Saturday Study Session 1
Theme of the Class
Enthalpy
Session 3 – Delta H Four Different
Ways
Enthalpy. Anyone not in
Chemistry thinks you’re
saying Empathy. I feel your
pain.
Energy
The capacity to do work or
to produce heat.
The Two Types of Energy
Potential: due to
position or composition can be converted to
kinetic.
Kinetic: due to motion
of the object
State Function
Depends only on the present
state of the system - not how
it arrived there.
It is independent of pathway.
Enthalpy is a state function
It’s like going to Vegas. On the
way home you know you lost
$400. It doesn’t matter what
casino or how many bets you
placed. You still lost money.
First Law
First Law of Thermodynamics:
The energy of the universe is
constant.
If you lose $$ in Vegas, someone
else made $$.
Enthalpy
H = energy flow as heat (at constant pressure and
volume)
Thus the change of enthalpy is the change in the
amount of energy of a system.
A block of wood burning has a negative change in
enthalpy.
A tree taking energy from the sun and building a tree
has a positive change in enthalpy.
If you lost money in Vegas, that is negative $.
If you won money, that is positive $ .
4 ways to calculate change in enthalpy
1. Calorimetry – use q=mct to find heat
gained or lost by water.
2. Hess’s Law –keep or flip equations. Add em
up.
3. Hess’s law #2– standard heats of formation.
4. Bond energies – more to come on that as
well.
Enthalpy Δ H
Calorimetry
q=mcΔt
Heats of
Formation
ΔH
Bond
Energy
Hess’s
Law
Calorimetry
• Use water in a device to measure the heat given
off or absorbed by a reaction or object.
• q= mct
• q is heat absorbed or lost by the water usually in
Joules
• m is the mass of the water in grams
• c is the specific heat of water (4.18 J/gxC)
• t is the change in temperature of the water K
or C doesn’t matter.
Applying mcat
• Find q for the water, don’t worry about the sign
of anything.
• If the water got warmer, the reaction was
exothermic and H will be negative.
• If the water got cooler, the reaction was
endothermic and H will be positive.
Hess’s Law
Reactants  Products
The change in enthalpy is the same whether
the reaction takes place in one step or a
series of steps.
You can lose one $10,000 bet or lose 10
$1,000 bets. Either way the end result is
the same. You’re out 10 large!
Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g)
2NO(g)  N2(g) + O2(g)
2.
H = 180 kJ
H = 180 kJ
If the coefficients of a reaction are multiplied
by an integer, H is multiplied by that same
integer.
6NO(g)  3N2(g) + 3O2(g)
H = 540 kJ
Hess law example
• Thermite is powdered aluminum plus iron
III oxide creating iron and aluminum oxide.
• 2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s)
• This is extremely exothermic.
• 2 Al + 3/2 O2  Al2O3 ∆H=-1676 kJ/mol
• 2 Fe +3/2 O2  Fe2O3 ∆H=-826 kJ/mol
• Which one gets flipped?
H using heats of formation
Standard heat of enthalpy for any element is zero.
Can be calculated from enthalpies of formation of
reactants and products.
Hrxn° = Hf(products)  Hf(reactants)
H using Bond Energies
Can be calculated from bond energies of reactants
and products.
Hrxn = (bond energies of reactants)
– (bond energies of products)
Notice this is opposite of the standard heats of formation.
In other words breaking all the bonds takes energy
(+ H) and building new molecules with new bonds releases
energy (-H).
1. C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(g)
For the reaction of ethylene represented above,
Hrxn is -1323 kJ.
What is the value of H if the combustion
produced liquid water H2O(l), rather than water
vapor H2O(g)?
(H for the phase change H2O(g)⇆ H2O(l) is -44
kJ mol-1.)
A) -1235 KJ
B) -1279 kJ
C) -1323 kJ
D) -1411 kJ
Question 1 Answer
D
Clue: Multiply one of the
equations to make it
work.
C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g)
Hrxn = -1323 kJ.
H2O(g) → H2O(l)
Hrxn = -44 kJ mol-1
Don’t flip anything. It works out for liquid water to be on the
right.
You do need to multiply the 2nd equation by 2 to get stuff to
cancel.
C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g) Hrxn = -1323 kJ.
2H2O(g) → 2H2O(l)
Hrxn = -44 kJ mol-1 x 2
= -88 kJ mol-1
Then add the two numbers together.
2. What is the standard enthalpy change H°,
for the reaction:
3C2H2(g)  C6H6(g)
H°f of C2H2(g) is 230 kJmol-1
H°f of C6H6(g) is 83 kJmol-1
A)
B)
C)
D)
-607 kJ
-147 kJ
-19 kJ
+19 kJ
Question 2 Answer
A
Clue: Products minus
reactants
Summation = add them together
•
•
•
•
•
•
3C2H2(g)  C6H6(g)
H°f of C2H2(g) is 230 kJmol-1
H°f of C6H6(g) is 83 kJmol-1
C6H6 – (3 x C2H2) = answer
83 kJ – (3 x 230 kJ) =
83 kJ – 690 kJ = -607 kJ
3. True for the evaporation of water at 1
atm and 25 ͦC
.
A)
B)
C)
D)
H>0
H<0
H=0
H is temperature dependent
Question 3 Answer
A
Clue: sweat
When water evaporate it absorbs
the energy and uses it to speed up
the molecules so they can break
free into a gas. Absorbing energy
is + H.
