Unit 6
The 6th planet in our solar system is Saturn
Ch. 5:
Thermodynamics: Study of heat and its relationship with other forms of energy
Two types of energy:
Kinetic: movement, active energy
Potential: stored energy, gravitational and electrostatic are two examples
KE=1/2mv2
`
W=Force (N)×Distance (m)
unit: Joules (J) =Newtons (N) times meters (m)
unit: Joules (J)
Energy is the capacity to do work or to transfer heat
A calorie is the energy required to raise 1g of water 1°C
1cal = 4.184 J
1Cal = 1000 cal
Energy is always conserved. It is never truly lost, only transferred.
When energy moves into a measured system, it is considered positive change in energy
When energy moves into the surroundings from the measured system, it is negative change.
ΔH=final-initial
+ΔH=endothermic reaction
-ΔH=exothermic reaction
State function: value of a state function depends on its present condition, not on the history of the
sample
Enthalpy: thermodynamic function accounting for the heat flow in chemical changes occurring at
constant pressure when no forms of work are performed other than pressure or volume work
H=enthalpy
ΔH=ΔE+PΔV
Calorimetry: measurement of flow
q= mCΔT
C=specific heat of the substance
The specific heat of toluene (C7H8) is 1.13J/gK. How many Joules of heat are needed to raise the
temperature of 62.0g of toluene from 16.3°C to 38.8°C?
This problem is simply a matter of plugging in numbers.
Q=(62.0g)(1.13J/gK)(38.8-16.3)
Q= 1576.35J
I found this problem in the textbook, ch. 5 number 48
Heat capacity: temperature change of an object when it absorbs a certain amount of energy
Dependent on amount of material present and the nature of the material
Specific heat is the heat capacity of 1g of the substance
CH2O = 4.184J/gK
Latent heat: energy released or absorbed during a process, like a phase change; temperature is constant
Enthalpy is an extensive property- it varies depending on the amount of reactant
The enthalpy of reverse reactions is the opposite of the enthalpy of the reaction; just flip the sign
Enthalpy also changes depending on the state of the reactants
Hess’s Law
If a reaction is expressed as the algebraic sum of two or more reactions, then the heat of the overall
reaction is the sum of the heats of the individual reactions
Use Hess’s Law to determine the ΔH for the reaction:
CH3OH(g)  CO(g) + 2H2(g)
Using these reactions and their enthalpies:
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l)
ΔH= -1453.12 kJ
2CO2(g)  O2(g) + 2CO(g)
ΔH= 566.0 kJ
2H2O(g)  2H2(g) + O2(g)
ΔH= 483.6 kJ
CH3OH(g)  CH3OH(l)
ΔH= -37.4 kJ
H2O(l)  H2O(g)
ΔH= 44.01 kJ
First begin with the combustion reaction, since you need methanol on the left side, but you only want
one, so divide the equation by 2, including the enthalpy.
CH3OH(l) + 3/2 O2(g)  CO2(g) + 2H2O(l)
ΔH= -726.56 kJ
Unfortunately, the methanol is in liquid form, and we need it to be a gas. So we can use the 4th equation
to find the enthalpy of vaporization.
CH3OH(g)  CH3OH(l)
ΔH= -37.4 kJ
Next, you need to convert CO2 to CO, so you would use the next equation, but you would divide this one
by 2 as well. Now, the CO2 cancels, and you are left with 1 O2 on the left and 1 CO on the right
CO2(g)  ½ O2(g) + CO(g)
ΔH= 283 kJ
The H2O can be converted to H2 and O2 next.
2H2O(g)  2H2(g) + O2(g)
ΔH= 483.6 kJ
However, the H2O doesn’t match up, as the one above is in gaseous form and in the combustion
reaction it is in liquid form. However, we need to multiply by 2.
2H2O(l)  2H2O(g)
ΔH= 88.02 kJ
Next, we need to add up all the enthalpies.
-726.56 kJ + -37.4 kJ + 283 kJ + 483.6 kJ + 88.02 kJ = ΔH
The Assyrians gave Saturn the
nickname “Lubadsagush,”
meaning “oldest of the old
ΔH= 90.7kJ
I found the resultant reaction in the textbook in ch. 5 number 34, and calculated each of the enthalpies
of the individual reactions.
