In the previous post, we talked about ICE tables which we use for determining the equilibrium concentrations based on the equilibrium constant, reaction quotient, and the initial concentrations of the reactants and products.
Below are some additional practice examples on the concept of ICE tables and equilibrium concentrations.
Practice
The equilibrium constant for the following reaction at 600 ^{o}C is determined to be Kc = 0.495:
H_{2}O(g) + CO(g) ⇆ H_{2}(g) + CO_{2}(g)
Calculate the number of H_{2 }moles that are present at equilibrium if a mixture of 0.400 mole CO and 0.500 mole H_{2}O is heated to 600°C in a 10.0-L container.
0.184 mol H_{2}
The concentrations of the gases are going to be moles over the container volume (10.0 L). So, [CO] = 0.0400 mol/L and [H_{2}O] = 0.0500 mol/L.
Let’s set up an ICE table assigning x mol/L as the depletion in the concentration of either CO or H_{2}O. The increase of H_{2 }and CO_{2 }concentration is going to be x which is their equilibrium concentration. The equilibrium concentration of CO and H_{2}O is going to be 0.0400-x and 0.0500-x respectively.
H_{2}O(g) + CO(g) ⇆ H_{2}(g) + CO_{2}(g)
[H_{2}O] | [CO] | [H_{2}] | [CO_{2}] | |
Initial | 0.0500 | 0.0400 | 0 | 0 |
Change | -x | -x | +x | +x |
Equil | 0.0500-x | 0.0400-x | x | x |
The equilibrium constat is:
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left[ {{{\rm{H}}_{\rm{2}}}} \right]\left[ {{\rm{C}}{{\rm{O}}_{\rm{2}}}} \right]}}{{\left[ {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right]\left[ {{\rm{CO}}} \right]}}\; = \,0.495\]
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left( {\rm{x}} \right)\left( {\rm{x}} \right)}}{{\left( {{\rm{0}}{\rm{.0500 – x}}} \right)\left( {{\rm{0}}{\rm{.0400 – x}}} \right)}}\;{\rm{ = }}\,{\rm{0}}{\rm{.495}}\]
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{\left( {{\rm{0}}.{\rm{0500 – x}}} \right)\left( {{\rm{0}}.{\rm{0400 – x}}} \right)}}\;{\rm{ = }}{\mkern 1mu} {\rm{0}}.{\rm{495}}\]
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\rm{x}}^{\rm{2}}}}}{{{{\rm{x}}^{\rm{2}}}\; – \;0.09x\; + \;0.002}}\;{\rm{ = }}{\mkern 1mu} {\rm{0}}.{\rm{495}}\]
0.505x^{2} + 0.04455x – 0.00099 = 0
\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]
a = 0.505, b = 0.04455, c = -0.00099
x = 0.0184 mol/L
The last step is to convert the concentration to the number of moles by multiplying by 10 since the reaction is carried out in a 10.0 L container.
n (H_{2}) = 0.0184 mol/L x 10.0 L = 0.184 mol
The equilibrium constant Kc for the following reaction at 800°C is 3.74 x 10^{5}
H_{2}(g) + I_{2}(g) ⇆ 2HI(g)
If 6.25 moles of HI were initially added to a 15.0-L empty vessel, what would the concentrations of H_{2}, I_{2}, and HI be at equilibrium.
[H_{2}] = [I_{2}] = 6.8 x 10^{-4 }mol/L
[HI] = 0.416 mol/L
First, determine the initial concentration of HI.
