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# THERMODYNAMICS

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by on 19 March 2014

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#### Transcript of THERMODYNAMICS

Calculate the H for the overall reaction given the following equations:

(Use "Big Mama"/ Hess's Law equation; products-reactants)

H = [1 mol (-1676) +1 mol(-7040+ 3 mol (90) + 6 mol (-242)] - [3 mol (-2950]
= [(-1676) + (-704) + (270) + (-1452)] - (-885)
= (-3562)-(-885)
= -2677 kJ/mol
Thermodynamics
What is Thermodynamics, anyway?
Energy
is the ability to do work or produce heat. The two main components of energy are
potential energy
(stored energy) and
kinetic energy
(energy of motion). Energy cannot be created nor destroyed; this is the
Law of Conservation of Energy
, which means that any change in energy within a system must be balanced by taking out or putting in more energy. That being said, the energy of the universe is always constant and this statement is the
First Law of Thermodynamics
.

Vocabulary
Thermodynamics
- is the study of energy and its interconversions

Heat (q)
- is the result of when two systems of differing temperatures exchange thermal energy

Temperature (T)
- is
not
the measure of energy. It is a measure of the
ability
of a system, to transfer heat energy to another physical system. Temperature is closely related to the average kinetic energy of its molecules, KE
avg

Enthalpy (
H
)
- is the flow of energy (heat exchange) at a constant pressure (AKA isobaric conditions) between two systems
Enthalpy of reaction ( H
rxn
)
- is the amount of heat released (negative magnitudes) or absorbed (positive magnitudes) from a reaction at constant pressure in kJ/mol
rxn

Enthalpy of combustion( H
comb
)
- is the heat absorbed or released by burning, often times with 02, in kJ/mol
rxn

Enthalpy of formation ( H
f
)
- is the heat absorbed or released when one mole of compound is formed from elements in their standard states to kJ/mol
rxn

System
- is the term to refer to the area of the universe we focus on when dealing with energy

Surroundings
- everything outside of the system

Endothermic
- is the total heat absorbed in an experiment. In situations like this, heat is a reactant.

Exothermic
- is the total heat released in an experiment. In situations like these, heat is a product.

Work
- is the force acting over distance and is expressed in, J, Joules.
How do Energy and Work relate to each other?
Valerie Tran
Per 2
3-9-14

Work is usually associated with energy, and when it's applied to measuring the energy of a system, the equation used is :

E= q (heat) + w (work)

If there is +q, then heat has been absorbed and the system is endothermic. So when there is -q, heat is being released and it is exothermic.
In the cases dealing with gases, work is the function of pressure. Pressure is the force over area and when volume changes, work is either done on the gas or by the gas. The equation to find how much work is done on/done by the gas is:

w= -P V

In this equation, +w means that work is being done
on
the system. -w means that work is being done
by
the system.
Example with Energy:
Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.
Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

Info given: - q= 15.6 kJ E = q + w
- w= 1.4 kJ = (15.6) +(1.4)
= 17.0 kJ

Pressure and Volume Example:
Calculate the work associated with the expansion of a gas from 46L to 64 L at a constant external pressure of 15 atm.
Answer to Pressure and Volume Example:
Calculate the work associated with the expansion of a gas from 46L to 64 L at a constant external pressure of 15 atm.

Info Given : - V
i
= 46L w = -P V
- V
f
= 64L = -(15)(18)
- P = 15 atm = -270 L atm
V= V
f
- V
i
= 64L - 46L
= 18L
Internal Energy, Heat and Work Example:
A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 10 L to 4.50 x 10 L by the addition of 1.3 x 10 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process. (To convert between L atm and J, use 1 L atm= 101.3 J)
6
Answer to Internal Energy, Heat and Work Example:
A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 10 L to 4.50 x 10 L by the addition of 1.3 x 10 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process. (To convert between L atm and J, use 1 L atm= 101.3 J)

