equilibrium entry task: jan 8 th tuesday at 1100 k, k p = 0.25 atm 1 for the reaction: 2 so 2 (g) +...

Equilibrium

Entry Task: Jan 8th Tuesday

At 1100 K, Kp = 0.25 atm1 for the reaction:

2 SO2(g) + O2(g) ↔ 2 SO3(g)

What is the value of Kc at this temperature?

Equilibrium

Agenda:

• Discuss Equilibrium constant Kp and Kc ws

• In-Class ICE table practice

• HW: Pre-Lab Determining Keq Constant

Equilibrium

Equilibrium

][

][][

3

22

OHCH

HCOp

P

PPK

1. Calculate Kp and Kc for the following reaction…..CH3OH (g) CO (g) + 2H2 (g)Given the following equilibrium pressures at 25oC: P CH3OH=6.10 x 10-4 atmPCO = 0.387 atmPH2= 1.34 atm

]1010.6[

]34.1[]387.0[

4

2

x

Kc Kc= 1.14 x 10-3

Equilibrium

1. Calculate Kp and Kc for the following reaction…..CH3OH (g) CO (g) + 2H2 (g)Given the following equilibrium pressures at 25oC: P CH3OH=6.10 x 10-4 atmPCO = 0.387 atmPH2= 1.34 atm

]1010.6[

]34.1[]387.0[

4

2

x

Kc

Kp= Kc(RT)∆n

∆n = 3-1 =2

Kp= (1.14 x 10-3)((0.0821)(298))2

Kp= 0.68

Equilibrium

2. A 127°C, Kc = 2.6 × 105 mol2/L2 for the reaction: 2 NH3(g) N2(g) + 3H2(g) Calculate Kp at this temperature.

Kp= Kc(RT)∆n

∆n = 4-2 =2

Kp= (2.6 × 105)((0.0821)(400))2

Kp= 2.8x10-2

Equilibrium

3. At 1100 K, Kp = 0.25 atm1 for the reaction: 2 SO2(g) + O2(g) 2 SO3(g) What is the value of Kc at this temperature?

Kp= Kc(RT)∆n

∆n = 2-3 =-1

0.25= (X)((0.0821)(1100))-1

Kc= 23

Equilibrium

4. Phosphorus pentachloride dissociates on heating: PCl5(g) PCl3(g) + Cl2(g) If Kc equals 3.28x10-2 at 191°C, what is Kp at thistemperature?

Kp= Kc(RT)∆n

∆n = 2-1 =1

Kp= (3.28x10-2)((0.0821)(464))1

Kp= 1.25

Equilibrium

5. Nitrogen dioxide dimerizes to form dinitrogen tetraoxide: 2 NO2(g) N2O4(g) Calculate the value of Kc, given that the gas phase equilibrium constant, Kp, for the reaction is 1.3 × 103 at 273 K. (R = 0.08206 L·atm/mol·K)

Kp= Kc(RT)∆n

∆n = 1-2 =-1

1.3 x10-2= (X)((0.08206)(273))-1

Kc= 0.29

Equilibrium

23

22

][

][][ 2

SO

OSOp

P

PPK

6. An equilibrium mixture of SO3, SO2, and O2 at 1000 K contains the gases at the following concentrations: [SO3] = 0.41 M, [SO2] = 0.032 M, and [O2] = 0.59 M. What is the equilibrium constant (Kc) and the Kp value for the decomposition of SO3? 2 SO3(g) 2 SO2(g) + O2(g)

Kc= 3.59 x 10-32

2

]41.0[

]59.0[]032.0[pK

Equilibrium

6. An equilibrium mixture of SO3, SO2, and O2 at 1000 K contains the gases at the following concentrations: [SO3] = 0.41 M, [SO2] = 0.032 M, and [O2] = 0.59 M. What is the equilibrium constant (Kc) and the Kp value for the decomposition of SO3? 2 SO3(g) 2 SO2(g) + O2(g)

Kp= Kc(RT)∆n

∆n = 3-2 =1

Kp= (3.59 x 10-3)((0.0821)(1000))1

Kp= 2.95 x 10-1

Equilibrium

Equilibrium

15.4- Calculating Equilibrium Constant

When a reaction has reached equilibrium, we often don’t know HOW the initial concentrations of the species have changed from the equilibrium concentrations.

Or we were given JUST the initial concentrations, how can we determine the equilibrium concentrations?

