Transcript Thermochemistry

Thermochemistry
By
Sumana Ramakrishnan
And
Reema Sil
Thermochemistry
is the study of heat change in
chemical reactions.
Exothermic
is any process that gives off
heat – transfers thermal
energy from the system to
the surroundings.
A  B  C  D  Heat
Endothermic
is any process in which heat has to
be supplied to the system from the
surroundings.
A  B  Heat  C  D
Enthalpy is used to quantify the heat flow into or out of a system in
a process that occurs at constant pressure.
H reaction  H  products  H reac tan ts
Hproducts < Hreactants
Hproducts > Hreactants
∆H < 0
Exothermic
(system gives off heat)
∆H > 0
Endothermic
(system absorbs heat)
Heat (q) absorbed or released:
q  mcT
m : mass of a given quantity in grams (g)
c : specific heat; the amount of heat (q) required to raise the temperature
of one gram of the substance by one degree Celsius (Jg-1°C-1)
∆t = temperaturefinal – temperatureinitial (°C)
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to
the equation above, 64.98 kilojoules of heat is released. Use the information in the
table below to answer the questions that follow.
Substance
Standard Heat of
Formation, ∆H˚; at 25˚C
(kJ/mol)
Absolute Entropy, S˚, at
25˚C (J/mol·K)
C(graphite)
0.00
5.69
CO2(g)
-393.5
213.6
H2(g)
0.00
130.6
H2O(l)
-285.85
69.91
O2(g)
0.00
205.0
C6H5OH(s)
?
144.0
(a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25˚C.
(b) Calculate the standard heat of formation, ∆H˚, of phenol in kilojoules per mole at 25˚C.
(c) Calculate the value of the standard free-energy change, ∆G˚, for the combustion of phenol
at 25˚C.
(d) If the volume of the combustion container is 10.0 liters, calculate the final pressure in the
container when the temperature is changed to 110˚C. (Assume no oxygen remains
unreacted and that all products are gaseous.)
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned
according to the equation above, 64.98 kilojoules of heat is released.
(a) Calculate the molar heat of combustion of
phenol in kilojoules per mole at 25˚C.
H comb 
kJ
mass
molarmass

 64.98kJ
2.000 g
kJ
 3058 mol
g
94.113 mol
The molar heat of combustion of phenol is
-3058 kilojoules.
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to
the equation above, 64.98 kilojoules of heat is released. Use the information in the table
below to answer the questions that follow.
Standard Heat of
(b) Calculate the standard heat
of formation, ∆H˚, of phenol in
kilojoules per mole at 25˚C.
Equation
Variables
Numbers
Substance
Formation, ∆H˚;
at 25˚C (kJ/mol)
C(graphite)
0.00
CO2(g)
-393.5
H2(g)
0.00
H2O(l)
-285.85
O2(g)
0.00
C6H5OH(s)
?
H reaction  H  products  H reac tan ts
H reaction  6  CO2   3  H 2 O   x
 3058  6 393.5  3 285.5  x
(From (a), ∆Hrxn = -3058 kJ)
kJ
x  161 mol
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to
the equation above, 64.98 kilojoules of heat is released. Use the information in the table
below to answer the questions that follow.
(c) Calculate the value of the standard free-energy change, ∆G˚, for the combustion of
phenol at 25˚C.
First, you must solve for ∆S
G  H   TS 
S   S ( products )  S ( reac tan ts )
 6  CO2   3  H 2 O  C6 H 5 OH  7O2 
 6213.6  369.91  144.0  7205.0 87.67J
Then plug values into
G  H   TS 
 3058  298 0.08767kJ
 3032kJ
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to
the equation above, 64.98 kilojoules of heat is released. Use the information in the table
below to answer the questions that follow.
(d) If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container
when the temperature is changed to 110˚C. (Assume no oxygen remains unreacted and that all
products are gaseous.)
PV  nRT
2.000gC6 H5OH 
1molC6 H 5 OH
6molCO2
 0.127molCO2

94.12 gC6 H 5 OH 1molC6 H 5 OH
1molC6 H 5 OH
3molH 2 O


 0.0637molH 2O
2.000 gC6 H 5OH
94.12 gC6 H 5 OH 1molC6 H 5 OH
PV  nRT
}
0.1907 moles of
gaseous products