Info Given: -Vi= 4.00 x 10 L
-Vf= 4.50 x 10 L w = -P V
-q= 1.3 x 10 J = -(1.0) (5 x 10 )
-p= 1.0 atm = -5 x 10 L atm
- (convert) 101.3 J/ 1L atm

-5 x 10 L atm (101.3 J/ 1 L atm)
w= -5.065 x 10 J

E= q + w
= ( 1.3 x 10 ) + (-5.065 x 10 )
= 8.0 x 10
8
6
8
6
6
(converting)
6
6
8
5
5
7
7
7
7
V= Vf - Vi
= (4.50 x 10 ) - (4.00 x 10 )
= 5 x 10 L
5
Enthalpy
Enthalpy is the heat within a system and is the difference of the potential energies between the products and the reactants. Enthalpy is focused on the initial and final conditions of the products and reactants.
E = activation energy
a
*
*
Enthalpy can be found through
stoichiometry
,
calorimetry
,
tables

with given standard value
s
,
Hess's Law
, and
Bond Energies
.

Stoichiometry
To find the enthalpy using this method, one must know the basics of stoichiometry. It's as simple as that. Just remember that
1 mole
=
GFM
=
22.4 L
=
6.02 x 10 particles
, and know how to do two dimensional analysis (picket fence) and/or proportions.

Calorimetry
This is the process of measuring heat based on observing the temperature change when a body absorbs or discharges energy as heat. Energy (
q
) is released or gained at a constant pressure and is found using this equation:

q= mC
p
T

To find the specific heat capacity, you switch the the components around in the equation above and you would get:

C
p
=

Also, when water is in its liquid form, its specific heat is 4.184J/g C. Water in its liquid form has one of the highest specific heat, which allows life on our planet to be possible.
Finding Enthalpy through Stoichiometry Example:
Upon adding solid potassium hydroxide pellets to water the following reaction takes place:

KOH KOH + 43kJ/mol

Answer the following questions regarding the addition of 14.0 g of KOH to water:
a) Does the beaker get warmer or colder?
b) Is the reaction endothermic or exothermic?
c) What is the enthalpy change for the dissolution of the 14.0 grams of KOH?
(
aq
)
(
s
)
Bond Energies
Hess's Law
6
6
5
Answer to Finding Enthalpy through Stoichiometry Example:
Upon adding solid potassium hydroxide pellets to water the following reaction takes place:

KOH KOH + 43kJ/mol

Answer the following questions regarding the addition of 14.0 g of KOH to water:
a) Does the beaker get warmer or colder?
b) Is the reaction endothermic or exothermic?
c) What is the enthalpy change for the dissolution of the 14.0 grams of KOH?

a) The heat that's given is on the reactant side which means the reaction is giving off heat.
Therefore, the beaker will get warmer
.

b) Because the reaction is releasing heat,
it is exothermic
.

c) Info given: -14.0 g of KOH K: 34g 14.0g
- 43 kJ/ mol O: 16g 56g/mol
H: 1g

43kJ x kJ
1 mol 0.25 mol

*
The enthalpy change is negative because
x= 10.75 kJ/mol -10.75 kJ/mol
the reaction is exothermic
*
56g/mol
= 0.25 mol
=
Vocabulary for Calorimetry
Heat capacity
- is the energy required to raise the temperature by 1 degree (Joules/ C)
Specific heat capacity (C )
- is the same as heat capacity, but it's more specific to 1 gram of substance and the experiment is under constant pressure
Calorie
- amount of heat needed to raise the temperature of 1.00 gram of water to 1.00 C
Kilocalorie
- 1,000 calories
Joules
- SI unit of energy;
1 cal=4.184J
o
o
p
q
m T
o
Calorimetry Example:
In a coffee cup calorimeter, 100.0 mL of 1.0
M
NaOH and 100.0 mL of 1.0
M
HCl are mixed. Both solutions were originally at 24.6 C. After the reaction, the final temperature is 31.3 C. Assuming that all solutions have a density of 1.0g/cm and a specific heat capacity of 4.184 J/g C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.
In a coffee cup calorimeter, 100.0 mL of 1.0
M
NaOH and 100.0 mL of 1.0
M
HCl are mixed. Both solutions were originally at 24.6 C. After the reaction, the final temperature is 31.3 C. Assuming that all solutions have a density of 1.0g/cm and a specific heat capacity of 4.184 J/g C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.

