1 “ice” calculations. 2 the equilibrium constant a a + b b ---> c c + d d (at a given t)....
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“ICE” CALCULATIONS
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THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTa A + b B ---> c C + d D
(at a given T)
K =[C]c [D]d
[A]a [B]b
conc. of products
conc. of reactantsequilibrium constant
Pure substances are not included in equation..
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H+ (aq)+ OH- (aq) H2O(l)
WRITE THE EQUILIBRIUM CONSTANT
EXPRESSION
CaO (s) + CO2 (g) CaCO3 (s)
K =
K =
4Typical CalculationsPROBLEM:
Place 1.00 mol each of H2 and I2 in a 1.0 L flask. Calc. equilibrium concentrations.
Kc = [HI]2
[H2 ][I2 ] = 55.3Kc =
[HI]2
[H2 ][I2 ] = 55.3
H2(g) + I2(g) 2 HI(g)
X X 2 X
Q = 0/12 = 0
- - +
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H2(g) + I2(g) 2 HI(g)Kc = 55.3
H2(g) + I2(g) 2 HI(g)Kc = 55.3
Step 1. Set up ICE table
[H2] [I2] [HI]
Initial 1.00 1.00 0
Change
Equilib
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[H2] [I2] [HI]
Initial 1.00 1.00 0
Change -x -x +2x
Equilib1.00-x 1.00-x 2x
where x is defined as am’t of H2 and
I2 consumed on approaching
equilibrium.
Step 2. define “X”
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HH22(g) + I(g) + I22(g) (g) ee 2 HI(g) 2 HI(g)KKcc = 55.3 = 55.3
Step 3. Put equilibrium concentrations into Kc expression.
Kc = [2x]2
[1.00 - x][1.00 - x] = 55.3Kc =
[2x]2
[1.00 - x][1.00 - x] = 55.3
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[H2] = [I2] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M
[H2] = [I2] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M
Step 4. Solve Kc expression - take square
root of both sides.
x = 0.79Therefore, at equilibrium
HH22(g) + I(g) + I22(g) (g) --->---> 2 HI(g) K 2 HI(g) Kcc = 55.3 = 55.3
Kc = [2x]2
[1.00 - x][1.00 - x] = 55.3Kc =
[2x]2
[1.00 - x][1.00 - x] = 55.3 7.44 =
2x1.00 - x
7.44 = 2x
1.00 - x
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Pg 132 CPSC
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Nitrogen Dioxide Equilibrium
N2O4(g) ---> 2 NO2(g)
Nitrogen Dioxide Equilibrium
N2O4(g) ---> 2 NO2(g)
X 2X
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Nitrogen Dioxide EquilibriumN2O4(g) ---> 2 NO2(g)
Nitrogen Dioxide EquilibriumN2O4(g) ---> 2 NO2(g)
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
Step 1. Set up an ICE table
[N2O4] [NO2]
Initial 0.50 0
Change
Equilib
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 KKc =
[NO2 ]2
[N2O4 ] = 0.0059 at 298 K
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Step 2. Define “X”
[N2O4] [NO2]
Initial 0.50 0
Change -x +2x
Equilib 0.50 - x 2x
Nitrogen Dioxide EquilibriumN2O4(g) ---> 2 NO2(g)
Nitrogen Dioxide EquilibriumN2O4(g) ---> 2 NO2(g)
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Step 2. Define “X”
[N2O4] [NO2]
Initial 0.50 0
Change -x +2x
Equilib 0.50 2x
APPROXIMATE: K is small: x will be small
APPROXIMATE: K is small: x will be small
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 KKc =
[NO2 ]2
[N2O4 ] = 0.0059 at 298 K
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Step 3. Substitute into Kc expression and solve.
Kc = 0.0059 = [NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x) Kc = 0.0059 =
[NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x)
Rearrange: 0.0059 (0.50) = 4x2
0.0029 = 4x2
0.0029/4 = x2
X = 0.027Step 4. check if assumption is correct
0.50 – 0.027 = 0.47 same?
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Step 3. Substitute into Kc expression and solve.
Kc = 0.0059 = [NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x) Kc = 0.0059 =
[NO2 ]2
[N2O4 ]=
(2x)2
(0.50 - x)
Rearrange: 0.0059 (0.50 - x) = 4x2
0.0029 - 0.0059x = 4x2
4x2 + 0.0059x - 0.0029 = 0
This is a QUADRATIC EQUATION ax2 + bx + c = 0
a = 4 b = 0.0059 c = -0.0029
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Solve the quadratic equation for x.
ax2 + bx + c = 0
a = 4 b = 0.0059 c = -0.0029
x = -b b2 - 4ac
2ax =
-b b2 - 4ac2a
x = -0.0059 (0.0059)2 - 4(4)(-0.0029)
2(4)x =
-0.0059 (0.0059)2 - 4(4)(-0.0029)2(4)
x = -0.00074 ± 1/8(0.046)x = -0.00074 ± 1/8(0.046)1/21/2 = = -0.00074 ± 0.027-0.00074 ± 0.027
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Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) --->---> 2 NO 2 NO22(g)(g)
Nitrogen Dioxide EquilibriumNitrogen Dioxide EquilibriumNN22OO44(g) (g) --->---> 2 NO 2 NO22(g)(g)
x = 0.026 or -0.028
negative value is not reasonable.
Conclusion: x = 0.026 M [N2O4] = 0.050 - x = 0.47 M
[NO2] = 2x = 0.052 M
x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027
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Solving Quadratic Solving Quadratic EquationsEquations
• Recommend you solve the
equation exactly on a calculator