Equilibrium

Equilibrium

To look at all these variables at the same time we need to create an ICE table.

I is initial concentration,

C is the change in concentration

E is the concentration at equilibrium.

Equilibrium

There are two types of ICE tablesType 1A. The initial or equilibrium concentration of some

substances must be determined.

B. Initial or equilibrium concentrations of some substances are given, but not both. Change is therefore treated as an unknown (x)

C. The equilibrium constant is given.

Equilibrium

There are two types of ICE tablesType 2A. The equilibrium constant or concentration

must be determined.B. Initial and equilibrium concentrations of at

least one substance are given so that change can be calculated directly.C. All other initial and equilibrium concentrations

of substances are determined directly from the table.

Equilibrium

Walkthrough-(See pg. 571-2 Sample Exercise)

A closed system initially containing

1.000 x 10−3 M H2 and 2.000 x 10−3 M I2

At 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448C for the reaction taking place, which is

H2 (g) + I2 (g) 2 HI (g)

Equilibrium

STEPS TO ICE

① Make Sure that all concentrations are in M- molarity!! (Done for you)

② Set up table- ICE (as you see it) and then species at top.

③Place the known concentrations provided in the question into table.

④Put in the CHANGE for HI (subtract equilibrium from initial)

Equilibrium

② Set up table- ICE (as you see it) and then species at top.

[H2], M [I2], M [HI], M

Initially

Change

At equilibrium

Equilibrium

③Place the known concentrations provided in the question into table.

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At equilibrium

1.87 x 10-3

Equilibrium

④Put in the CHANGE for HI (subtract equilibrium from initial)

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium

1.87 x 10-3

Equilibrium

⑤NOW we have to use the stoichiometry of the reaction to get the change of H2 and I2. Put a negative sign in front because they are reactants.

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium

1.87 x 10-3

1.87 x 10-3mol L

1 mol H2

2 mol of HI

= 0.935 x 10-3

Same goes for iodine

H2 (g) + I2 (g) 2 HI (g)

Equilibrium

⑤ Stoichiometry tells us [H2] and [I2]decrease by half as much

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

1.87 x 10-3

The change MUST be in the negative because they are reactants!!

Equilibrium

⑥ Subtract the initial concentrations from the change which will provide the equilibrium value.

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

1.87 x 10-3

1.000 x10-3 – 9.035 x10-4) =6.5 x 10-5

6.5 x 10-5 1.065 x 10-3

2.000 x10-3 – 9.035 x10-4) =1.065 x 10-3

Equilibrium

⑦ Finally, provide the equilibrium expression for this reaction.

Substitute the equilibrium values from chart into the expression to solve for Kc.

=(1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

Kc =[HI]2

[H2] [I2]

Kc = 51

Equilibrium

Let’s Try anotherSulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g) 2SO2(g) + O2(g). Initially, the vessel is charged at 1000 K with SO3(g) at a concentration of 6.09 x 10-3 M. At equilibrium the SO3 concentration is 2.44 x10-3M. Calculate the value of Kp at 1000 K.

       INITIAL      CHANGE      EQUILIBRIUM      

2SO3 2SO2 O2

6.09 x 10-3 M 0 M 0 M

2.44 x10-3M

Equilibrium

Let’s Try another

What information can we fill in with what we are given?

       INITIAL      CHANGE      EQUILIBRIUM      

2SO3 2SO2 O2

6.09 x 10-3 M 0 M 0 M

2.44 x10-3M

We can fill in the change of SO3.

6.09 x 10-3 M - 2.44 x10-3M = -3.65 x 10-3 M

-3.65 x 10-3 M

Equilibrium

Let’s Try another

If SO3 went down by -3.65 x 10-3 M we have to use stoichiometry to find out the relationship the products have with the reactant.

       INITIAL      CHANGE      EQUILIBRIUM      

2SO3 2SO2 O2

6.09 x 10-3 M 0 M 0 M

2.44 x10-3M

3.65 x 10-3 M of SO3= +3.65 x 10-3 M

of SO2

-3.65 x 10-3 M

2 moles SO2

2 mole of SO3

+3.65 x 10-3 M

3.65 x 10-3 M of SO3= +1.83 x 10-3 M

of O2

1 moles O2

2 mole of SO3

+1.83 x 10-3 M

Equilibrium

Let’s Try another

Now subtract the initial from the change to get the equilibrium.

