Assignment

There are two correct answers for #9 on the exam

choice b and c. If you have one of these and it is

marked wrong--submit your exam and get 6 points.

Thermochemical Equations

• 1. The stoichiometric coefficients always refer to the number of moles of a substance

• 2. If you reverse a reaction, the sign of !H changes

• 3. 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 = 2 x 6.01 = 12.0 kJ

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

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

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

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

H2O (l) H2O (g) !H = 44.0 kJ

!H˚ Is The Standard Thermodynamic State

In thermodynamics we also define a standard state of where conditions are rigorously defined. They are different than a gas at STP (1 atm and 0˚C).

--1 atmosphere pressure--25˚C = 298.15K--1 Molar concentration for solutions containing solutes

THERMODYNAMIC STANDARD STATE

WE ADD A SUPERSCRIPT !H˚ TO DENOTE STANDARD STATE CONDITIONS OR MEASUREMENT

!H˚ Is The Standard Thermodynamic State

Example of Reaction

N2 (g) + 3H2 (g) -------> 2NH3 (g) !H˚ = -92.38 kJ

1 mol of N2 reacts with 3 mol H2 to produce 3NH3 and

evolves -92.38 kJ at 1 atm, 25˚C

2N2 (g) + 6H2 (g) -------> 4NH3 (g) !H˚ = 2 X -92.38 kJ

1/3N2 (g) + H2 (g) ----> 2/3NH3 (g) !H˚ = 1/3 X -92.38 kJ

Types of Heats of Reaction, !Hrxn

WHEN MEASURED UNDER STANDARD CONDITIONS THEY

ALL BECOME !H˚rxn

Thermochemical Equations

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

123.9 g/mol.

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

266 g P4

1 mol P4

123.9 g P4

x-3013 kJ

1 mol P4

x = -6470 kJ!H =

!H = H (products) – H (reactants)

Using the Heat of Reaction (!Hrxn) to Find Amounts

SOLUTION:

The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by

If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 103 kJ of heat is transferred?

Al2O3(s) 2Al(s) + 3/2O2(g) !Hrxn = 1676 kJ

heat produced = 1.000! 103 kJ ! 2 mol Al

1676 kJ! 26.98 g Al

1 mol Al= 32.20 g Al

Why Do We Care About State Functions?

State 2

State 1

• A change in a functions of state, ! has a unique value between any two arbitrary states.

mgh2

h = 1

Hf

Hi

!H!h

Enthalpy Energy Level Diagrams

En

thalp

y, H

En

thalp

y, H

CH4 + 2O2

CO2 + 2H2O

Hinitial

Hfinal

H2O(l)

H2O(g)

heat out heat in!H < 0 !H > 0

A Exothermic process B Endothermic process

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)

Hfinal

Hinitial

Why Do We Care About State Functions?

Data From

A Table

• 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. Property of a State Function!

Suppose we want to calculate the enthalpy for the following reaction and can not measure it directly. Thermodynamics let’s us calculate it.

CH4 (g) + 2O2 ==> CO2 + 2H2O !H1 = ? kJ

CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O !H3 = - 283 kJ

CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 !H2 = - 607 kJ

Hess Law Example 2

CH4 (g) + 2O2 ==> CO2 + 2H2O !H1 = ? kJ

CO(g) + 2H2O + 1/2 O2(g) ==> CO2 + 2H2O !H3 = - 283 kJ

CH4 (g) + 2O2 ==> CO + 2H2O + 1/2 O2 !H2 = - 607 kJ

Tabulated Standard Heats of Reactions Are Used To Predict Any !H Using Hess’s law.

Using the characteristics of Hess’s Law just described,

calculate the enthalpy of the following reaction, using the

reactions and their associated enthalpies from below.

1. Fe(s) + Cl2(g) FeCl2(s) !H = -341.8 kJ/mol

2. Fe(s) + 3/2 Cl2(g) FeCl3(s) !H = -399.5 kJ/mol

FeCl2(s) + 1/2 Cl2(g) FeCl3(s) !H = ?

Standard Enthalpy of Formation !Hf˚Standard Enthalpy of Formation, !Hf

°, is the enthalpy change associated

with the formation of one mole of a compound from its elements in their standard states----298.15K (25˚C) and 1 atmosphere pressure.

•Examples:

H2(g) + O2(g) "# H2O(l ) !Hf° = -285.8 kJ/mol

3C(s) + 4H2(g) "# C3H8(g) !Hf° = -103.85 kJ/mol

Ag(s) + 1/2Cl2(g) ----------> AgCl(s)

Always For One Mole of Product!---thus, we see fractional molesas Reactants on Occasion.

