Chemical Equilibrium

Chemical EquilibriumChapter 6E-mail: [email protected]: http://clas.sa.ucsb.edu/staff/terri/

Chemical Equilibrium – Ch. 61. Consider the following reaction:

N2 (g) + 3 H2 (g) 2 NH⇌ 3 (g) a. Write the equilibrium expression (K and Kp)b. What is the value of Kp at 25°C if the equilibrium pressures for

N2, H2 and NH3 were 1.8 atm, 0.5 atm and 0.017 atm respectively. c. What is the value of K at 25°C? Kp = K(RT)Δn

d. Does the reaction favor reactants or products?

Chemical Equilibrium – Ch. 6

Where does the equilibrium lie?K > 1 ⇒ The products are favored

K < 1 ⇒ The reactants are favored

Equilibrium Expression

jA + kB ⇌ lC + mD

K =

Pure solids and liquids

are given the value of 1 in the

expression

Chemical Equilibrium – Ch. 6

Chemical Equilibrium – Ch. 62. Which reaction does the K = Kp?

a. P4 (s) + 6 Cl2 (g) 4 PCl⇌ 3 (g)b. H2O (l) H⇌ 2O (g)c. H2 (g) + Cl2 (g) 2 HCl (g)⇌d. C3H8 (g) + 5 O2 (g) 3 CO⇌ 2 (g) + 4 H2O (g)

Kp = K(RT)Δn

Chemical Equilibrium – Ch. 63. Given the following information:

H2(g) + I2(g) 2 HI(g)⇌ K = 54.0N2(g) + 3 H2(g) 2 NH⇌ 3(g) K = 96.0

Determine the value of the equilibrium constant for the following reaction:

2 NH3(g) + 3 I2(g) 6 HI(g) + N⇌ 2(g)

Chemical Equilibrium – Ch. 6Since 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 B then K= for B 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. 64. Consider the following reaction:

UO2 (s) + 4HF (g) UF⇌ 4 (g) + 2H2O (g) Kp = 11If you fill a 5 L container with 15 g of UO2, 1.1 atm of HF, 0.75 atm of UF4 and 5.9 atm of H2O, what will be the mass of the solid UO2 at equilibrium?a. 15 gb. > 15 gc. < 15 gd. not enough information

Chemical Equilibrium – Ch. 6Reactions 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

Q =

K KK Q

Q

QK > Q

System has too

much reactant

K = Q

System is at

equilibrium

K < Q

System has too

much product

Shift reaction

to the right

Shift reaction

to the left

causes Q to

increase

causes Q to

decrease

Chemical Equilibrium – Ch. 65. Consider the following reaction:

N2 (g)  +  O2 (g)  2  NO (g)    ⇌ Kp = ? Initially the partial pressures of N2 and O2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the Kp for this reaction?

Chemical Equilibrium – Ch. 66. Pure PCl5 is introduced into an empty rigid 5-L flask at 24 °C with a

pressure of 0.45 atm.  The PCl5 is heated to 90 °C upon which it decomposes into solid P and gaseous Cl2. 

2 PCl5 (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 (Kp) at 90 °C? What percentage of the PCl5 decomposed? How many grams of P are in the flask at equilibrium?

Chemical Equilibrium – Ch. 67. At a certain temperature the partial pressures of an equilibrium mixture of N2O4 (g) and NO2 (g) are 0.34 atm and 1.2 atm respectively.

N2O4 (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. 68. Consider the following reaction: 

2 NOBr (g) 2 NO (g) + Br⇌ 2 (g)     Kp = 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. 69. At a particular temperature Kp = 5.7x10-8 for the following reaction: 

2 C (s) + 3 H2 (g) C⇌ 2H6 (g) What is the pressure of C2H6 at equilibrium if initially there's 5 g of C and 3 atm of H2?

Chemical Equilibrium – Ch. 610. Consider the following endothermic reaction at equilibrium: 

CCl4 (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 increased?c. What happens to K as the temperature is increased?d. Which way will the reaction shift if the partial pressure of CCl4 is

increased?e. Which way will the reaction shift if you add more C?f. Which way will the reaction shift if you remove Cl2?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. 61. Consider the following reaction:

N2 (g) + 3 Cl2 (g) 2 NCl⇌ 3 (g) a. Write the equilibrium expression (K and Kp)

and Kp = b. What is the value of Kp at 25°C if the equilibrium pressures for N2, Cl2 and NCl3 were 1.8 atm, 0.50 atm and 0.017 atm respectively.

Kp = = 0.0013c. What is the value of K at 25°C? K = ⇒

K = = 0.77

Answer Key – Ch. 61. …continued

d. Does the reaction favor reactants or products?Since Kp < 1 and/or K < 1 the reaction favors the reactants

2. Which reaction does the K = Kp? a. P4 (s) + 6Cl2 (g) 4PCl⇌ 3 (g)b. H2O (l) H⇌ 2O (g)c. H2 (g) + Cl2 (g) 2HCl (g)⇌d. C3H8 (g) + 5O2 (g) 3CO⇌ 2 (g) + 4H2O (g)

If Δn=0 ⇒ Kp=K

Δn=0 when there are the same number of molesof gaseous reactants as gaseous products

3. Given the following information:H2(g) + I2(g) 2 HI(g)⇌ K = 54.0N2(g) + 3 H2(g) 2 NH⇌ 3(g) K = 96.0

Determine the value of the equilibrium constant for the following reaction:

2 NH3(g) + 3 I2(g) 6 HI(g) + N⇌ 2(g)

Answer Key – Ch. 6

Rxn 2

Flipped

Rxn 1

x 3

Rxn 1

x 3

Rxn 2

Flipped

3 H2(g) + 3 I2(g) 6 HI(g)⇌ K = 54.03

2 NH3(g) N2(g) + 3 H2(g) K = 96.0⇌-1

2 NH3(g) + 3 I2(g) 6 HI(g) + N2(g) K = (54.0⇌3

)(96.0-1

) = 1640

Answer Key – Ch. 64. Consider the following reaction: UO2 (s) + 4HF (g) UF⇌ 4 (g) + 2H2O (g) Kp = 11

If you fill a 5 L container with 15 g of UO2, 1.1 atm of HF, 0.75 atm of UF4 and 5.9 atm of H2O, what will be the mass of the solid UO2 at equilibrium?a. 15 gb. > 15 gc. < 15 gd. not enough information

Qp = = = 17.8

Since K < Q the reaction will go backwards thereby

increasing the mass of the UO2

Answer Key – Ch. 65. Consider the following reaction:

N2 (g)  +  O2 (g)  2  NO (g)    K⇌ p = ? Initially the partial pressures of N2 and O2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the Kp for this reaction?

N2 O2 ⇌ 2 NO

I 1 atm 3 atm 0

Δ - x - x + 2x

Eq 1 – x 3 – x 2x

Since the P of NO at equilibrium is 1.5

2x = 1.5 => x = 0.75

So at equilibrium the

PN2 and PO2 are 0.25 atm and 2.25 atm respectively

Kp= = 4

Answer Key – Ch. 66. Pure PCl5 is introduced into an empty rigid flask at 24 °C with a

pressure of 0.45 atm.  The PCl5 is heated to 90 °C upon which it decomposes into solid P and gaseous Cl2. 

2 PCl5 (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 (Kp) at 90 °C? What percentage of the PCl5 decomposed?

2 PCl5 (g) ⇌ 2 P (s) + 5 Cl2 (g)

I 0.55 atm N/A 0

Δ - 2x N/A + 5x

Eq 0.55 – 2x N/A 5x

Continue to next slide…

Using the combined gas law

=

Where n and V are constant

=

P2 = 0.55 atm

Answer Key – Ch. 66. …continued

Ptotal = PPCl5 + PCl2

0.93 = (0.55 – 2x) + (5x)x = 0.13

PPCl5 = 0.55 – 2(0.13) = 0.29

PCl2 = 5(0.13) = 0.65

Kp = Kp = 1.38

% decomposed = x100% decomposed = x100 = 47%

Answer Key – Ch. 67. At a certain temperature the partial pressures of an equilibrium mixture

of N2O4 (g) and NO2 (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.

N2O4 (g) ⇌ 2 NO2 (g) Since the equilibrium pressures are given we can find Kp⇒

Kp = = = 4.24According to P1V1 = P2V2 ⇒ if the volume is doubled the pressures will be

halved ⇒ PN2O4 = 0.17 atm and PNO2

= 0.6Calculate Q to know the direction the reaction proceedsQ = = = 2.12 since K > Q the reaction goes forward in

order to re-establish equilibrium

…continue to next slide

Answer Key – Ch. 6

N2O4 (g) ⇌ 2 NO2 (g)

I 0.17 0.6

Δ - x + 2x

Eq 0.17 - x 0.6 + 2x

Use K to solve for x

4.24 =

Manipulate to polynomial format

0 = 4x2 + 6.64x – 0.3608

x = 0.053 or -0.71

So the new equilibrium pressures ⇒

PN2O4 = 0.17 – 0.053

PNO2 = 0.6 + 2(0.053)

7. …continued

PN2O4 = 0.12 atm

PNO2 = 0.71 atm

Answer Key – Ch. 68. Consider the following reaction: 

2 NOBr (g) 2 NO (g) + Br⇌ 2 (g)     Kp = 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?

2 NOBr (g) ⇌ 2 NO (g) + Br2 (g)

I x 0 0

Δ -0.86x + 0.86x +0.43x

Eq 0.14x 0.86x 0.43x…continue to

next slide

Answer Key – Ch. 68. …continued

Use K to solve for x34 =

x = 2.2 atm ⇒the initial pressure of NOBr is 2.2 atm

The total pressure in the flaskat equilibrium

Ptotal = 0.14x + 0.86x + 0.43x = 3.1 atm

Answer Key – Ch. 69. At a particular temperature Kp = 5.7x10-8 for the following reaction: 

2 C (s) + 3 H2 (g) C⇌ 2H6 (g) What is the pressure of C2H6 at equilibrium if initially there's 5 g of C and 3 atm of H2?

3 H2 ⇌ C2H6

I 3 0Δ - 3x + x

Eq 3 - 3x x

Since Kp is so small we can make the assumption that x is negligible with

respect to a non-zero value

Kp =

5.7x10-8

=

x = 1.5x10-6

atm

Negligible b/c Kp is

so small

10. Consider the following endothermic reaction at equilibrium:  CCl4 (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 increased? As the temperature is increased the system will have more heat causing the reaction to shift right

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 CCl4 is increased? More reactant pushes the reaction to the right

… 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 Cl2? 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

Answer Key – Ch. 6

Answer Key – Ch. 6