Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
Solving Equilibrium Problems
♦
♦
A reaction (ICE) table shows
−
−
−
−
116_10_PSV_equil_01
In a study of carbon oxidation, an evacuated vessel containing a small amount of powdered graphite (a
form of solid carbon) is heated to 1080 K. Gaseous CO2 is added to a pressure of 0.458 atm and CO
forms. At equilibrium, the total pressure is 0.757 atm. Calculate Kp.
Kp= 2.25
116_10_PSV_equil_02
In order to study hydrogen halide decomposition, a researcher fills an evacuated 2.00-L flask with
0.200 mol of HI gas and allows the reaction to proceed at 453°C
2 HI(g) ←→ H2 (g) + I2(g)
At equilibrium, [HI] = 0.078 M. Calculate Kc.
Kc=0.020
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Page 1 of 8
Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
116_10_PSV_equil_03
In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous CH4 and
H2O in a 0.32-L flask at 1200 K. At equilibrium the flask contains 0.26 mol of CO, 0.091 mol of H2,
and 0.041 mol of CH4. What is the [H2O] at equilibrium? Kc = 0.26 for this process at 1200 K
CH4(g) + H2O(g) ←→ CO(g) + 3H2(g)
[H2O]eq = 0.56 M
116_10_PSV_equil_04
Fuel engineers use the extent of the change from CO and H2O to CO2 and H2 to regulate the
proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.250 mol of H2O are placed in a 125mL flask at 900 K, what is the composition of the equilibrium mixture? At this temperature, Kc is 1.56
CO(g) + H2O(g) ←→ CO2(g) + H2(g)
[CO]eq=[H2O]eq= 0.89 M
[CO2]eq= [H2]eq = x = 1.11 M
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Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
Concept Check
The following pictures represent reaction mixtures that contain A2 molecules (shaded) and B2
molecules (unshaded), and AB molecules. Which reaction mixture is at equilibrium?
A2 + B2 ←→ 2 AB
Kc = 1.8
116_10_PSV_equil_05
The research and development unit of a chemical company is studying the reaction of CH4 and H2S,
two components of natural gas as shown below. In one experiment, 1.00 mol of CH4, 1.00 mol of CS2,
2.00 mol of H2S, and 2.00 mol of H2 are mixed in a 250-mL vessel at 960°C.
CH4(g) + 2H2S(g) ←→ CS2(g) + 4H2(g)
At this temperature, Kc = 0.036
If [CH4] = 5.56 M at equilibrium, what are the equilibrium concentrations of the other substances?
[H2S]eq = 11.1 M
[CS2]eq = 2.44 M
[H2]eq = 1.76 M
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Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
The Method of Successive Approximations (one variable)
♦
assume an approximate value for the variable that will simplify the equation
− for most instances in Chem 116, an initial guess of x0=0 is the best place to start, if this
does not yield a reasonable answer, you may want to switch to the quadratic equation
♦
solve for the variable
♦
use the answer as the second approximate value and solve the equation again
♦
repeat this process until a results converge to a single value
Solve for x using the method of successive approximations: 8.4 × 10 −5 =
x2
0.200 - x
x = 0.0041
♦ This method will work with most polynomials.
♦ In this example, a consistent value has been obtained after making only two approximations. A
consistent value is often obtained in less than 3 to 4 successive approximations. With the aid of
a calculator, the method of successive approximations can be done quickly.
Solve for x using the method of successive approximations; 7.3 × 10 −3 =
x2
0.060 - x
x = 0.018
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Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
116_10_PSV_equil_06
Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It
decomposes by the reaction shown below. Calculate [CO], [Cl2], and [COCl2] when the 0.100 moles of
phosgene gas decompose and reach equilibrium in a 10.0-L flask.
COCl2(g) ←→ CO(g) + Cl2(g)
Kc = 8.3×10−4 at 360°C
[COCl2]eq = 0.0075 M
[CO]eq = [Cl2]eq = x= 0.0025 M
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Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
116_10_PSV_equil_07
Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It
decomposes by the reaction shown below. Calculate [CO], [Cl2], and [COCl2] when the 5.00 moles of
phosgene gas decompose and reach equilibrium in a 10.0-L flask.
COCl2(g) ←→ CO(g) + Cl2(g)
Kc = 8.3×10−4 at 360°C
[COCl2] = 0.480 M
[CO]=[Cl2] = x = 0.020 M
The Simplifying Assumption
[A]eq = [A]initial − [A]reacting
We try assumption that
or [A]eq = [A]initial − x
[A]initial − [A]reacting ≅ [A]initial this is justified if…..
− Kc is relatively small and/or
− [A]initial is relatively large
o General rule
Spring 2016 (Ratcliff)
If [A]initial > K×1000 then the simplifying assumption is OK
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Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
Steps in solving equilibrium problems
Preliminary Construct the ICE table
♦ Write the balanced equation.
♦ Write the reaction quotient, Q (or K as appropriate)
♦ Convert all amounts into the correct units (M or atm)
♦ Construct a reaction table
Solving for x and equilibrium quantities
♦ Substitute the quantities into Q (or K as appropriate)
♦ Solve for x. (The tricky part)
− Look for perfect square
− Method of Successive approximation
− Quadratic equation
− Simplifying assumption (only if justified)
♦ Find the equilibrium quantities (plug x into reaction table)
Sanity check
Extra Practice problems
116_10_PSV_equil_08
Consider the theoretical equilibrium shown below. Given that the equilibrium concentration of B is
2.20 M, determine the numerical value of KC. Hint: create an ICE table
initial
2 A(aq)
+ B(aq) ←→
3C
1.00 M
2.00 M
5.00 M
Kc = ?
change
equil
KC = 19.8
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Chem 116 Lecture Note Outlines
Set 10: Equilibrium Problems
116_10_PSV_equil_09
Consider the theoretical equilibrium shown below. What is the equilibrium concentration of O2(g), if
initially only 0.300 M NO(g) was present?
N2(g) + O2(g)
←→
2 NO(g)
KC = 81.0
[O2]eq = 0.0273 M
116_10_PSV_equil_10
Consider the the reaction shown below. What is the equilibrium partial pressure of SO2 if the
equilibrium partial pressures of O2 and SO3 are 0.43 atm and 1.56 atm respectively
2 SO2(g)
+ O2(g)(g) ←→
2 SO3(g)
KP = 48.2
PSO2 = 0.34 atm
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Page 8 of 8