equilibrium calculations
DESCRIPTION
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
![Page 1: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/1.jpg)
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
![Page 2: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/2.jpg)
Problem type #1
• Given some information about the conditions before equilibrium and after equilibrium, find the equilibrium constant.
![Page 3: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/3.jpg)
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
![Page 4: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/4.jpg)
N2 + 3H2 2NH3
N2 H2 NH3
Initial 0 0 0.500 M
Change
At equilibrium
0.106 M
![Page 5: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/5.jpg)
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
![Page 6: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/6.jpg)
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
![Page 7: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/7.jpg)
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
![Page 8: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/8.jpg)
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
![Page 9: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/9.jpg)
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
![Page 10: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/10.jpg)
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
![Page 11: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/11.jpg)
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
![Page 12: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/12.jpg)
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
![Page 13: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/13.jpg)
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
![Page 14: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/14.jpg)
KN H
N H
[ ]
[ ][ ]
32
2 23
S o K ( . )
( . )( . ).
0 106
0 197 0 5910 276
2
3
![Page 15: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/15.jpg)
Problem type #2a
• Given information about conditions before equilibrium and the equilibrium constant, find the concentrations when equilibrium is reached.
![Page 16: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/16.jpg)
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
? ? ?
![Page 17: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/17.jpg)
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
![Page 18: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/18.jpg)
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
![Page 19: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/19.jpg)
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
![Page 20: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/20.jpg)
![Page 21: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/21.jpg)
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
![Page 22: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/22.jpg)
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
![Page 23: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/23.jpg)
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
? ?
![Page 24: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/24.jpg)
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
![Page 25: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/25.jpg)
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
![Page 26: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/26.jpg)
KF
F
x
x
[ ]
[ ]
( )
( . ).
2
2
232
1 02 7 10
F2 2F
At equilibrium
1.0 – x 2x
![Page 27: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/27.jpg)
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
![Page 28: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/28.jpg)
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
![Page 29: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/29.jpg)
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
![Page 30: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/30.jpg)
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. ( . ) ( )( . )
( )
![Page 31: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/31.jpg)
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 )
![Page 32: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/32.jpg)
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
![Page 33: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/33.jpg)
Don’t memorize. . . .
![Page 34: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/34.jpg)
Don’t memorize
• UNDERSTAND
![Page 35: Equilibrium calculations](https://reader035.vdocuments.us/reader035/viewer/2022062422/56813a4b550346895da24312/html5/thumbnails/35.jpg)
Problem type 2c
• Maybe next time . . . . . . .