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Chemical Equilibrium
Chapter 6 E-mail: [email protected]
Web-site: http://clas.sa.ucsb.edu/staff/terri/
Chemical Equilibrium – Ch. 6
N
2
(g) + 3 H
2
(g)
⇌
2 NH
3
(g)
Chemical Equilibrium – Ch. 6
Chemical Equilibrium – Ch. 6
1. Consider the following reaction: N 2 (g) + 3 Cl 2 (g) ⇌ 2 NCl 3 (g) a. Write the equilibrium expression (
K and K p
) b. What is the value of
K p
N 2 , Cl 2 and NCl 3 at 25°C if the equilibrium pressures for were 1.8 atm, 0.5 atm and 0.017 atm respectively. c. What is the value of K at 25°C?
K
p
= K(RT)
Δn
d. Does the reaction favor reactants
or
products?
Chemical Equilibrium – Ch. 6
Equilibrium Expression jA + kB
⇌
lC + mD K =
𝐶 𝐴 𝑙 𝑗 𝐷 𝐵 𝑚 𝑘
Pure solids and liquids are given the value of 1 in the expression Where does the equilibrium lie?
If the products are favored
⇒
K > 1 If the reactants are favored
⇒
K < 1
Chemical Equilibrium – Ch. 6
2. Which reaction does the
K p
=
K
?
a. P 4 (s) + 6 Cl 2 (g) ⇌ 4 PCl 3 (g) b. H 2 O (l) ⇌ H 2 O (g) c. H 2 (g) + Cl 2 (g) ⇌ d. C 3 H 8 (g) + 5 O 2 (g) ⇌ 2 HCl (g) 3 CO 2 (g) + 4 H 2 O (g)
K
p
= K(RT)
Δn
Chemical Equilibrium – Ch. 6
3. Given the following information: H 2 (g) + I 2 (g) ⇌ 2 HI(g) N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) K = 54.0
K = 96.0
Determine the value of the equilibrium constant for the following reaction: 2 NH 3 (g) + 3 I 2 (g) ⇌ 6 HI(g) + N 2 (g)
Chemical Equilibrium – Ch. 6
Since reversible reactions can be written in either direction or balanced with any multiple of coefficients you must consider how this affects the equilibrium constant – K 1. If you flip a reaction take the reciprocal of the constant ex: if K=5 for 2A
⇌ ⇌
2A 2. If you multiply the reaction coefficients raise the constant to the same multiple ex: if K=5 for 2A
⇌
B then K=5 ½ for A
⇌
½B 3. If you add two or more reactions take the product of the constants ex: if K=5 for 2A
⇌
B and if K= 8 for B
⇌
3C then K = (5)(8) = 40 for 2A
⇌
3C
Chemical Equilibrium – Ch. 6
4. Consider the following reaction: UO 2 (s) + 4HF (g) ⇌ UF 4 (g) + 2H 2 O (g)
K p
= 11 If you fill a 5 L container with 15 g of UO 2 , 1.1 atm of HF, 0.75 atm of UF 4 and 5.9 atm of H 2 O, what will be the mass of the solid UO 2 at equilibrium?
a. 15 g b. > 15 g c. < 15 g d. not enough information
Chemical Equilibrium – Ch. 6
Reactions favor a state of equilibrium - so if a reaction mixture is not at equilibrium it will automatically shift one way or the other in order to achieve equilibrium status The equation for Q looks identical to the equation for K - however you can apply Q at any stage of a reaction whereas you can only apply K at equilibrium
𝑐𝑜𝑒𝑓 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
Q =
𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑐𝑜𝑒𝑓
K > Q System has too much reactant K K = Q System is at equilibrium K Q K Q K < Q System has too much product Q Shift reaction to the right causes Q to increase Shift reaction to the left causes Q to decrease
Chemical Equilibrium – Ch. 6
5. Consider the following reaction: N 2 (g) + O 2 (g) ⇌ 2 NO (g)
K p
= ? Initially the partial pressures of N 2 and O 2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the
K p
for this reaction?
