Equilibrium calculations

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Problem type #1

• Given some information about the conditions before equilibrium and after equilibrium, find the equilibrium constant.

Suppose a flask was made to be 0.500 M in NH3 initially, and when equilibrium was

reached, the [NH3] had dropped to 0.106 M. Find the value of K for:

N2 + 3H2 2NH3

Initially,

S o reaction m ust shift to the left and form

reactants

QN H

N HK

[ ]

[ ][ ]

(. )!3

2

2 23

2500

0

N2 + 3H2 2NH3

N2 H2 NH3

Initial 0 0 0.500 M

Change

At equilibrium

0.106 M

N2 + 3H2 2NH3

N2 H2 NH3

Initial 0 0 0.500 M

Change 0.106 – 0.500 M

= -0.394 M

At equilibrium

0.106 M

N2 + 3H2 2NH3

N2 H2 NH3

Initial 0 0 0.500 M

Change +

0.394/2 M

0.106 – 0.500 M

= -0.394 M

At equilibrium

0.106 M

N2 + 3H2 2NH3

N2 H2 NH3

Initial 0 0 0.500 M

Change +

0.394/2 M

0.106 – 0.500 M

= -0.394 M

At equilibrium

0 + .197 M

= .197 M

0.106 M

N2 + 3H2 2NH3

N2 H2 NH3

Initial 0 0 0.500 M

Change +

0.394/2 M

+

3(.394/2) M

0.106 – 0.500 M

= -0.394 M

At equilibrium

0 + .197 M

= .197 M

0.106 M

N2 + 3H2 2NH3

N2 H2 NH3

Initial 0 0 0.500 M

Change +

0.394/2 M

+

3(.394/2) M

0.106 – 0.500 M

= -0.394 M

At equilibrium

0 + .197 M

= .197 M

0 + .591 M 0.106 M

N2 + 3H2 2NH3

N2 H2 NH3

Initial 0 0 0.500 M

Change +

0.394/2 M

+

3(.394/2) M

0.106 – 0.500M

= -0.394 M

At equilibrium

0 + .197 M

= .197 M

0 + .591 M

= .591 M

0.106 M

N2 + 3H2 2NH3

• Let 2x be the amount of NH3 that reacts

• Use stoichiometry of reaction!

N2 H2 NH3

Initial 0 0 0.500 M

Change +x +3x -2x

At equilibrium

N2 + 3H2 2NH3

• Let 2x be the amount of NH3 that reacts

• 2x = 0.500 – 0.106 = 0.394

N2 H2 NH3

Initial 0 0 0.500 M

Change +x +3x -2x

At equilibrium

0+x 0+3x 0.500 – 2x

= 0.106 M

N2 + 3H2 2NH3

• Let 2x be the amount of NH3 that reacts

• 2x = 0.500 – 0.106 = 0.394

N2 H2 NH3

Initial 0 0 0.500 M

Change +x +3x -2x

At equilibrium 0+x =

0.394/2 =

0.197 M

0+3x =

0.197 x 3 =

0.591 M

0.500 – 2x

= 0.106 M

KN H

N H

[ ]

[ ][ ]

32

2 23

S o K ( . )

( . )( . ).

0 106

0 197 0 5910 276

2

3

Problem type #2a

• Given information about conditions before equilibrium and the equilibrium constant, find the concentrations when equilibrium is reached.

For the reaction N2 + O2 2NO at 1700oC, K = 3.52x10-4. If 0.0100 mole NO is placed in a 1 – L flask at 1700o, what will [N2] be at

equilibrium?

N2 O2 2NO

Initial 0 0 0.0100 M

Change

At equilibrium

? ? ?

For the reaction N2 + O2 2NO at 1700oC, K = 3.52x10-4. If 0.0100 mole NO is placed in a 1 – L flask at 1700o, what will [N2] be at

equilibrium?

N2 O2 2NO

Initial 0 0 0.0100

Change +x +x -2x

At equilibrium

For the reaction N2 + O2 2NO at 1700oC, K = 3.52x10-4. If 0.0100 mole NO is placed in a 1 – L flask at 1700o, what will [N2] be at

equilibrium?

