalpy of formation: the enthalpy change for a reaction in which the reactants in their standard state yield products in their standard states. Standard state denoted with a superscript (º)

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Standard enthalpy of formation ΔHºf (also called heat of formationenthalpy change that occurs in the formation of 1 mole of the substance from its elements when both products and reactants are in their standard states (f stands for formation).

Reference forms: standard enthalpy of formation of a pure element in its reference form is 0.standard enthalpy of formation (ΔH0) as a reference point for all enthalpy expressions is the formation of H2(g) = 0 kJ/mol

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Standard enthalpy of formation (H0f) is the heat change

that results when one mole of a compound is formed from its elements at a pressure of 1 atm and 25 ºC .

The standard enthalpy of formation of any element in its most stable form is zero.

H0 (O2) = 0f

H0 (O3) = 142 kJ/molf

H0 (C, graphite) = 0f

H0 (C, diamond) = 1.90 kJ/molf


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standard

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Determination of Heat of Formation

1. Direct method: From elements when it can be easily measured.

C(graphite) + O2(g) →CO2(g) ΔHrxn = -393.5 kJ

ΔHrxn= ΔHoCO2 – ΔH0O2 – ΔHgraphite = -393.5 kJ

ΔH0CO2 = -393.5 kJ

2. Indirect method: Using the heat of reaction and heats of formations

Given CH4(g) +2O2(g) → CO2(g) + 2H2O(g) ΔHrxn=-890.3 kJ

Calculate the heat of formation of CH4

[-74.8kJ/mol]

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The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm and 25ºC.

rxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn nH0 (products)f= mH0 (reactants)f-

Standard Enthalpy of Reaction

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Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn nH0 (products)f= mH0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

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Standard Enthalpy of Reaction

18. Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds, such as methanol. One reaction for producing synthesis gas is shown here:

3CH4(g) + 2H2O(l) + CO2(g) → 4CO(g) + 8H2(g) ΔHº = ?

[747.5 kJ]

19. The combustion of isopropyl alcohol, common rubbing alcohol, is represented by this equation:

2(CH3)2CHOH(l) + 9O2(g) → 6CO2(g) + 8H2O(l)ΔHº = -4011 kJuse this equation and data from Thermodynamic tables to establish the standard enthalpy of formation for isopropyl alcohol.

[ -318 kJ/mol]

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Heat of Formation: Examples

20. Pentaborane-9, B5H9, is a colorless liquid. It is highly reactive substance that will burst into flame or even explode when exposed to oxygen:

2B5H9(l) + 12O2 → 5B2O3(s) + 9H2O(l)

It was considered a potential rocket fuel as it release large amount of heat per gram. Calculate the kJ of heat released per gram of the compound reacted with oxygen. The standard enthalpy of formation of B5H9 is 73.2 kJ/mol

[-71.58 kJ/gB5H9]

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Heat of Reactions (Formations): Hess’s Law of Constant heat of Summations

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

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Hess’ Law: Example

Given the following reactions:

CO(g) + ½ O2(g) → CO2(g) ΔH = -283.0 kJ

C(graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ

Calculate the enthalpy for the formation of CO from its elements

C(graphite) + ½ O2(g) → CO(g) ΔH = ?

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Hess’s Law: Example

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Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

H0 = -393.5 + 2(-296.1) + 1072 = 86.3 kJrxn6.6

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Hess’ Law: Examples

21. GivenC(graphite) + O2 (g)→ CO2(g) ΔHºrxn= -393.5 kJ

H2(g) + ½ O2(g) → H2O(l) ΔHºrxn = -285.8 kJ

2C2H2(g) + 5O2(g) → CO2(g) + 2H2O(l) ΔHºrxn = -2598.8 kJ

Calculate the heat of formation of C2H2(g)

2C(graphite) + H2(g) → C2H2(g)

[52.3 kJ]

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Hess’ Law: Conceptual Example

Without performing acalculation, determinewhich of the twosubstances should yield the greater quantity of heat uponComplete combustion,on a per mole basis:ethane, C2H6(g), or ethanol, CH3CH2OH(l)

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The Solution Process for NaCl

Hsoln = Step 1 + Step 2 = 788kJ – 784kJ = 4 kJ/mol

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The enthalpy of solution (Hsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

Hsoln = Hfinal soln - Hcomponents

Which substance(s) could be used for melting ice?

Which substance(s) could be used for a cold pack?

Which substances could be used for hot pack?

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Heat of Solution and Dilution

The process: NaCl(s) → Na+(aq) + Cl-(aq)

The steps:

1. Separate ions in the crystal into ions in gaseous state

2. Hydration of the gaseous ions

Energy involved:

Lattice Energy, U

Hydration energy, ΔH hydration

ΔH solution= U + ΔH hydration

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Heat of Solution and Dilution

Lattice energy: energy required to completely separate one mole of solid ionic compound into gaseous ions (endothermic process)

U + NaCl(s) →Na+(g) + Cl-(g)

Hydration of the gaseous ionsHeat of hydration: energy released when ions interact with water molecules.

Na+(g) + Cl-(g) → Na+(aq) + Cl-(aq) + Energy

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Heat of Dilution

The heat change associated with the

dilution of a solution.

Can be exothermic or endothermic

ch 5 thermochemistry power point

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