0.1907mol 0.08206 KLatm
nRT
mol 383K 
P

 0.599atm
V
10L
2 NO(g) + O2(g) → 2 NO2(g)
∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of
photochemical smog.
(a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g).
(b) For the reaction at 25˚C, the value of the standard free-energy change, ∆G˚, is -70.4
kJ.
(i) Calculate the value of the equilibrium constant, Keq, for the reaction at 25˚C.
(ii) Indicate whether the value of ∆G˚ would become more negative, less negative, or
remain unchanged as the temperature is increased. Justify your answer.
(c) Use the data in the table to calculate the value of the standard molar entropy, S˚, for
O2(g) at 25˚C.
Standard Molar
Entropy, S˚
(J K-1 mol-1)
NO(g)
210.8
NO2(g)
240.1
(d) Use the data in the table to calculate the bond energy, in kJ mol-1, of the nitrogenoxygen bond in NO2. Assume that the bonds in the NO2 molecule are equivalent (i.e.,
they have the same energy).
Bond Energy
(kJ mol-1)
Nitrogen-oxygen bond in NO
607
Oxygen-oxygen bond in O2
495
Nitrogen-oxygen bond in NO2
?
2 NO(g) + O2(g) → 2 NO2(g)
∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of
photochemical smog.
(a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g).
1molNO
 114.1kJ
73.1gNO 

 138.97kJ
30.01gNO
2molNO
The reason our answer is negative is because the reaction is exothermic.
Even though it results in a negative change in heat for the reaction,
the amount of heat released is 138.97kJ (a positive quantity).
So the answer to the question in part (a) is the positive quantity
138.97 kJ.
2 NO(g) + O2(g) → 2 NO2(g)
∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of
photochemical smog.
(b) For the reaction at 25˚C, the value of the standard free-energy change, ∆G˚, is -70.4 kJ.
(i) Calculate the value of the equilibrium constant, Keq, for the reaction at 25˚C.
G   RT ln K eq
 70,400 J  8.314 molJ K 298 K  ln K eq
28.4  ln K eq
e 28.4  K eq
K eq  2.158  1012
2 NO(g) + O2(g) → 2 NO2(g)
∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of
photochemical smog.
(b) For the reaction at 25˚C, the value of the standard free-energy change, ∆G˚, is -70.4 kJ.
(ii) Indicate whether the value of ∆G˚ would become more negative, less negative, or remain
unchanged as the temperature is increased. Justify your answer.
G  H  TS
As T increases, the quantity T∆S becomes more
and more negative.
According to the equation, subtracting a large
negative number (T∆S) from another negative
number (∆H) is the same as adding a value to the
negative ∆H, making ∆G˚ less negative.
∆G˚ will become less negative because of the roles
of ∆H, T, and ∆S in the equation.
2 NO(g) + O2(g) → 2 NO2(g)
∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of
photochemical smog.
(c) Use the data in the table below to calculate the value of the standard molar entropy,
S˚, for O2(g) at 25˚C.
Standard Molar
Entropy, S˚
(J K-1 mol-1)
∆S˚ = ∑ S˚products - ∑ S˚reactants
NO(g)
210.8
NO2(g)
240.1
S  2  S NO2   2  S NO  S O2 
S   146.5  2  240.1  2  210.8  x 
x  146.5  480.2  421.6  205.1 K Jmol
The standard molar entropy for O2(g) is
205.1 J K-1 mol-1.
2 NO(g) + O2(g) → 2 NO2(g)
∆H˚= -114.1 kJ, ∆S˚= -146.5 J K-1
The reaction represented above is one that contributes significantly to the formation of
photochemical smog.
(d) Use the data in the table below to calculate the bond energy, in kJ mol-1, of the
nitrogen-oxygen bond in NO2. Assume that the bonds in the NO2 molecule are
equivalent (i.e., they have the same energy).
Bond Energy
(kJ mol-1)
Nitrogen-oxygen bond in NO
607
Oxygen-oxygen bond in O2
495
Nitrogen-oxygen bond in NO2
?
H   H bondsbroken  H bondsformed
(bonds of reactants are broken, bonds of products are formed)
H   H bondsreac tan ts  H bondsprodu cts
H   114.1kJ  2  607  495  2 x 
→
In this equation, x represents the total bond energy
in an NO2 molecule. Since there are two N-O bonds
in NO2, x divided by two equals the bond energy of
each N-O bond in the molecule.
kJ
x  911.5 mol
kJ
x 911.5 mol
kJ