N2 O2 2NO

Initial 0 0 0.0100

Change +x +x -2x

At equilibrium

x x 0.0100-2x

KN O

N O

x

x x

x

x

[ ]

[ ][ ]

( . )

( )( ).

.. .

2

2 2

24

4 2

0 0100 23 52 10

0 0100 23 52 10 1 88 10

T ake square root of both sides:

N2 O2 2NO

At equilibrium

x x 0.0100-2x

0 0100 23 52 10 1 88 104 2.. .

x

x

• 0.0100 - 2x = (1.88 x 10-2)x• 0.0100 = 2.02 x

• x = 4.95 x 10-3 M = [N2] (also = [O2])

Note that because K was small, most of the NO became N2 and O2

Final [NO] = 0.0100 – 2(4.95 x 10-3) =1.00 x 10-4 M

Problem type #2b

• Given information about conditions before equilibrium and the equilibrium constant, find the concentrations when equilibrium is reached.

• . . . .But the math doesn’t work out as nicely

For the reaction F2 2F at 1000oC, K = 2.7 x10-3. If 1.0 mole F2 is placed in a 1 – L

flask at 1000o, what will [F] be at equilibrium?

F2 2F

Initial 1.0 M 0 M

Change

At equilibrium

? ?

For the reaction F2 2F at 1000oC, K = 2.7 x10-3. If 1.0 mole F2 is placed in a 1 – L

flask at 1000o, what will [F] be at equilibrium?

F2 2F

Initial 1.0 M 0 M

Change - x + 2x

At equilibrium

For the reaction F2 2F at 1000oC, K = 2.7 x10-3. If 1.0 mole F2 is placed in a 1 – L

flask at 1000o, what will [F] be at equilibrium?

F2 2F

Initial 1.0 M 0 M

Change - x + 2x

At equilibrium

1.0 – x 2x

KF

F

x

x

[ ]

[ ]

( )

( . ).

2

2

232

1 02 7 10

F2 2F

At equilibrium

1.0 – x 2x

KF

F

x

x

[ ]

[ ]

( )

( . ).

2

2

232

1 02 7 10

• 4x2 = 2.7 x 10-3(1.0 – x) =

• 4x2 = 2.7 x 10-3 – 2.7 x 10-3 x

KF

F

x

x

[ ]

[ ]

( )

( . ).

2

2

232

1 02 7 10

• 4x2 = 2.7 x 10-3(1.0 – x) =

• 4x2 = 2.7 x 10-3 – 2.7 x 10-3 x

• This is a quadratic equation

• Rearrange to the form ax2 + bx + c = 0

4x2 + 2.7x10-3x – 2.7 x 10-3 = 0

• a = 4 b = 2.7 x 10-3 c = -2.7 x 10-3

For ax + bx + c = 0 ,2

x = -b b ac

a

2 4

2

4x2 + 2.7x10-3x – 2.7 x 10-3 = 0• a = 4 b = 2.7 x 10-3 c = -2.7 x 10-3

For ax + bx + c = 0 ,2

x = -b b ac

a

2 4

2

4x2 + 2.7x10-3x – 2.7 x 10-3 = 0• a = 4 b = 2.7 x 10-3 c = -2.7 x 10-3

x = - 2 7 10 2 7 10 4 4 2 7 10

2 4

3 3 2 3. ( . ) ( )( . )

( )

x = - 2 7 10 2 7 10 4 4 2 7 10

2 4

3 3 2 3. ( . ) ( )( . )

( )

x 0 . 0256 M o r

x 0 . 211 M

O n ly o ne r esu l t i s phy si cal l y po ssib le!

(R ejec t negativ e co ncentr atio n )

We found x = 0.0256 M

F2 2F

Initial 1.0 M 0 M

Change - x + 2x

At equilibrium

1.0 – x = 0.974 M

2x = 0.051 M

Don’t memorize. . . .

Don’t memorize

• UNDERSTAND

Problem type 2c

• Maybe next time . . . . . . .