Chemical Equilibrium – Ch. 6
6. Pure PCl 5 is introduced into an empty rigid 5-L flask at 24 °C with a pressure of 0.45 atm. The PCl 5 is heated to 90 °C upon which it decomposes into solid P and gaseous Cl 2 .
2 PCl 5 (g) ⇌ 2 P (s) + 5 Cl 2 (g) At equilibrium the total pressure is measured to be 0.93 atm. What is the equilibrium constant (
K p
) at 90 °C? What percentage of the PCl 5 decomposed? How many grams of P are in the flask at equilibrium?
Chemical Equilibrium – Ch. 6
7. At a certain temperature the partial pressures of an equilibrium mixture of N 2 O 4 (g) and NO 2 (g) are 0.34 atm and 1.2 atm respectively. N 2 O 4 (g) ⇌ 2 NO 2 (g) The volume of the container is doubled. Find the new partial pressures when the equilibrium is re-established.
Chemical Equilibrium – Ch. 6
8. Consider the following reaction: 2 NOBr (g) ⇌ 2 NO (g) + Br 2 (g)
K p
= 34 An unknown pressure of NOBr is put into a rigid container at 40 °C, equilibrium is reached when 86 % of the original partial pressure of NOBr has decomposed. What was the initial pressure of the NOBr? What is the total pressure in the flask at equilibrium?
Chemical Equilibrium – Ch. 6
9. At a particular temperature
K p
= 5.7x10
-8 for the following reaction: 2 C (s) + 3 H 2 (g) ⇌ C 2 H 6 (g) What is the pressure of C 2 H 6 of C and 3 atm of H 2 ? at equilibrium if initially there's 5 g
Chemical Equilibrium – Ch. 6
10. Consider the following endothermic reaction at equilibrium: CCl 4 (g) ⇌ C (s) + 2 Cl 2 (g) a. Is heat absorbed or released as the reaction goes forward?
b. Which way will the reaction shift if the temperature is decreased?
c. What happens to
K
as the temperature is increased?
d. Which way will the reaction shift if the partial pressure of CCl 4 increased?
e. Which way will the reaction shift if you add more C?
is f. Which way will the reaction shift if you remove Cl 2 ?
g. Which way will the reaction shift if you compress the system?
h. Which way will the reaction shift if you add a catalyst?
Answer Key – Ch. 6
You have completed ch. 6
Answer Key – Ch. 6
1. Consider the following reaction: N 2 (g) + 3 Cl 2 (g) ⇌ 2 NCl 3 (g) a. Write the equilibrium expression (K
and
K
p
)
K
= 𝑁𝐶𝑙3 𝑁2 𝐶𝑙2 2 3
and K
p
= 𝑃 𝑁𝐶𝑙 3 𝑃 𝑁 2 𝑃 𝐶𝑙 2 2 3 b. What is the value of K
p
and NCl 3 at 25°C if the equilibrium pressures for N were 1.8 atm, 0.50 atm and 0.017 atm respectively.
2 , Cl 2
K
p
=
0.017
2
1.8 0.50
3
= 0.0013
c. What is the value of K at 25°C?
K = K p Δn
⇒
K = (0.0013) (0.08206x298)(2−4) = 0.77
Answer Key – Ch. 6
1.
…continued
d. Does the reaction favor reactants
or
products?
Since K p < 1 and/or K < 1 the reaction favors the reactants
2. Which reaction does the K
p
=K?