Info Given: - cup 200mL= 200g q = mC
p
T
- C
p
= 4.184 J/g C = (200)(4.184)(6.7)
- T
i
= 24.6 C = 5,606.56J/1000= 5.60656 kJ/mol
- T
f
= 31.3 C

o
o
o
o
3
o
T= T
f
- T
i
= 31.3-24.6
= 6.7
To find enthalpy through this method, you must use the equation:

H = H - H

When dealing with finding enthalpy with Hess's Law, a table with standard enthalpies of formation for the compounds mentioned in the reaction is provided. Using the given information from the table, you plug in the enthalpy of formations that corresponds with each compound. After plugging the enthalpies, you multiply it with the use the number of moles of each
o
o
o
3
rxn
f (products)
f (reactants)
Hess's Law Example 1:
Given the information above, calculate the H for the following chemical reaction.
Substance:

NH ClO
Al O
AlCl
NO
H O
H (kJ/mol)

-297
-1676
-704
+90
-242
4 4
3
2
3
(g)
(s)
(s)
(s)
2
(g)
f
o
3 Al + 3 NH ClO Al O + AlCl + 3NO + 6 H O
(s)
(s)
4
4
2
3
(s)
3 (s)
(g)
2
(g)
Answer to Hess's Law Example 1:
Given the information above, calculate the H for the following chemical reaction.
Substance:

NH ClO
Al O
AlCl
NO
H O
H (kJ/mol)

-297
-1676
-704
+90
-242
4 4
3
2
3
(g)
(s)
(s)
(s)
2
(g)
f
o
3 Al + 3 NH ClO Al O + AlCl + 3NO + 6 H O
(s)
(s)
4
4
2
3
(s)
3 (s)
(g)
2
Hess's Law (cont.)
When you aren't given a table with the enthalpies of formation for each compound, but are given one final equation in the problem and a set of chemical equations and their designated enthalpies, H, you have to use the set of equations. You take the equations you are given and try to rearrange it so that the outcome is the final equation, with the right compounds on the product and reactant sides.
When trying to get the right amount of moles of reactants/products to match that of the final equation, you may need to multiply the equation to get there. When you do that though, you must also multiply the enthalpy of the reaction of that particular equation.
When you need to reverse an equation so that the equations reactants become the products and vice versa, the enthalpy of that reaction must be negated (put a negative sign in front of it).
There are other compounds and/or elements that don't show up in the final equation, so you have to make sure that once all the equations are combined, the compounds that don't appear are canceled out. These compounds/elements cancel out when the same compound/element is on the reactant side of one equation, and the product side of another. And when they are on the appropriate sides, they must share the same amount of moles.
Hess's Law (cont.)
Hess's Law Example 2:
Calculate the H for this overall reaction given the following equations:

2H BO B O + 3H O
H BO HBO + H O
H B O + H O 4HBO
H B O 2B O + H O
H= -0.02 kJ/mol
H= 11.3 kJ/mol
H= 17.5 kJ/mol
rxn
rxn
rxn
3 3
(aq)
2 3
(s)
2
(l)
2H BO B O + 3H O
3 3 (aq)
2 3 (s)
2 (l)
3 3
(aq)
2
(aq)
2
(l)
2 4 7
(aq)
2
(l)
2
(aq)
2 4 7
(aq)
2 3
(s)
2
(l)
H BO HBO + H O
H B O + H O 4HBO
H B O 2B O + H O
3 3 (aq)
2 (aq)
2 (l)
2 4 7 (aq) 2 (l) 2 (aq)
2 4 7 (aq) 2 3 (s) 2 (l)
H= -0.02 kJ/mol
H= 11.3 kJ/mol
H= 17.5 kJ/mol
Answer to Hess's Law Example 2:
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