4. A 10 g sample of a metal was heated to
100°C and then quickly transferred to an
insulated container holding 100 g of water at
20°C. The temperature of the water rose to
reach a final temperature of 50°C. Calculate
the heat absorbed by the water. Specific heat of
water is 4 J/(gx°C)
A)
B)
C)
D)
400 kJ
12 kJ
8 kJ
1.2 kJ
Question 4 Answer
B
Clue: You don’t always
use every number given.
q = m c t for the water since that
is what the question asks.
q = 100 g x 4 J/(gx ⁰ C) x (50-20 ⁰C)
q = 12000 J or 12 kJ
5. The dissolution of an ionic solute in a polar solvent
can be imagined as occurring in three steps, as shown
in the figure above. In step 1, the separation between
ions in the solute is greatly increased, just as will occur
when the solute dissolves in the polar solvent. In step
2, the polar solvent is expanded to make spaces that
the ions will occupy. In the last step, the ions are
inserted into the spaces in the polar solvent. Which of
the following best describes the enthalpy change, H,
for each step?
A) Step 1 and Step 3 are exothermic and Step 2 takes no
energy so is neither endothermic nor exothermic.
B) Step 1 and 2 add together to be 2 X Step 3.
C) Steps 2 and 3 add together to be Step 1
D) Step 3 releases energy while Step 1 and 2 absorb energy
Question 5 Answer
D
Clue: Read carefully.
Anytime you dissolve anything it takes energy to pull apart the
solvent molecules and pull apart the solute particles (step 1 and 2)
When you combine them together energy is released. Thus D is the
only statement that makes sense.
The steps do not have to add up to each other.
If there is more energy released than absorbed
the temperature of the solution would increase.
If more energy is absorbed than released the
temperature of the solution would decrease.
But the question says nothing about the
temperature so we don’t care.
Process
H (kJ/molrxn)
Na(s)  Na(g)
m
Na(g)  Na+(g) + en
Br2(g)  2 Br(g)
p
Br (g) + e-  Br -(g)
q
Na+(g) + Br -(g)  NaBr(s)
r
Na(s) + ½ Br2(l)  NaBr(s) H =-361 kJ/molrxn
The elements Na and Br react directly to form the
compound NaBr according to the equation above. Refer to
the information above and the table below to answer the
questions 6-8.
6. How much heat is released or absorbed when 0.050 mol
of Br2(g) is formed from NaBr(s)?
A) 72.2 kJ is released
B) 36.1 kJ is released
C) 36.1 kJ is absorbed
D) 72.2 kJ is absorbed
Question 6 Answer
C
Clue: Stoichiometry
0.050 mol Br2 x
361 kJ
=36.1 kJ
0.5 mole Br2
Realize the equation must be reversed
to produce Br2 so the sign of H must
be flipped.
Since the reaction has a positive H,
then heat is absorbed so 36.1 kJ is
absorbed.
7. Which of the values of H° for a
process in the table is (are) less than
zero (i.e., indicate(s) an exothermic
process) ?
A) r only
C) p, q, and r only
B) q and r only D) n, p, q, and r
Question 7 Answer
A
Clue: Which one of these
is NOT like the others.
Knocking off an electron from a neutral
atom or squishing another electron on
a neutral atom takes energy to make it
happen. ENDOthermic
Turning a liquid into a gas or splitting
apart a diatomic element take energy to
make it happen. ENDOthermic.
8. Br2(g) + 2e- 2Br -(g)
Which of the following expressions is
equivalent to H° for the reaction
represented above?
A) p + q
B) p - q
C) p + 2q
D) (p/2)- q
Question 8 Answer
C
Clue: Hess’s law
Value p is all set to match the
target equation.
Value q must be multiplied x 2 to
match.
Nothing needs to be flipped. Just
add.
9. 4 NH3(g) + 3O2(g) 2 N2(g) + 6 H2O(g)
If the standard molar heats of formation of
ammonia, NH3(g), and gaseous water,
H2O(g), are -46 kJ/mol and -242 kJ/mol,
respectively, what is the value of H°298
for the reaction represented above?
A)
B)
C)
D)
-190 kJ/molrxn
-290 kJ/molrxn
-580 kJ/molrxn
-1,270 kJ/molrxn
Question 9 Answer
D
Clue: What is the heat of
formation for an
element?
Heats of formation products
-Heats of formation of reactants
(6 x -242 kJ) – (4x -46 kJ) =
-1268 kJ
The nitrogen and oxygen have
heats of formation of zero.
10. ½ H2(g) + ½ I2(s)  HI(g) H = 26 kJ/mol
rxn
½ H2(g) + ½ I2(g)-> HI(g) H = -5.0 kJ/mol rxn
Based on the information above, what is the
enthalpy change for the sublimation of
iodine, represented below?
I2(s)  I2(g)
A)
B)
C)
D)
62 kJ/mol rxn
21 kJ/mol rxn
31 kJ/mol rxn
42 kJ/mol rxn
Question 10 Answer
A
Clue: flip and multiply
½ H2(g) + ½ I2(s)  HI(g) H = 26 kJ/mol rxn
½ H2(g) + ½ I2(g)  HI(g) H = -5.0 kJ/mol rxn
Multiply each equation by 2 to get the
coefficients to match.
The 2nd equation must be flipped.
H2(g) + I2(s)  2 HI(g) H = 52 kJ/mol rxn
2 HI(g)  H2(g) + I2(g) H = +10. kJ/mol rxn
Short Free Response 1
(3 points possible)
Free Response 2 (9 pts possible)