a. P 4 (s) + 6Cl 2 (g) ⇌ 4PCl 3 (g) b. H 2 O (l) ⇌ H 2 O (g)
c. H 2 (g) + Cl 2 (g)
⇌
2HCl (g)
d. C 3 H 8 (g) + 5O 2 (g) ⇌ 3CO 2 (g) + 4H 2 O (g)
If Δn=1
⇒
K p =K Δn=1 when there are the same number of moles of gaseous reactants as gaseous products
Answer Key – Ch. 6
3. Given the following information: H 2 (g) + I 2 (g) ⇌ 2 HI(g) N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) K = 54.0
K = 96.0
Determine the value of the equilibrium constant for the following reaction: 2 NH 3 (g) + 3 I 2 (g) ⇌ 6 HI(g) + N 2 (g)
Rxn 2 Flipped Rxn 1 x 3 Rxn 1 x 3 Rxn 2 Flipped 3 H 2 (g) + 3 I 2 (g)
⇌
2 NH 3 (g)
⇌
6 HI(g) K = 54.0
3 N 2 (g) + 3 H 2 (g) K = 96.0
-1 2 NH 3 (g) + 3 I 2 (g)
⇌
6 HI(g) + N 2 (g) K = (54.0
3 )(96.0
-1 ) = 1640
Answer Key – Ch. 6
4. Consider the following reaction: UO 2 (s) + 4HF (g) ⇌ UF 4 (g) + 2H 2 O (g) K
p
= 11 If you fill a 5 L container with 15 g of UO 2 , 1.1 atm of HF, 0.75 atm of UF 4 and 5.9 atm of H 2 O, what will be the mass of the solid UO 2 at equilibrium?
a. 15 g b. > 15 g c. < 15 g d. not enough information
Q p = P UF4 P H2O P HF
4 2
= 0.75 5.9
1.1
4 2
= 17.8
Since K < Q the reaction will go backwards thereby increasing the mass of the UO 2
Answer Key – Ch. 6
I Δ Eq
5. Consider the following reaction: N 2 (g) + O 2 (g) ⇌ 2 NO (g) K
p
= ? Initially the partial pressures of N 2 and O 2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the K p for this reaction?
Since the P of NO at equilibrium is 1.5
N 2 O 2
⇌
2 NO 2x = 1.5 => x = 0.75
1 atm 3 atm 0 - x 1 – x - x 3 – x + 2x 2x So at equilibrium the P N2 K p = and P O2 are 0.25 atm and 2.25 atm respectively P P NO N2 P 2 O2
=
(1.5) 2 (0.25)(2.25) = 4
Answer Key – Ch. 6
6. Pure PCl 5 is introduced into an empty rigid flask at 24 °C with a pressure of 0.45 atm. The PCl 5 is heated to 90 °C upon which it decomposes into solid P and gaseous Cl 2 .
2 PCl 5 (g) ⇌ 2 P (s) + 5 Cl 2 (g) At equilibrium the total pressure is measured to be 0.93 atm. What is the equilibrium constant (K
p
) at 90 °C? What percentage of the PCl 5 decomposed?
Using the combined gas law
𝑃 1 𝑉 1
=
𝑃 2 𝑉 2 𝑛 2 𝑇 2 𝑛 2 𝑇 2
Where n and V are constant
0.45𝑎𝑡𝑚 24+273𝐾
=
𝑃 2 90+273𝐾
P 2 = 0.55 atm I Δ Eq 2 PCl 5 (g) 0.55 atm - 2x 0.55 – 2x
⇌
2 P (s) N/A N/A N/A + 5 Cl 2 (g) 0 + 5x 5x Continue to next slide…
Answer Key – Ch. 6
6.
…continued P total = P PCl5 + P Cl2 0.93 = (0.55 – 2x) + (5x) x = 0.13
P PCl5 = 0.55 – 2(0.13) = 0.29
P Cl2 = 5(0.13) = 0.65
K p = (0.65) (0.29) 5 2 K p = 1.38
% decomposed =
𝑎𝑚𝑜𝑢𝑛𝑡 𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑠𝑒𝑑 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡
x100 % decomposed = 2(0.13) 0.55 x100 = 47%
Answer Key – Ch. 6
7. At a certain temperature the partial pressures of an equilibrium mixture of N 2 O 4 (g) and NO 2 (g) are 0.34 atm and 1.2 atm respectively. The volume of the container is doubled. Find the new partial pressures when the equilibrium is reestablished.