       INITIAL      CHANGE      EQUILIBRIUM      

2SO3 2SO2 O2

6.09 x 10-3 M 0 M 0 M

2.44 x10-3M

0 - 3.65 x 10-3 M = 3.65 x 10-3 M

-3.65 x 10-3 M +3.65 x 10-3 M

0 – 1.83 x 10-3 M = 1.83 x 10-3 M

1.83 x 10-3 M+1.83 x 10-3 M

3.65 x 10-3 M

Equilibrium

Let’s Try another

Use the equilibrium concentrations to find Kc- plug n chug

       INITIAL      CHANGE      EQUILIBRIUM      

2SO3 2SO2 O2

6.09 x 10-3 M 0 M 0 M

2.44 x10-3M-3.65 x 10-3 M +3.65 x 10-3 M

1.83 x 10-3 M+1.83 x 10-3 M

3.65 x 10-3 M

23

22

2

][

][][

SO

OSOKc

23-

-32-3

] x102.44[

]10 x 1.83[]10 x 3.65[cK

(1.33225 x10-5)(1.83 x10-3) = 2.4380175 x10-8

5.9536 x10-6

Kc= 4.11 x 10-3

Equilibrium

In many situations we will know the value of the equilibrium constant

and the initial concentrations of all species. We must then solve for

the equilibrium concentrations. We have to treat them as variables or

“x”

Equilibrium

A chemist has a container of A2 and B2 and they react as given: A2 (g) + B2 (g) 2 AB (g)

Kc = 9.0 at 100°CIf 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0

L container, what are the equilibrium concentrations of A2, B2, and AB?

① Convert to molarity!!

1 mol2.0 L

= 0.50 M

Equilibrium

A chemist has a container of A2 and B2 and they react as given: A2 (g) + B2 (g) 2 AB (g)

Kc = 9.0 at 100°CIf 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0

L container, what are the equilibrium concentrations of A2, B2, and AB?

[A2], M [B2], M [AB], M

Initially

Change

At equilibrium

0.50 M 0.50 M 0.0 M

Equilibrium

[A2], M [B2], M [AB], M

Initially

Change

At equilibrium

0.50 M 0.50 M 0.0 M

Fill in the chart with variables!!

You need to (for this ICE table) to factor in the stoich relationship)

A2(g) + B2(g) 2AB (g)

-x -x + 2x

0.50-x 0.50-x 2x

Equilibrium

We have to work backwards!!

Provide the Kc expression

9.0=(2x)2

(0.50-x)(0.50-x)

Kc =[AB]2

[A2] [B2]

[A2], M [B2], M [AB], M

Initially 0.50M 0.50 M 0 M

Change

At equilibrium

-x

-x + 2x

0.50-x 0.50-x 2x

Equilibrium

Get 0.50-x out by multiply on both sides

Root both sides9.0=

(2x)2

(0.50-x)(0.50-x)Or

(0.50-x)2

3.0=(2x)

(0.50-x)

Equilibrium

Get 0.50-x out by multiply on both sides

3.0=(2x)

(0.50-x)

3.0(0.50-x)=2x Multiply 3 through

1.5-3x= 2x Add 3x to both sides to get rid of -3x

Equilibrium

1.5-3x= 2x Add 3x to both sides to get rid of -3x

1.5 = 5x Divide both sides by 5 to get 5 out of there

0.3 = x SO back to the chart with our x value of 0.3

Equilibrium

[A2], M [B2], M [AB], M

Initially

Change

At equilibrium

0.50 M 0.50 M 0.0 M

Now plug in our “x” value of 0.3

-x -x + 2x 2(0.3) =0.6M

0.50-0.3 = 0.2M

0.50-0.3 = 0.2M

Equilibrium

Sample Exercise 15.11

A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the

equilibrium constant (Kc) is 50.5. H2(g) + I2(g) 2HI (g)

What are the concentrations of H2, I2, and HI in the flask at equilibrium?

Equilibrium

[H2], M [I2], M [HI], M

Initially 1.000M 2.000 M 0 M

Change

At equilibrium

Fill in the chart with variables!!

-x

-x + 2x

1.000-x 2.000-x 2x

You need to (for this ICE table) to factor in the stoich relationship)

H2(g) + I2(g) 2HI (g)

Equilibrium

[H2], M [I2], M [HI], M

Initially 1.000M 2.000 M 0 M

Change

At equilibrium

Fill in the chart with variables!!

