HONORS CHEMISTRY - CHAPTER 18
NAME:
REACTION RATES AND EQUILIBRIUM
DATE:
LE CHATELIER’S PRINCIPLE WKST - ANS - V10
PAGE #:
Write The Equilibrium Constant Equation For Each Reaction And Then Complete Each Chart By
Writing Left, Right, Increase, Decrease, Or No Change In Each Section
I. The Synthesis of Phosphorus Pentachloride - See Explanations on the Following Page
PCl3 (g) + Cl2 (g)  PCl5 (g)
Write the Equilibrium
Constant Equation
ΔH = - 111 kJ
K=
!" PCl5 #$
!" PCl3 #$ !"Cl2 #$
Stress
Equilibrium
Shift
[PCl3]
[Cl2]
[PCl5]
K
1.
Add PCl3
Right
Increase
Decrease
Increase
No Change
2.
Remove PCl3
Left
Decrease
Increase
Decrease
No Change
3.
Add Cl2
Right
Decrease
Increase
Increase
No Change
4.
Remove Cl2
Left
Increase
Decrease
Decrease
No Change
5.
Add PCl5
Left
Increase
Increase
Increase
No Change
6.
Remove PCl5
Right
Decrease
Decrease
Decrease
No Change
7.
Increase Temp
Left
Increase
Increase
Decrease
Decrease
8.
Decrease
Temp
Right
Decrease
Decrease
Increase
Increase
9.
Decrease
Volume
Right
Increase
Increase
Increase
No Change
10.
Increase
Volume
Left
Decrease
Decrease
Decrease
No Change
1
I. The Synthesis of Phosphorus Pentachloride Continued - Thanks to Mrs. Doyle!!!
Explanations To The Above
The concentration of PCl3 increases and the system is no longer at equilibrium when the product
of the denominator increases with the increased [PCl3]. The value of Q instantly becomes lower
1. than the value of K. The system immediately reacts to reduce the disturbance. Hence,
equilibrium shifts to the right reacting to consume some of the added PCl3. In doing so, PCl3
reacts with Cl2 thereby reducing [Cl2] and increasing [PCl5].
The concentration of PCl3 decreases and the system is no longer at equilibrium when the product
of the denominator decreases with the decreased [PCl3]. The value of Q instantly becomes
2. higher than the value of K. The system immediately reacts to reduce the disturbance. Hence,
equilibrium shifts to the left reacting to replace some of the PCl3. PCl5 decomposes to form PCl3
and Cl2.
The concentration of Cl2 increases and the system is no longer at equilibrium when the product of
the denominator increases with the increased [Cl2]. The value of Q instantly becomes lower than
3. the value of K. The system immediately reacts to reduce the disturbance. Hence, equilibrium
shifts to the right reacting to consume some of the added Cl2. In doing so, Cl2 reacts with PCl3
thereby reducing [PCl3] and increasing [PCl5].
The concentration of Cl2 decreases and the system is no longer at equilibrium when the product
of the denominator decreases with the decreased [Cl2]. The value of Q instantly becomes higher
4. than the value of K. The system immediately reacts to reduce the disturbance. Hence,
equilibrium shifts to the left reacting to replace some of the Cl2. PCl5 decomposes to form PCl3
and Cl2.
The concentration of PCl5 increases and the system is no longer at equilibrium when the value of
the numerator increases with the increased [PCl5]. The value of Q instantly becomes higher than
5. the value of K. The system immediately reacts to reduce the disturbance. Hence, equilibrium
shifts to the left reacting to consume some of the added PCl5. In doing so, PCl5 decomposes to
form PCl3 and Cl2.
The concentration of PCl5 decreases and the system is no longer at equilibrium when the value of
the numerator decreases with the decreased [PCl5]. The value of Q instantly becomes lower than
6.
the value of K. The system immediately reacts to reduce the disturbance. Hence, equilibrium
shifts to the right as PCl3 and Cl2 react to replace some of the PCl5.
When the temperature of this exothermic reaction is increased, the endothermic side of the
equation is favored where heat can be absorbed and the effect of the disturbance is reduced. PCl5
7.
decomposes to form PCl3 and Cl2. An overall lower ratio is established, resulting in a decreased
value for K.
When the temperature of this exothermic reaction is decreased, the exothermic side of the
equation is favored where heat can be released and the effect of the disturbance is reduced. PCl3
8.
and Cl2 react to form PCl5. An overall higher ratio is established, resulting in an increased value
for K.
