Equilibrium Examples

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# Equilibrium Examples - PowerPoint PPT Presentation

Equilibrium Examples. Ch. 15. Variations in Concentration problems.

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### Equilibrium Examples

Ch. 15

Variations in Concentration problems
• If all components are present, find Q before proceeding so that you know which direction the reaction proceeds. This is not necessary if one side of reaction has 0 initial concentration since you then know which way reaction must go.
• If K is very small, then it is possible to approximate the term (yy – x) as yy if yy is the initial concentration and x is the change.
Calculating Equilibrium Concentrations

The reaction of hydrogen and fluorine to form HF gas has an equilibrium constant of 1.15 x 102 at a given temp. 3.000 mol of each component was added to 1.500-L flask. Calculate the equilibrium concentrations of each.

• Equation: H2(g) + F2(g)   2HF (g)
• K = 1.15 x 102 = [HF]2(product is favored)

[H2][F2]

HF Problem, continued (K = 1.15 x 102)
• [HF]0 = [H2]0 = [F2]0 = 3.000 mol = 2.000M

1.500 L

• Q = [HF]02 ___= 2.0002= 1.000

[H2]0[F2]0 (2.000)(2.000)

Q < K so system shifts to products to reach equilibrium (to the right).

(K = 1.15 x 102)

Kc = 1.15 x 102 = (2.000 + 2x)2

(2.000-x)2

How convenient! 2 squared terms, so take square root of each side.

√1.15 x 102 = 2.000 + 2x

2.000 – x

10.7238(2-x)=2+2x

19.4476 = 12.7238x x = 1.528

[H2] = [F2] = 2.000M – 1.528 = 0.472 M

[HF] = 2.000M + 2(1.528) = 5.056 M

Calculating Equilibrium Concentrations

The reaction of hydrogen and fluorine to form HF gas has an equilibrium constant of 1.15 x 102 at a given temp. 3.000 mol H2 and 6.000 mol F2 are added to a 3.000-L flask. Calculate the equilibrium concentrations of each.

• Equation: H2(g) + F2(g)   2HF (g)
• K = 1.15 x 102 = [HF]2(product is favored)

[H2][F2]

2nd HF Problem, continued (K = 1.15 x 102)
• [H2]0 = 3.000 mol = 1.000M [F2]0 = 6.000 mol = 2.000 M

3.000 L 3.000 L

• System shifts to products to reach equilibrium (to the right) since there is no HF at start.
2nd HF Problem, cont. (K = 1.15 x 102)

Kc = 1.15 x 102 = (2x)2 _____________

(1.000 – x)(2.000-x)

115(2-3x+x2) = 4x2

111x2 – 345x + 230 = 0

X = -b ± √(b2 – 4ac)= 345 ± √ (3452-4(111)(230)

2a 2(111)

X = 2.14 and x = .968

[H2] = 1.000M-2.14 = not possible, so x = .968

[H2] = 1.000M-.968 = 0.032 M

[F2] = 2.000M – .968 = 1.032 M

[HF] = 2(0.968) = 1.936 M

Small Equilibrium Constants

Gaseous NOCl decomposes to form the gases NO and Cl2. At 35º C, the equilibrium constant is 1.6 x 10-5. In an experiment in which 1.0 mol of NOCl is placed in a 2.0-L flask, what are the equilibrium concentrations?

• 2NOCl (g)   2NO (g) + Cl2 (g)
• K = [NO]2[Cl2] = 1.6 x 10-5

[NOCl]2

Small Equilibrium Constant, cont.
• [NOCl]0 = 1.0 mol/2.0 L = 0.50 M
• System moves to right since no products were present in the beginning.
Small Kc, continued

Kc = 1.6 x 10-5 = (2x)2 (x)___ Yikes!

(0.50 – 2x)2

But--- the system does not have to proceed far to reach equilibrium since K is so small. X must be very small compared to 0.50, so ignore x in the subtraction term (but not in the cubed term.)

0.50 – 2x ~ 0.50

(0.50)2(1.6 x 10-5) = 4x3

x3 = 1x 10-6 x = 0.01

[NOCl] = 0.50 -2x ~ 0.50 M (recall we approximated)

[NO] = 2x = 2(0.01) = 0.02M

[Cl2] = x = 0.01M

### More Equilibrium Practice

(Zumdahl 13.6)

The reaction between nitrogen and oxygen to form nitric oxide has an equilibrium constant value of 4.1 x 10-4 at 2000 K. If 0.50 moles of nitrogen and 0.86 moles of oxygen are put into a 2.0 L container at 2000 K, what are the equilibrium concentrations of each species?

Prob. 13-6K=4.1 x 10-4
• What is the extent of this reaction? K is small, so reaction stays far to left.
• [N2]0 = 0.50 mole/2.0 L = 0.25 M
• [O2]0 = 0.86 mole/2.0 L = 0.43 M
Prob. 13-6K=4.1 x 10-4
• K is small, so assume x is negligible compared to 0.25 and 0.43.
• K = [NO]2 = 4x2 = 4.1 x 10-4

[N2]0 [O2]0 (.25)(.43)

• X = 3.3 x 10-3 M
• [N2] = 0.25 M [O2] = 0.43 M [NO]= 6.6 x 10-3M
Intermediate K: Practice 13.6 B
• Sulfurous acid dissociates in water:

H2SO3 (aq)   H+ (aq) + HSO3- (aq)

If [H2SO3]0 = 1.5 M and the other concentrations are 0 M, find the equilibrium concentrations of all species at 25º C if K = 1.20 x 10-2.

• Since K is small, try considering x negligible in the reduction of [H2SO3]0.
Prob. 13-6 BK=1.20 x 10-2
• K = [H+]0 [HSO3-]0 = x2 = 1.20 x 10-2

[H2SO3] 1.50

• X = 0.134 M
• This is 9% of 1.50—cannot consider x negligible.
Prob. 13-6 BK=1.20 x 10-2
• K = [H+]0 [HSO3-]0 = x2 = 1.20 x 10-2

[H2SO3] 1.50 - x

• x2 + 0.0120x – 0.0180 = 0
• X = 0.128 M
• [H+] = [HSO3-] = 0.128 M
• [H2SO3] = 1.50 – 0.128 = 1.37 M
Le Châtelier’s Principle

N2(g) + O2(g)   2NO(g) ΔH = +1.81 kJ

Kp = 3.3 x 1030

What happens to the direction of equilibrium if the following changes were made?

• The container is made larger
• The system is cooled
N2(g) + O2(g)   2NO(g)
• N2 is added Q = [NO]2

[N2][O2]

This makes Q smaller. When Q<K, more product must be made. Shift to right.

Helium is inert and does not affect any partial pressures of these gases. No change to the equilibrium.

N2(g) + O2(g)   2NO(g)
• The container is made larger:

This favors the side with more gas molecules because they can exist with fewer collisions per unit time. Both sides have same number of molecules at equilibrium, so no change occurs in equilibrium.

• The system is cooled: This is an endothermic reaction which requires heat to go to right. Cooling shifts equilibrium to left.
2H2 (g) + O2 (g)   2H2O(g) ΔH= -484 kJ

Predict the effect of each of the following changes to the system on the direction of the equilibrium.

• H2O is removed as it is being generated.

Moves to right (Q < K if P removed).