-x

-x + 2x

1.000-x 2.000-x 2x

You need to (for this ICE table) to factor in the stoich relationship)

H2(g) + I2(g) 2HI (g)

Equilibrium

50.5=(2x)2

(1.000-x)(2.000-x)

50.5 =4x2

2-3x+x2

Factor the denominator

Get the denominator out of there multiply both sides

50.5 (2-3x+x2)=4x2 Multiply 51 though

Equilibrium

50.5 (2-3x+x2)= 4x2 Multiply 51 though

101-151.5x+50.5x2= 4x2Subtract 4x2

from both sides

101-151.5x+46.5x2= 0 Quadratic equation

C - Bx + Ax2 = 0

You have program your calculator OR do it the LONG WAY

Equilibrium

101-151.5x+46.5x2= 0C - Bx + Ax2 =

X= -(-151.5)± √151.52 – 4(46.5)(101)

2(46.5)

Equilibrium

101-151.5x+46.5x2= 0C - Bx + Ax2 =

X= -(-151.5)± √22952.25 –18786

93

Equilibrium

101-151.5x+46.5x2= 0C - Bx + Ax2 =

X= -(-151.5)± √4166.25

93

Equilibrium

101-151.5x+46.5x2= 0C - Bx + Ax2 =

X= -(-151.5)+ 64.5

93= 2.32

X= -(-151.5)- 64.5

93= 0.935

Equilibrium

Which is the correct

X= -(-151.5)+ 64.5

93= 2.32

X= -(-151.5)- 64.5

93= 0.935

Substitute the value for X. If we get a negative concentration that value would

NOT be the correct on.

Equilibrium

[H2], M [I2], M [HI], M

Initially 1.000M 2.000 M 0 M

Change

At equilibrium

-x

-x + 2x

1.000-x 2.000-x 2x

[H2] 1.000- 2.32= -1.32

From Quadratic 1

NOT correct value!!

Equilibrium

[H2], M [I2], M [HI], M

Initially 1.000M 2.000 M 0 M

Change

At equilibrium

-0.935

-0.935 2(0.935)

0.065M 1.065 M 1.87 M

[H2] 1.000- 0.935= 0.065M

From Quadratic 2

Correct value!!

Equilibrium

[H2], M [I2], M [HI], M

Initially 1.000M 2.000 M 0 M

Change

At equilibrium

-0.935

-0.935 2(0.935)

0.065M 1.065 M 1.87 M

Plug the equilibrium concentrations into Kc expression see if its 50.5

]065.1[]065.0[

]87.1[ 2

cK069225.0

4969.3cK Kc= 50.5

Equilibrium

Equilibrium

Equilibrium

Equilibrium

HW: PreLab Determining Kc constant and ice tables

ws

Equilibrium

ICE TABLES

Equilibrium

Sample Exercise 1• An equilibrium was established after

0.100 mol of hydrogen gas and 0.100 mol of iodine gas were added to an empty 1.00 L reaction vessel and heated to 700 K. The color intensity of the mixture changed from deep purple to a lighter purple color. At equilibrium, concentration of iodine was 0.0213 mol/L.

• Calculate the equilibrium concentrations of hydrogen gas and hydrogen iodide gas.

Equilibrium

Sample Exercise 1-Plan a Strategy

• You will need to write the balanced chemical equation for the reaction, set up an ICE table, and enter the given data.

• Then, using the given data and your stoichiometric skills, you will find the change in the concentration of each species. This will involve finding change in [I2] by subtraction and changes in [H2] and [HI] by stoichiometry using the change in [I2]. Update the ICE table as values are calculated.

• Calculate the equilibrium concentrations by addition or subtraction. Since H2 and I2 react to produce HI, their concentrations should be lower at equilibrium.

• Communicate the answers.

Equilibrium

Sample Exercise 1-

• Step 1: Write the equation and build the ICE table.

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.100 M 0.100 M

0.0213 M

0.000 M

Equilibrium

Sample Exercise 1-

• Step 2: Calculate the changes in [I2], [H2], and [HI].

• Use the given change with the # moles

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.100 M 0.100 M

0.0213 M

0.0787 M

0.000 M

Equilibrium

Sample Exercise 1-

• Step 2: Calculate the changes in [I2], [H2], and [HI].