As the volume of this gaseous system is decreased, pressure is increased (Boyle’s Law). The
system reacts to relieve the effect of increased pressure by shifting in the direction that reduces
9. the number of moles of gas. Since two moles of reactants are consumed for every one mole of
product produced, this reaction will shift to the right. The decrease in the volume of this system
results in an overall increase in the concentration of reactants and the product.
As the volume of this gaseous system is increased, pressure is reduced. The system reacts to
relieve the effect of reduced pressure by shifting in the direction that increases the number of
10. moles of gas. Since two moles of reactants are consumed for every one mole of product
produced, this reaction will shift to the left. The increase in the volume of this system results in
an overall decrease in the concentration of reactants and the product.
2 - HC - Chapter 18 - Le Chatelier’s Principle Worksheet - Answers - V10
II. The Haber Process
N2 (g) + 3 H2 (g)  2 NH3 (g) + 92 kJ
Write the Equilibrium
Constant Equation
K=
!" NH 3 #$
2
!" N 2 #$ !" H 2 #$
3
Stress
Equilibrium
Shift
[N2]
[H2]
[NH3]
K
1.
Add N2
Right
Increase
Decrease
Increase
No Change
2.
Add H2
Right
Decrease
Increase
Increase
No Change
3.
Add NH3
Left
Increase
Increase
Increase
No Change
4.
Remove N2
Left
Decrease
Increase
Decrease
No Change
5.
Remove H2
Left
Increase
Decrease
Decrease
No Change
6.
Remove NH3
Right
Decrease
Decrease
Decrease
No Change
7.
Increase
Temperature
Left
Increase
Increase
Decrease
Decrease
8.
Decrease
Temperature
Right
Decrease
Decrease
Increase
Increase
9.
Increase
Pressure
Right
Increase
Increase
Increase
No Change
10.
Increase
Volume
Left
Decrease
Decrease
Decrease
No Change
3 - HC - Chapter 18 - Le Chatelier’s Principle Worksheet - Answers - V10
III. The Synthesis of Hydrogen Iodide
H2 (g) + I2 (g) + 12.6 kcal  2 HI (g)
2
! HI #
K= " $
!" H 2 #$ !" I 2 #$
Write the Equilibrium
Constant Equation
Stress
Equilibrium
Shift
[H2]
[I2]
[HI]
K
1.
Add H2
Right
Increase
Decrease
Increase
No Change
2.
Add I2
Right
Decrease
Increase
Increase
No Change
3.
Add HI
Left
Increase
Increase
Increase
No Change
4.
Remove H2
Left
Decrease
Increase
Decrease
No Change
5.
Remove I2
Left
Increase
Decrease
Decrease
No Change
6.
Remove HI
Right
Decrease
Decrease
Decrease
No Change
7.
Increase
Temperature
Right
Decrease
Decrease
Increase
Increase
8.
Decrease
Temperature
Left
Increase
Increase
Decrease
Decrease
9.
Decrease
Volume
No Change
Increase
Increase
Increase
No Change
10.
Decrease
Pressure
No Change
Decrease
Decrease
Decrease
No Change
4 - HC - Chapter 18 - Le Chatelier’s Principle Worksheet - Answers - V10
IV. The Dissolution of Sodium Hydroxide
NaOH (s)  Na+ (aq) + OH - (aq)
Write the Equilibrium
Constant Equation
ΔH = - 44.5 kJ
K = !" Na + #$ !"OH - #$
Stress
Equilibrium
Shift
Amount
NaOH (s)
[Na+]
[OH-]
K
1.
Add NaOH (s)
No Change
Increase
No Change
No Change
No Change
2.
Add NaCl
(Adds Na+)
Left
Increase
Increase
Decrease
No Change
3.
Add KOH
(Adds OH -)
Left
Increase
Decrease
Increase
No Change
4.
Add HCl
(Adds H +
which then
removes OH -)
Right
Decrease
Increase
Decrease
No Change
5.
Increase
Temperature
Left
Increase
Decrease
Decrease
Decrease
6.
Decrease
Temperature
Right
Decrease
Increase
Increase
Increase
7.
Increase
Pressure
No Change
No Change
No Change
No Change
No Change
8.
Decrease
Pressure
No Change
No Change
No Change
No Change
No Change
5 - HC - Chapter 18 - Le Chatelier’s Principle Worksheet - Answers - V10