Enthalpy of formation: tabulated change in the enthalpy that is the result of the formation of one mole
of a compound from its constituent elements at standard conditions
ΔH=ΣnΔHf(products) – ΣnΔHi(reactants)
Standard conditions: 1atm and 298K
Above, when using Hess’s Law, I calculated each of the enthalpies of the individual reactions using that
equation. Find the enthalpy of the combustion of methanol, as I had done previously.
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l)
ΔH=ΣnΔHf(products) – ΣnΔHi(reactants)
ΔH=[4(-285.83) + 2(-393.5)] – [2(-238.6) + 3(0)]
ΔH=(-1930.32) – (-477.12)
ΔH=-1453.2
Find the enthalpies of formation for the substance, and multiply by the coefficient in the BCE. Then add
together the
Ch. 19:
Spontaneous: a process that will proceed without any outside interventions
What makes a reaction spontaneous:
Change in enthalpy (H)
Change in entropy (S)
Enthalpy: reactions prefer to be exothermic; -ΔH
Entropy is the degree of randomness. To calculate entropy:
ΔS = ΣnSproducts –ΣnSreactants
Systems prefer chaos (+ΔS)
Gases have greater entropy than liquids, and liquids have greater entropy than solids
When S=0, this is a perfect crystalline shape at 0K, molecules can’t move at all
Entropy of the universe is increasing; it increases with any spontaneous reaction
Unlike energy, enthalpy is not conserved.
ΔH˂0
ΔH˂0
ΔS˃0
Always spontaneous
ΔH˂0
ΔS˂0
Spontaneous at low temps.
ΔS˃0
ΔS˂0
ΔH˃0
ΔH˃0
ΔS˃0
Spontaneous at high temps.
ΔH˃0
ΔS˂0
Never spontaneous*
*at equilibrium
Given the ΔS and the ΔH, determine whether the reaction is sometimes, always, or never spontaneous,
and describe the effect of temperature on the reaction’s spontaneity.


ΔS˃0, ΔH˃0
☼ Sometimes spontaneous; as temperature increases, the likelihood of spontaneity
increases
ΔS˂0, ΔH˃0
☼ Never spontaneous; temperature has no effect
Determine if the reaction is spontaneous at standard conditions using the tabulated values of the
substances and the equation ΔG=ΣnΔGf(products) – ΣnΔGi(reactants).
2HNO3(g) + Cl2(g)  N2(g) + 3O2(g) + 2HCl(g)
Simply plug in the numbers and solve.
ΔGo= [2(-95.27)] – [2(-73.94)]
ΔGo= -42.66
Spontaneous
I created this question.
Gibb’s Free Energy Equation: relates entropy temperature and enthalpy to predict spontaneity at
standard conditions; the “o” indicates at standard conditions
ΔGo=ΔHo-TΔSo
If ΔGo˂0, the reaction is spontaneous
If ΔGo˃0, the reaction is not spontaneous*
(*the forward reaction isn’t spontaneous, the reverse reaction is spontaneous)
If ΔGo=0, the reaction is at equilibrium
Determine whether the reaction is spontaneous at 247°C using Gibb’s free energy equation, given that
ΔH= -92.38kJ/mol and ΔS= -198.24 and doesn’t vary with changes in temperature.
N2(g) + 3H2(g)  2NH3(g)
247 + 273 = 520 K
ΔGo=ΔHo-TΔSo
ΔG= (-92.38) – (520)(-.19824)
ΔG= 10.7048 kJ
Not spontaneous
I created this question and calculated the change in enthalpy and entropy used.
To find ΔG not at standard conditions:
ΔG=ΔGo + RTlnQ
R=8.314 J/molK
Q=reaction quotient
This can also relate to equilibrium:
0=ΔGo + RTlnKeq
Keq = e-ΔG/RT
Jupiter was named after the titan
depicted here on the left. He was the
titan of time, and the oldest and most
powerful. Legend has it that he ate
his children.