![RADIATIVE -CONVECTIVE EQUILIBRIUM CALCULATIONS FOR A TWO-LAYER MARS ATMOSPHERE · 2017-06-26 · NASr-21(07] MEMORANDUM RM-5017-NASA MAP 1900 RADIATIVE -CONVECTIVE EQUILIBRIUM CALCULATIONS](https://static.fdocuments.in/doc/165x107/5e87f0affe2db0256f08a922/radiative-convective-equilibrium-calculations-for-a-two-layer-mars-atmosphere-2017-06-26.jpg)
Languages
Pages
Legal
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
• 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 . . . . . . .
Top Related