•Oxygen exists as O2 gas at 25 °C

•Carbon exists as solid graphite at 25 °C.

•Sulfur exits as S8 as a solid at 25˚C

•Water is H2O(l ) in its standard state (not ice or water vapor).

Writing Formation Equations

PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include !H0

f.

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Use the table of heats of formation for values.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

Writing Formation Equations PROBLEM:

Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include !H0

f.

SOLUTION:

PLAN:

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

Use the table of heats of formation for values.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

!H0f = -127.0 kJ(a) Ag(s) + 1/2Cl2(g) AgCl(s)

!H0f = -1206.9 kJ(b) Ca(s) + C(graphite) + 3/2O2(g) CaCO3(s)

!H0f = 135 kJ(c) 1/2H2(g) + C(graphite) + 1/2N2(g) HCN(g)

Calculating !Hf° from tabulated data

Applications of Enthalpies of Formation.

Example: calculate the !Hcomb for the combustion of benzene, C6H6(l )

from enthalpies of formation taken from a table.

2 C6H6(l ) + 15 O2(g) "# 12 CO2(g) + 6 H2O(l ) !Hcomb = ?

C(s) + O2(g) "# CO2(g) !Hf° = -393.5 kJ

H2(g) + O2(g) "# H2O(l ) !Hf° = -285.8 kJ

6C(s) + 3H2(g) "# C6H6(l ) !Hf° = 49.04 kJ

1. From a

Table

2. Rearrange thermochemical equations to satisfy desired equation

C6H6(l ) + O2(g) "# 6CO2(g) + 3H2O(l) !Hcomb° = ?

1/2

Heats of Reaction From !Hf°

!Hcomb° = 6(-393.5 kJ) + 3(-285.8 kJ) + (-49.04 kJ) = -3,267.4 kJ

6C(s) + 6O2(g) "# 6CO2(g) !Hf° = 6(-393.5 kJ)

3H2(g) + O2(g) "# 3H2O(l ) !Hf° = 3(-285.8 kJ)

C6H6(l ) "# 6C(s) + 3H2(g) !Hf° = (-49.04 kJ)

C6H6(l ) + O2(g) "# 6CO2(g) + 3H2O(l) !Hcomb° = ?

2 C6H6(l ) +15 O2(g) # 12 CO2(g) + 6 H2O(l ) !Hcomb = 2 X -3267 kJ/mol

that’s for one mole of benzene only!

where $ means “the sum of”

ni is the respective molar coefficient for ith product

mi is the respective molar coefficient for each ith reactant

Major Riff.......

We can obtain the same answer by using a balanced equation for one mole of product (combustion, neutralization, dissolution) a table of standard enthalpies and the following easy formula:

!H°rxn

= $ ni !Hif°(products) - $ mi !Hif

°(reactants)

Heats of Reaction From !Hf°

• !H°rx for any chemical reaction:

!H° = !H°f (Products) – !H°f (Reactants

!H° = [c!H°f (C) + d!H°f (D)] – [a!H°f (A) + b!H°f (B)]

aA + bB cC + dD

Calculating the Heat of Reaction from Heats of Formation

PLAN:

Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate !H0rxn from !H0

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Look up the !H0f values and use Hess’s Law to find !Hrxn.

!H0f NH3(g) = -46.3 kJ/mol

!H0f H2O = -241.8 kJ/mol !H0

f NO(g) = 90.4 kJ/mol

!H0f O2(g) = 0

Calculating the Heat of Reaction from Heats of Formation

Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

Calculate !H0rxn from !H0

f values.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

SOLUTION: !Hrxn = $ m!H0f (products) - $ n!H0

f (reactants)

!Hrxn = [4(!H0f NO(g) + 6(!H0

f H2O(g)] - [4(!H0f NH3(g) + 5(!H0

f O2(g)]

= (4 mol)(90.4 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-46.3 kJ/mol) + (5 mol)(0 kJ/mol)]

!Hrxn = -904 kJ

Heats of Reaction From !Hf°

Energy Diagram Depiction !H˚f

The general process for determining !H0rxn from !H0

f values.

Enth

alp

y, H

Elements

Reactants

Products

!H˚rxn = $ m!H˚

f(products) - $ n!H˚f(reactants)

de

co

mp

os

itio

n

-!H0f !H0

f

form

ati

on

!H0rxn

Hinitial

Hfinal