! Lecture 16 Practice Worksheet
ANSWERS
"
A. Coefficient/ Reciprocal /Multiple Equilibrium Rules
1. 2NO(g) + O (g) ⟺ 2NO (g)
2
2
5
Kc= 6.44 x 10 @ 230 ℃
a. Find the Kc raised to the fifth power.
[2NO(g) + O2 (g) ⟺ 2NO2 (g)]5
Kc= [6.44 x 105 ]5 = 1.11 x 1029
-----------------------------------------------------------b. Find the reciprocal of the Kc using the reciprocal rule.
[2NO(g) + O2 (g) ⟺ 2NO2 (g)]
[2NO2 (g) ⟺ 2NO(g) + O2 (g)] FLIPPED
Kc= 1/ [6.44 x 105 ]= 1.55 x 10-6
-----------------------------------------------------------c. Find the inverse of the Kc solved for letter a.
a. Kc= 1.11 x 1029
1/Kc
1/1.11 x 1029 = 9.01 x 10 -30=Kc
-----------------------------------------------------------d. Find the Kc solved for in letter b raised to the third power.
b. Kc= 1.55 x 10-6
[1.55 x 10-6]3 = 3.72 x 10-18 = Kc
__________________________________________________________________
B. The Reaction Quotient
2. Decipher whether the reactions are at equilibrium. Explain.
a. 0.035 moles of SO , 0.500 moles of SO Cl , and 0.080 moles of Cl are
2
2 2
2
combined in an evacuated 5.00 L flask and heated to 100℃. Given that:
SO Cl (g)⟺ SO (g) + Cl (g) Kc = 0.078 @ 100℃
2 2
2
2
is the reaction at equilibrium? If not, in which direction must the reaction proceed in order to
establish equilibrium? What is the value of Q for this reaction? Which direction will the
reaction shift?
Given: SO2 Cl2 (g)⟺ SO2 (g) + Cl2 (g )
[SO2Cl2]= 0.500M
[SO2]= 0.035M
[Cl2]= 0.080M
K= 0.078
Solution:
Step 1: Write the Q formula. Note that because we are
dealing with gases we use parenthesis rather than the
brackets.
Qc= (SO2 ) (Cl2 ) / (SO2 Cl2 )
Step 2: Plug in Values. Since the concentrations were given, we
can plug those right in to the equation.
Qc= (0.035) (0.080) / (0.500)
Step 3: Solve for Q.
Q=0.0056
* sigfig >>> Q= 0.006
Step 4: Compare Q to K. Since K=0.078 and Q is 0.006, K is
greater than Q, meaning that the reaction will shift right in order
to recreate equilibrium.
Answer: Q=0.006, shifts right.
------------------------------------------------------------------------
b. 0.81 moles of H gas and 0.44 moles of I ,react together to form 0.58 of HI. Given
2
2
that:
2
H (g) + I ⟺ 2HI (g) Kc = 7.1 x 10 @ 25℃
2
2
is the reaction at equilibrium? If not, in which direction must the reaction proceed in order to
establish equilibrium?
2
Given: H2 (g) + I2 ⟺ 2HI (g) Kc = 7.1 x 10 @ 25℃
[H2]= 0.81M
[I2]= 0.44M
[2HI]= 0.58M
K= 7.1 x 102
Solution:
Step 1: Write the Q formula. Note that because we are
dealing with gases we use parenthesis rather than the
brackets.
Qc= (HI)2 / (H2 ) (I2 )
Step 2: Plug in Values. Since the concentrations were given, we
can plug those right in to the equation.
Qc= (0.58)2 / (0.81) (0.44)
Step 3: Solve for Q.
Q=0.94388
* sigfig >>> Q= 0.94
Step 4: Compare Q to K. Since K=7.1 x 102 and Q is 0.94, K is
much greater than Q, so the reaction will shift right in order to
recreate equilibrium.
Answer: Q=0.006, shifts right.
--------------------------------------------------------------------
c. 1.0 moles of CO gas and 1.0 moles of H O react together to form 2.0 moles of CO and
2
2
2.0 moles of H . Given that:
2
CO(g) + H O(g) ⟺ CO (g) + H (g) Kc = 1.0 @ 25℃
2
2
2
is the reaction at equilibrium? If not, in which direction must the reaction proceed in order to
establish equilibrium?