N 2 O 4 K p =
⇒
(g)
⇌
P P NO2 N2O4
⇒
P N2O4 2 NO 2 (g) Since the equilibrium pressures are given we can find K p 2 = (1.2) 2 (0.34) = 4.24
According to P 1 V 1 = P 2 V 2 halved = 0.17 atm and P NO2 = 0.6
Calculate Q to know the direction the reaction proceeds
⇒
if the volume is doubled the pressures will be Q = P P NO2 N2O4 2 = (0.6) 2 (0.17) = 2.12 since K > Q the reaction goes forward in order to re-establish equilibrium …continue to next slide
Answer Key – Ch. 6
7.
…continued I Δ Eq N 2 O 4 (g) 0.17
- x 0.17 - x
⇌
2 NO 2 (g) 0.6
+ 2x 0.6 + 2x Use K to solve for x 4.24 = (0.6 + 2x)2 (0.17 − x) Manipulate to polynomial format 0 = 4x 2 + 6.64x – 0.3608
−𝑏 ± 𝑏 2 − 4𝑎𝑐 𝑥 = 2𝑎
x = 0.053 or -0.71
So the new equilibrium pressures
⇒
P N2O4 P NO2 = 0.17 – 0.053 = 0.6 + 2(0.053) P N2O4 P NO2 = 0.12 atm = 0.71 atm
Answer Key – Ch. 6
8. Consider the following reaction: 2 NOBr (g) ⇌ 2 NO (g) + Br 2 (g) K
p
= 34 An unknown pressure of NOBr is put into a rigid container at 40 °C, equilibrium is reached when 86 % of the original partial pressure of NOBr has decomposed. What was the initial pressure of the NOBr? What is the total pressure in the flask at equilibrium?
I Δ Eq 2 NOBr (g) x -0.86x
0.14x
⇌
2 NO (g) 0 + 0.86x
0.86x
+ Br 2 (g) 0 +0.43x
0.43x
…continue to next slide
Answer Key – Ch. 6
8.
…continued Use K to solve for x 34 = (0.86x)2(0.43x) (0.14x)2 x = 2.2 atm
⇒
the initial pressure of NOBr is 2.2 atm The total pressure in the flask at equilibrium P total = 0.14x + 0.86x + 0.43x = 3.1 atm
Answer Key – Ch. 6
9. At a particular temperature K
p
= 5.7x10
-8 for the following reaction: 2 C (s) + 3 H 2 (g) ⇌ C 2 H 6 (g) What is the pressure of C 2 H 6 C and 3 atm of H 2 ? at equilibrium if initially there's 5 g of
I Δ Eq 3 H 2 3 - 3x 3 - 3x
⇌
C 2 H 6 0 + x x Negligible b/c Kp is so small Since K p is so small we can make the assumption that x is negligible with respect to a non-zero value K p = P C2H6 P H2 3 5.7x10
-8
(
3) 3 x = 1.5x10
-6 atm
Answer Key – Ch. 6
10. Consider the following endothermic reaction at equilibrium: CCl 4 (g) ⇌ C (s) + 2 Cl 2 (g) a. Is heat absorbed or released as the reaction goes forward?
Since the reaction is endothermic heat is absorbed if the reaction goes forward and vice versa
b. Which way will the reaction shift if the temperature is decreased?
As the temperature is decreased the system is losing heat causing the reaction to shift left
c. What happens to K as the temperature is increased?
As the temperature is increased heat is being added causing the reaction to shift right producing more products thereby causing an increase in K
d. Which way will the reaction shift if the partial pressure of CCl 4 increased?
More reactant pushes the reaction to the right
is
… continue to next slide
Answer Key – Ch. 6
e. Which way will the reaction shift if you add more C?
Adding more solid will have no effect on the equilibrium
f. Which way will the reaction shift if you remove Cl 2 ?
Removing a product causes the reaction to shift to the right
g. Which way will the reaction shift if you compress the system?
Decreasing the volume of the system causes the reaction to shift to the side with fewer moles of gas – in this case the left
h. Which way will the reaction shift if you add a catalyst?
Catalysts will speed up the forward and reverse direction therefore having no effect on the equilibrium