 455.775 mol
2
2
Answer the following questions that relate to the
chemistry of nitrogen.
(a) Two nitrogen atoms combine to form
a nitrogen molecule, as represented
by the below equation. Using the
table of average bond energies,
determine the enthalpy change, ∆H,
for the reaction.
2N ( g )  N 2 ( g )
Bond
Average Bond
Energy (kJ mol-1)
N-N
160
N=N
420
N=N
950
(b) The reaction between nitrogen and hydrogen to from ammonia is represented
below. Predict the sign of the standard entropy change, ∆S˚, for the reaction.
Justify your answer.
N 2( g )  3H 2( g )  2NH 3( g ) ∆H˚= -92.2 kJ
(c) The value of ∆G˚ for the reaction represented in part (b) is negative at low
temperatures but positive at high temperatures. Explain.
(d) When N2(g) and H2(g) are placed in a sealed contained at a low temperature,
no measurable amount of NH3(g) is produced. Explain.
(a) Two nitrogen atoms combine to form a nitrogen molecule, as
represented by the below equation. Using the table of average bond
energies, determine the enthalpy change, ∆H, for the reaction.
Bond
Average Bond
Energy (kJ mol-1)
N-N
160
N=N
420
N=N
950
2N ( g )  N 2 ( g )
H   H bondsbroken  H bondsformed
kJ
H   0  950 mol
 950kJ
The enthalpy change when two nitrogen atoms combine is -950 kJ.
The reaction only forms bonds, which requires heat energy.
This makes the equation endothermic because it requires a heat
supply to form N2(g).
(b) The reaction between nitrogen and hydrogen to from ammonia is
represented below. Predict the sign of the standard entropy change,
∆S˚, for the reaction. Justify your answer.
∆H˚= -92.2 kJ
N 2( g )  3H 2( g )  2NH 3( g )
∆S˚ (change in entropy) is the measure of disorder in a reaction. This measure
of entropy can be judged by the change in moles of gaseous particles.
In this reaction, since there are fewer gaseous moles of products than there
are gaseous moles of reactants, there is a decrease in disorder - meaning,
∆S is negative for this reaction.
MolesGas reac tan ts  MolesGas products
 MolesGas  inEntropy  S
MolesGas reac tan ts  MolesGas products
 MolesGas  inEntropy  S
(c) The value of ∆G˚ for the reaction represented in part (b) is
negative at low temperatures but positive at high temperatures.
Explain.
G  H  TS




Small
L arg e


The value of ∆G, free energy, is determined by the value of ∆H – T(∆S).
We determined that both ∆H and ∆S are negative (from part b),
and T will always be positive (because temperature in Kelvin is
never negative.
If T is a small value, T(∆S) will not be large enough to overcome ∆H.
Thus, ∆G will be negative.
If T is a large value, then T(∆S) will be large enough to overcome ∆H and
make ∆G positive.
So, subtracting the negative value of T(∆S) from ∆H can result in either a
negative or positive number.
(d) When N2(g) and H2(g) are placed in a sealed contained at
a low temperature, no measurable amount of NH3(g) is
produced. Explain.
N 2( g )  3H 2( g )  2NH 3( g )
As we determined in Part C, when at a low temperature, ∆G is
negative. When ∆G is negative, it means that the reaction is
spontaneous, proceeding in the forward direction on its own.
However, if the reaction does not seem to visibly proceed, it does not
mean that the reaction is not spontaneous.
There are two reasons for why a measurable amount of NH3(g) is not
observed.
- All reactions require a certain activation energy to initiate the
reaction. If a reaction does not have that required energy, the
reaction will remain inert, even if it is spontaneous.
- The reaction might require an extensive amount of time to proceed.
Because the reaction is so slow, the amount of NH3(g) produced may not
be immediately visible, but the reaction continues to occur.
H  (aq)  OH  (aq)  H 2O(l )
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction
represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an open
polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation
q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of water.
(a) Give appropriate units for each of the terms in the equation q = mc ∆T.
(b) List the measurements that must be made in order to obtain the value of q.
(c) Explain how to calculate each of the following.
I. The number of moles of water formed during the experiment
II. The value of the molar enthalpy of neutralization, ∆Hneut, for the reaction between HCl(aq)
and NaOH(aq)
(d) The student repeats the experiment with the same equal volumes as before, but this time uses
2.0 M HCl and 2.0M NaOH.
I. Indicate whether the value of q increases, decreases, or stays the same when compared to
the first experiment. Justify your prediction.
II. Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut, increases,
decreases or stays the same when compared to the first experiment. Justify your prediction.
(e) Suppose that a significant amount of heat were lost to the air during the experiment. What
effect would that have on the calculated value of the molar enthalpy of neutralization, ∆Hneut?
Justify your answer.
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the
reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M
NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is
determined by using the equation q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of
water.
(a) Give appropriate units for each of the terms in the equation
q = mc ∆T.
q  mcT
Kilojoules / Calories
KiloGrams
J
J
or
g  C g  K
C / K
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the
reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M
NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is
determined by using the equation q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of
water.
(b) List the measurements that must be made in order to obtain the
value of q.
1.
2.
3.
Volume or mass of the HCl and NaOH solutions
Initial temperature of HCl and NaOH before mixing
Final temperature of HCl and NaOH after mixing
H  (aq)  OH  (aq)  H 2O(l )
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the
reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M
NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is
determined by using the equation q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of
water.
(c) Explain how to calculate each of the following.
I. The number of moles of water formed during the experiment
1.0molHCl 1molH 2O
volumeHCl 