Given:
= 1.0
[CO2(g)]= 2.0 M
[H2(g)]= 2.0 M
[CO(g)]= 1.0 M
[H2O(g)]= 1.0 M
Solution:
Step 1: Write the Q formula
Step 2: Plug in given concentration values.
Q= 4.0
Step 3: Compare Q to K.
Since 4.0 is greater than 1.0, Q > K. This means that
the reaction shifts left.
Answer: Q= 4.0 and shifts left.
--------------------------------------------------------------------
d.3.2 moles of HCl gas and 4.3 moles of I react together to form 6 moles of NaCl and
2
water. Given that:
HCl(g) + NaOH (aq) ⟺ NaCl (aq) + H O ( l ) Kc = 0.5 @ 25℃
2
is the reaction at equilibrium? What is the value of Q for this reaction? Which direction will
the reaction shift?
Given K=0.5
[HCl]= 3.2M
[NaOH]= 4.3M
[NaCl]=6M
Solution:
Step 1: Write Q formula. Since we know that the activity
of a liquid is 1, we can omit the water component in the
equation.
Step 2: Plug in given concentrations into the Q formula.
Step 3: Calculate using the given concentrations
Q = 0.436
Step 4: Compare Q to K. Now that Q is found to be
0.436, we can compare it to the given K value of 0.5, so
Q is less than K.
Since Q < K, the reaction is not at equilibrium and will
proceed to the products side to reach dynamic
equilibrium once again.
Answer: Q= 0.436 and the reaction favors the products.
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e. 0.04 moles of N and 0.09 moles of H react together to form NH gas. Given that:
2
2
3
N (g) + 3H (aq) ⟺ 2NH (g) Kc = 0.040 @ 25℃
2
2
3
is the reaction at equilibrium? What is the value of Q for this reaction? Which direction will
the reaction shift?
Given:
[N2]= 0.04M
[H2]= 0.09M
K= 0.040
Solution:
Step 1: Write the Q formula
Step 2: Plug in Values. Since the concentrations for N2 and H 2
were given, we can plug those right in to the equation. However,
no concentration value was given for NH3 so we assume that
there is no concentration, which is denoted with a zero.
Step 3: Solve for Q.
Q=0
Step 4: Compare Q to K. Since K=0.04 and Q is zero, K is
greater than Q, meaning that the reaction will shift right in order
to recreate equilibrium.
Answer: Q=0, shifts right.
__________________________________________________________
C. Le Chatelier
3. Determine where the equilibrium of each reaction would shift.
a.which way will the equilibrium shift if more H is added to this reaction at equilibrium:
2
N + 3 H ⟺ 2 NH
2
2
3
The H2 amount goes up (by adding it), therefore according to LeChatelier's
Principle, the reaction will try and use up the added H2 . It does so by shifting
the position of equilibrium to the right. This makes more NH3 by using up N2
and H2
-------------------------------------------------------------------b.which way will the equilibrium shift if the system temperature goes up (heat is added):
2 SO + O ⟺ 2 SO + heat
2
2
3
Even though heat is not a chemical substance, for the purposes of
LeChatelier's Principle, you can treat it as if it has physical existence. Since
heat is added, the reaction will shift to try and use up some of the added heat.
In order to do this, the reaction must shift to the left.*Note that, since heat is
not part of the equilibrium expression, the value of Keq would change when
the chemical system is heated or cooled.
--------------------------------------------------------------------
c.the container holding the following reaction (already at equilibrium) has its volume suddenly
reduced by half. Which way will the equilibrium shift to compensate?
PCl + Cl ⟺ PCl
3
2
5
Since the volume went down, this means the pressure went up. The reaction
will try to lessen the pressure by shifting to the side with the lesser number of
gas molecules. This means a shift to the right because for every PCl5 molecule
made, two molecules are used up. The lesser the total number of gas phase
molecules in the container, the lesser the pressure.
--------------------------------------------------------------------
d.the container holding the following reaction (already at equilibrium) has its volume suddenly
increased. Which way will the equilibrium shift to compensate?
H + Cl ⟺ 2 HCl
2
2
Neither side is favored over the other since both sides have the same number
of total molecules (two). No matter which way the reaction shift, the total
number of molecules would remain unchanged. In cases like this, where there
is an equal number of molecules on each side, the equilibrium would remain
unchanged by the change in pressure (in either direction).
