1

Applications of Equilibrium ConstantsApplications of Equilibrium ConstantsPredicting the Direction of ReactionPredicting the Direction of Reaction• We define Q, the reaction quotient, for a reaction at

conditions NOT at equilibrium

as

where [A], [B], [P], and [Q] are molarities at any time.• Q = K only at equilibrium.

aA + bB(g) pP + qQ

ba

qpQ

BA

QP

Applications of Equilibrium ConstantsApplications of Equilibrium Constants

Predicting the Direction of ReactionPredicting the Direction of ReactionIF

• Qc > Kc system proceeds from right to left to reach equilibrium

• Qc = Kc the system is at equilibrium

• Qc < Kc system proceeds from left to right to reach equilibrium

14.4

3

Applications of Equilibrium ConstantsApplications of Equilibrium ConstantsPredicting the Direction of ReactionPredicting the Direction of Reaction

• If Q > K then the reverse reaction must occur to reach equilibrium (go left)

• If Q < K then the forward reaction must occur to reach equilibrium (go right)

Calculating Equilibrium Concentrations

1. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.

2. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.

3. Having solved for x, calculate the equilibrium concentrations of all species.

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Calculating Equilibrium ConstantsCalculating Equilibrium Constants• Steps to Solving Problems:

1. Write an equilibrium expression for the balanced reaction.

2. Write an ICE table. Express the equilibrium concentrations of all species in terms of the initial concentrations.

3. Use stoichiometry (mole ratios) to express change in concentration with respect to the unknown x on the change in concentration line.

4. Solve for x and calculate the equilibrium concentrations of all species.

Usually, the initial concentration of products is zero. (This is not always the case.)

At 12800C the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)

Br2 (g) 2Br (g)

Let x be the change in concentration of Br2

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

[Br]2

[Br2]Kc = Kc =

(0.012 + 2x)2

0.063 - x= 1.1 x 10-3 Solve for x

14.4

Kc = (0.012 + 2x)2

0.063 - x= 1.1 x 10-3

4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x

4x2 + 0.0491x + 0.0000747 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

Br2 (g) 2Br (g)

Initial (M)

Change (M)

Equilibrium (M)

0.063 0.012

-x +2x

0.063 - x 0.012 + 2x

x = -0.00178x = =.00015

At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M

At equilibrium, [Br2] = 0.062 – x = 0.0648 M

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Example Problem: Calculate Keq

This type of problem is typically tackled using the “three line” approach:2 NO + O2 2 NO2

Initial:

Change:

Equilibrium:

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Example Problem: Calculate Concentration

Note the moles into a 10.32 L vessel stuff ... calculate molarity.Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M

2 HI H2 + I2

222

][

]][[

HI

IHKeq

Initial:Change:Equil:

0.242 M 0 0

-2x +x +x

0.242-2x x x

32

2

21026.1

]2242.0[]2242.0[

]][[

xx

x

x

xxKeq

What we are asked for here is the equilibrium concentration of H2 ... ... otherwise known as x. So, we need to solve this beast for x.

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Example Problem: Calculate Concentration

32

2

1026.1]2242.0[

xx

x

232 ]2242.0[1026.1 xxx

]4968.00586.0[1026.1 23 xxx

2335 1004.51022.11038.7 xxxxx

01038.71022.1995.0 532 xxxx

And yes, it’s a quadratic equation. Doing a bit of rearranging:

a

acbbx

2

42

x = 0.00802 or –0.00925Since we are using this to model a real, physical system,we reject the negative root.The [H2] at equil. is 0.00802 M.

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Approximating

If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants.

0.20 – x is just about 0.20 is x is really dinky.

If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

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Example

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 MI2 2 I

Initialchangeequil:

0.20 0-x +2x0.20-x 2x

102

10

2

2

1094.2]20.0[

]2[

1094.2][

][

xx

x

xI

IKeq

With an equilibrium constant that small, whatever x is, it’s neardink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars).

0.20 – x is the same as 0.20

102

1094.220.0

]2[ xx

x = 3.83 x 10-6 M

More than 3orders of mag.between thesenumbers. The simplification willwork here.

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Example

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 MI2 2 I

Initialchangeequil:

0.20 0-x +2x0.20-x 2x

209.0]20.0[

]2[

209.0][

][

2

2

2

x

x

I

IKeq These are too close to

each other ... 0.20-x will not betrivially close to 0.20here.

Looks like this one has to proceed through the quadratic ...