 molH 2O
1L
1molHCl
volumeNaOH 
- Or -
1.0molNaOH 1molH 2O

 molH 2 O
1L
1molNaOH
You could also reason that both HCl and NaOH react in a 1:1 ratio to form
the same number of moles of water.
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the
reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M
NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is
determined by using the equation q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of
water.
(c) Explain how to calculate each of the following.
II. The value of the molar enthalpy of neutralization, ∆Hneut, for the reaction between
HCl(aq) and NaOH(aq)
First, you must solve for the heat produced using q=cm ∆T. Then, dividing this value by
the moles of H2O (from part I).
H neut
q

molH 2 O
This gives you the molar enthalpy of neutralization for the reaction.
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction
represented above. The student combines equal volumes of 1.0M HCl and 1.0 M NaOH in an
open polystyrene cup calorimeter. The heat released by the reaction is determined by using the
equation q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of
water.
(d) The student repeats the experiment with the same equal volumes as before, but this
time uses 2.0 M HCl and 2.0M NaOH.
I. Indicate whether the value of q increases, decreases, or stays the same when
compared to the first experiment. Justify your prediction.
q  mcT
You can reason that since more moles of HCl and NaOH are reacting,
the final temperature of the mixture will be higher.
So ∆T will be greater, which will increase q.
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the
reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M
NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is
determined by using the equation q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of
water.
(d) The student repeats the experiment with the same equal volumes as before, but this
time uses 2.0 M HCl and 2.0M NaOH.
II. Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut,
increases, decreases or stays the same when compared to the first experiment.
Justify your prediction.
As done in part C II, the molar enthalpy is found by dividing q by the moles
of water. In the reaction, both q and water increase proportionally by x.
H neut
q
xq


x  molH 2 O molH 2 O
Since x drops out, you can see that the value of the molar enthalpy of
neutralization remains the same as in the first experiment.
A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the
reaction represented above. The student combines equal volumes of 1.0M HCl and 1.0 M
NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is
determined by using the equation q=mc ∆T.
Assume the following.
Both solutions are at the same temperature before they are combined.
The densities of all the solutions are the same that of water.
Any heat lost to the calorimeter of to the air is negligible.
The specific heat capacity of the combined solutions is the same as that of
water.
(e) Suppose that a significant amount of heat were lost to the air during the experiment.
What effect would that have on the calculated value of the molar enthalpy of
neutralization, ∆Hneut? Justify your answer.
Heat lost to the air will decrease ∆T –
which in turn equals a smaller q.
Thus, when you divide q by the
number of moles of H2O, ∆Hneut will
be a smaller value.
 q  mcT 
 H neut
q

molH 2 O
The End
Voices by Sumana Ramakrishnan
and Reema Sil
Powerpoint by Reema Sil
AP Problems painstakingly solved by both.
Special Thanks to:
Mr. Gangluff
Mr. Amendola
Eric Liu
Jack Wang