--------------------------------------------------------------------
e.the system below is already at equilibrium when a catalyst is added to the system. What
happens to the position of the equilibrium? Does it shift right, left, or no change?
PCl + Cl <===> PCl
3
2
5
There will be no change in the equilibrium. BOTH (with emphasis on both) the
forward and the reverse reactions are speeded up. A catalyst just gets you to
equilibrium faster, it doesn't affect the final position of equilibrium like
changing the concentration would
__________________________________________________________________
D. TheReaction Quotient & Le Chatelier
4. Phosgene, COCl was formed by mixing 0.90 atm of CO and 1.9 atm of Cl in a
2,
2
reaction vessel @ 700 K. At equilibrium, the Kp is 1.35 atm.
CO(g) + Cl (g) ⟺ COCl (g)
2
2
a. Use the ICE chart to solve for x.
(PCO)
I
C
E
0.90 atm
-x
0.90 - x
(PCl )
2
1.9atm
-x
1.9 - x
{Kp= 1.35atm}
Kp = (PCOCl )
2
(PCO)(PCl )
2
(PCOCl )
2
0
+x
x
1.35 = ____(x)______
(0.90 - x)(1.9 - x)
1.35 = ____(x)______
(0.90) (1.9)
1.35 = _(x)_
(1.71)
(1.71)(1.35) = (x)
x=2.31
b. Find the partial pressure for each gas.
✩(PCO)= 0.90 - x = 0.90 - 2.31 = - 1.41 atm
✩(PCl )=1.9 - x= 1.9 - 2.31 = -0.41 atm
2
✩(PCOCl )=x= 2.31 atm
2
c. Convert the Kp to Kc.
𝚫
Kp=Kc(RT) n
Kp= 1.35 atm
Kc=?
R= 0.0821Latm/molK
T=700K
𝚫n=(moles of reactants - moles of products) 2-1 = 1
1
(1.35)=Kc(0.0821 Latm/molK)(700K)
1
_________(1.35) _______=Kc(0.0821 Latm/molK)(700K)
(0.0821 Latm/molK)(700K)
(0.0821 Latm/molK)(700K)
0.023=Kc
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5.Nitric oxide and bromine at initial pressures of 98.4 and 41.3 torr, respectively, were
allowed to react at 300. K. At equilibrium the Kp was 0.145 atm. The reaction is as
follows.
2 NO(g) + Br (g) <==> 2 NOBr(g)
2
{*NOTE: convert torr to atm. << 760 torr = 1 atm>>}
a. Use the ICE chart to solve for x.
NO= 98.4torr x 1 atm
1
760 torr
Br2 = 41.3 torr x 1 atm
1
760 torr
= 0.129atm
=
=
0.054atm
=
(PBr )
2
0.129atm
0.054atm
-2x
-x
0.129 - 2x
0.054 - x
{Kp= 1.35atm}
(PNO)
I
C
E
Kp = (PNOBr)2
(PNO)2 (PBr )
2
0.145 = _______(x)2 _______
(0.129 - 2x)2 (0.054 - x)
0.145 = ____(x)2 ______
(0.129) (0.054)
0.145 = ____(x)2 ______
(0.00697)
(0.00697)(0.145) = (x)2
0.00101=(x)2
√(0.00101)=x
x=0.0318
(PNOBr)
0
+2x
2x
b. Find the partial pressure for each gas.
✩(PNO)= 0.129 - 2x = 0.129 - 2(0.0318) = 0.0654 atm
✩(PBr )=0.054 - x= 0.054 - 0.0318 = 0.508 atm
2
✩(PNOBr)=2x= 2(0.0318) = 0.0636 atm
c. Convert the Kp to Kc.
𝚫
Kp=Kc(RT) n
Kp= 0.145 atm
Kc=?
R= 0.0821Latm/molK
T=300.K
𝚫n=(moles of reactants - moles of products) 2-1 = 1
1
(1.35)=Kc(0.0821 Latm/molK)(700K)
1
_________(0.145) _______=Kc(0.0821 Latm/molK)(300K)
(0.0821 Latm/molK)(300K)
(0.0821 Latm/molK)(300K)
0.00589=Kc
+NOTE: Some of these problems I made up so don't be surprised is a few of your answers
turn out a little wacky. (Example, the answers to problem 4 are impossible.)