assignment - chemistry 7 · assignment there are two correct answers for #9 on the exam choice b...
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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
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