• Use mole ratios with the CHANGE in I2

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.100 M 0.100 M

0.0213 M

0.0787 M0.0787 M

0.000 M

Equilibrium

Sample Exercise 1-

• Step 2: Calculate the changes in [I2], [H2], and [HI].

• Use mole ratios with the CHANGE in I2

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.100 M 0.100 M

0.0213 M

0.0787 M0.0787 M 0.157 M

0.000 M

Equilibrium

Sample Exercise 1-• Step 3: Calculate the equilibrium concentrations.• Subtract the initial from the change (reactants)• Add the initial from the change (product)

[H2] Equilibrium = 0.100 M - 0.0787 M = 0.0213 M

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.100 M 0.100 M

0.0213 M

0.0787 M0.0787 M 0.157 M

0.0213 M

0.000 M

Equilibrium

Sample Exercise 1-• Step 3: Calculate the equilibrium concentrations.• Subtract the initial from the change (reactants)• Add the initial from the change (product)

[HI] Equilibrium = 0.000 + 0.1574 = 0.1574 M

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.100 M 0.100 M

0.0213 M

0.0787 M0.0787 M 0.157 M

0.0213 M

0.000 M

0.157 M

Equilibrium

Step 4: ANSWER

At equilibrium the concentrations of hydrogen gas and hydrogen iodide gas are 0.0213 M and 0.157 M respectively

Equilibrium

Things to note:

• You should see that the initial concentrations of the species placed in the vessel are greater than their equilibrium concentrations. That's because these species have been consumed to make the product.

• The product species have higher concentrations at equilibrium because initially they were not present in the vessel.

• All aqueous/gas species have non-zero concentrations at equilibrium.

• It is always a good idea to review the final numbers in an ICE table to ensure that the values make sense. For example, it would not make sense in the above example for the concentration of hydrogen gas to be higher at equilibrium than it was initially.

Equilibrium

Sample Exercise 2

• An equilibrium was established in a 1.00 L container when 0.120 mol of phosphorus pentachloride gas was decomposed at 500 K. At equilibrium, the concentration of chlorine gas was 0.0540 mol/L.

• Calculate the equilibrium constant for this system and state a conclusion about the position of the equilibrium

Equilibrium

Sample Exercise 2-

• Step 1: Write the equation and build the ICE table.

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.120 M 0.000 M

0.0540 M

0.000 M

Equilibrium

Sample Exercise 2-

• Step 2: Calculate the changes in given species. • Use the given change with the # moles

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.120 M 0.000 M

0.0540 M

0.000 M

0.0540 M

Equilibrium

Sample Exercise 2-

• Step 2: Calculate the changes in given species. • Now we have 1 for the change can use stoich to figure out the

other changes

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.120 M 0.000 M

0.0540 M

0.000 M

0.0540 M0.0540 M

Equilibrium

Sample Exercise 2-

• Step 2: Calculate the changes in given species. • Now we have 1 for the change can use stoich to figure out the

other changes

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.120 M 0.000 M

0.0540 M

0.000 M

0.0540 M0.0540 M0.0540 M

Equilibrium

Sample Exercise 2-• Step 3: Calculate the equilibrium concentrations.• Subtract the initial from the change (reactants)• Add the initial from the change (product)

[PCl5] Equilibrium = 0.120 M - 0.0540 M = 0.066 M

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.120 M 0.000 M

0.0540 M

0.000 M

0.0540 M0.0540 M0.0540 M

0.0066 M

Equilibrium

Sample Exercise 2-• Step 3: Calculate the equilibrium concentrations.• Subtract the initial from the change (reactants)• Add the initial from the change (product)

[PCl3] Equilibrium = 0.000 + 0.0540 = 0.0540 M

Equilibrium

Fill in information from question

Initial

Change

Equilibrium

0.120 M 0.000 M

0.0540 M

0.000 M

0.0540 M0.0540 M0.0540 M

0.0066 M 0.0540 M

Equilibrium

Sample Exercise 2-• Step 4: Write the K expression and solve.

Equilibrium

Sample Exercise 2-• Calculate the equilibrium constant for this

system and state a conclusion about the position of the equilibrium

The equilibrium constant for the decomposition of phosphorus pentachloride at 500 K is 4.4 x 10-2. The equilibrium position lies to the left, or in other words, the reactant is favored.