equilibrium calculations
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Equilibrium calculations. To view as a slide show, Click on the “Slideshow” on the menu bar, and then on “View Show” Click the mouse or press Page Down to go to the next slide Press the Escape key to leave the presentation Press Page Up to go back. Problem type #1. - PowerPoint PPT PresentationTRANSCRIPT
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 . . . . . . .