11/30/2009
Quantitative
Changes in
Equilibrium
Systems
The Reaction Quotient, Q
• All ratios of product to reactant other than ratios equal
to K.
• Q is used to test a reaction mixture to determine if
equilibrium has been achieved.
aA + bB
Q
dD + eE
[D]d [E]e
[A]a [B]b
Chapter 7.5
PCl3(g) + Cl2(g) @ 250oC Kc= 4.0 x 10-2
Example: PCl5(g)
If: [Cl2] and [PCl3] = 0.30M and [PCl5] = 3.0M, is the system
at Equilibrium? If not, which direction will it proceed?
Find Q and compare to Keq to decide.
[PCl3 ][Cl2 ]
Q=
[PCl5 ]
[0.30][0.30]
= 3.0 x 10-2
[3.0]
Q < Keq therefore not at Equilibrium
Which way must the RXN go to achieve Equilibrium?
If Q = Keq ?
If Q < Keq ?
ratio
ratio
If Q > Keq ?
ratio
Prod.
React.
Prod.
React.
Prod.
React.
Remember ratio is
is equal, so RXN @ Equilibrium
[Products]
[Reactants]
more products makes the number bigger
is too small, so RXN
RXN goes
is too large, so RXN
Calculations Using Keq: (1st case....Perfect Square)
Comparing Q to K
• At 2000K the equilibrium constant K for the
formation of NO
• N2(g) + O2(g)
2 NO(g)
• is 4.0 x 10-4. If the reaction vessel is sampled and
• [N2] = 0.50, [O2] = 0.25, [NO]=4.2 x 10-3M
• Has the reaction reached equilibrium?
@ 699K H2(g) + I2(g)
2HI(g)
Keq = 55.17
Experiment: 1.00 mol of each H2 and I2 in a 0.500 L flask.
Find [ ] of products and reactants @ Equilibrium.
C.
H2(g) + I2(g)
[ ]
[ ]
2.00
2.00
-x
-x
E.
2.00 -x
conc. in mol/L
K eq
K eq
I.
[HI]2
[H 2 ][I 2 ]
[2x]2
[2.00-x][2.00-x]
2.00 - x
2HI(g)
[ ]
0
2x
2x
[2x]2
= 55.17
[2.00- x]2
“perfect square”
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11/30/2009
Calculations Using Keq: (1st case....Perfect Square--continued)
@ 699K H2(g) + I2(g)
K eq
[HI]2
[H 2 ][I 2 ]
E.
[2x]2
[2.00-x][2.00-x]
K eq
2HI(g)
Keq = 55.17
H2(g) + I2(g)
[ ]
[ ]
2.00 -x 2.00 - x
Calculations Using Keq: (1st case....Perfect Square--continued)
@ 699K H2(g) + I2(g)
2HI(g)
[ ]
2x
[HI]2
[H 2 ][I 2 ]
K eq
E.
2HI(g)
[ ]
2x
7.428(2.00 - x) = 2x
1.58 = x
[2x]2
= 55.17
[2.00- x]2
“perfect square”
[2x]
55.17
[2.00- x]
[H2] = 2.00 - 1.58 = 0.42M
[I2] = 0.42M
7.428(2.00 - x) = 2x
1.58 = x
[HI] = 2(1.58) = 3.16M
Keq Problems with Quadratic Equation
ax2
Keq = 55.17
2HI(g)
H2(g) + I2(g)
[ ]
[ ]
2.00 -x 2.00 - x
Problem:
H2(g) + I2(g)
2HI(g) @ 458 oC Keq = 49.7
Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask.
+ bx +c = 0
Find: conc. of the equilibrium mixture.
If equilibrium expression is not a perfect square must use
quadratic equation.
2
b
x
K eq
b
2a
4ac
[HI]2
[H 2 ][I 2 ]
I.
K
C.
H2(g) + I2(g)
[ ]
[ ]
2.00
1.00
-x
-x
1.00 - x
E.
2.00 - x
2HI(g)
[ ]
0
2x
2x
2
K eq
49.7
[2x]
[1.00-x][2.00-x]
0.920x2 - 3.00x + 2.00 = 0
Problem:
H2(g) + I2(g)
2HI(g) @ 458 oC Keq = 49.7
Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask.
For the reaction A B
Find: conc. of the equilibrium mixture.
2
K eq
K eq
49.7
[HI]
[H 2 ][I 2 ]
I.
C.
[2x]2
[1.00-x][2.00-x]
E.
– K = [B]eq/[A]eq where all concentrations are obtained at
H2(g) + I2(g)
[ ]
[ ]
2.00
1.00
-x
-x
1.00 - x
2.00 - x
2HI(g)
[ ]
0
2x
2x
0.920x2 - 3.00x + 2.00 = 0
2 solutions for x: 1.63
x = 2.33 or 0.93
Working With K Values
0.70
will give positive solution for Eq. Conc.
equilibrium
A
+B
B
C
– When adding reactions, the K value for the reaction sum
will be K1 x K2
For the reactions A
B vs 2A 2B n = 2
– If a reaction is multiplied by a number, n the new K
value will be Kn
For the reaction A B vs B
A
– Kforward = 1/Kreverse
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11/30/2009
Steps For Solving
Equilibrium Problems
1) Write the balanced equation.
2) Convert all amounts given to either M (molarity) or
pressure
3) Construct a Reaction Table (ICE Table)
4) Substitute numbers into equilibrium expression and
solve for x
5) Initially, neglect +/- x values.
6) If x is > 5% of the number from which it is being
subtracted, or to which it is being added, use the
quadratic equation to solve for x.
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