M (HI) = 6.25 mol ÷ 15.0 L = 0.417 mol/L
Then, set up the ICE table, assuming x mol/L of HI decomposed to form H_{2 }and I_{2. }From the stoichiometric ratio, we know that 0.5x mol/l of H_{2 }and I_{2 }will be formed:
H_{2}(g) + I_{2}(g) ⇆ 2HI(g)
[H_{2}] | [I_{2}] | [HI] | |
Initial | 0 | 0 | 0.417 |
Change | +0.5x | +0.5x | -x |
Equil | 0.5x | 0.5x | 0.417-x |
The equilibrium constant is:
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left[ {{\rm{HI}}} \right]}^2}}}{{\left[ {{{\rm{H}}_{\rm{2}}}} \right]\left[ {{{\rm{I}}_{\rm{2}}}} \right]}}\; = \,3.74{\rm{ }}x{\rm{ }}{10^5}\]
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}^2}}}{{\left( {{\rm{0}}{\rm{.5x}}} \right)\left( {{\rm{0}}{\rm{.5x}}} \right)}}\; = \,3.74{\rm{ }} \times {\rm{ }}{10^5}\]
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}^{\rm{2}}}}}{{{\rm{0}}{\rm{.25}}{{\rm{x}}^{\rm{2}}}}}\;{\rm{ = }}\,{\rm{3}}{\rm{.74 \times 1}}{{\rm{0}}^{\rm{5}}}\]
Taking the square root of both sides we obtain:
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left( {{\rm{0}}{\rm{.417 – x}}} \right)}}{{{\rm{0}}{\rm{.5x}}}}\;{\rm{ = }}\,{\rm{612}}\]
x = 0.00136
Therefore, the equilibrium concentrations are:
[H_{2}] = [I_{2}] = 0.5 x 0.00136 = 0.00068 = 6.8 x 10^{-4 }mol/L
[HI] = 0.417 – 0.00136 = 0.416 mol/L
Check:
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{0.416}^2}}}{{{{0.00068}^2}}}\; = \,3.74{\rm{ }} \times {\rm{ }}{10^5}\]
✔
The equilibrium constant for the following reaction at 700 K is Kp = 6.7 x 10^{-3}
CO(g) + 2H_{2}(g) ⇆ CH_{3}OH(g)
A reaction mixture contains 0.248 atm of H_{2}, 0.085 atm of CO, and 0.598 atm of CH_{3}OH. Is the reaction mixture at equilibrium? If not, in what direction will the reaction proceed?
The reaction is not at equilibrium and will proceed to the left.
The equilibrium constant for this reaction is:
\[{K_p}\; = \;\frac{{{P_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}}}{{{P^2}_{{{\rm{H}}_{\rm{2}}}}{P_{{\rm{CO}}}}}}\; = \;6.7{\rm{ }} \times {\rm{ }}{10^{ – 3}}\]
Now, if the mixture is at equilibrium, then plugging the concentration values in this equation, we should get the value for Kp.
\[{K_p}\; = \;\frac{{0.598}}{{{{0.248}^2}0.085}}\; = \;114\; > \;6.7{\rm{ }}x{\rm{ }}{10^{ – 3}}\]
This is greater than Kp (6.7 x 10^{-3}), and therefore, the reaction will proceed to the left forming more CO and H_{2. }This can also be predicted when we compare the partial pressures of the gases. Despite a small equilibrium constant, CH_{3}OH has a higher partial pressure than CO and H_{2. }The system, cannot stay in this state, and thus some of the CH_{3}OH will break into CO and H_{2.}
For the reaction shown below, K_{c} = 0.654 at 600 K.
N_{2}O_{4}(g) ⇆ 2NO_{2}(g)
If initially, 0.0600 M of N_{2}O_{4} are present in the reaction vessel, what are the equilibrium concentrations of the gases at 600 K?
[N_{2}O_{4}] = 0.0133 mol/L
[NO_{2}] = 0.0934 mol/L
To set up the ICE table, we need to determine which gas will increase and which one will decrease. Because only N_{2}O_{4 }is initially present in the system, then its concentration is going to decrease, while the concentration of NO_{2 }will increase.
Let’s suppose x mol/L of N_{2}O_{4 }decomposes (reacts) to form 2x mol/L NO_{2. }This is based on the stoichiometric ratio of the gases – every one mole of N_{2}O_{4} produces 2 mol NO_{2}.
N_{2}O_{4}(g) ⇆ 2NO_{2}(g)
[N_{2}O_{4}] | [NO_{2}] | |
Initial | 0.0600 | 0 |
Change | -x | +2x |
Equil | 0.0600-x | 2x |
The equilibrium constant for this reaction is:
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{\left[ {{\rm{N}}{{\rm{O}}_{\rm{2}}}} \right]}^2}}}{{\left[ {{{\rm{N}}_{\rm{2}}}{{\rm{O}}_4}} \right]}}\; = \,0.654\]
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{{\rm{(2x)}}}^2}}}{{0.0600 – x}}\; = \,0.654\]
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{{{{\rm{4x}}}^2}}}{{0.0600 – x}}\; = \,0.654\]
4x^{2 }+ 0.654x – 0.003924 = 0
\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]
a = 4, b = 0.654, c = -0.003924
x=0.0467 or x=−0.210 (doesn’t work)
Therefore, the equilibrium concentrations are:
[N_{2}O_{4}] = 0.0600 – 0.0467 = 0.0133 mol/L
[NO_{2}] = 2 x 0.0467 = 0.0934 mol/L
For the reaction shown below, Kc = 255 at 800 K.
PCl_{3}(g) + Cl_{2}(g) ⇆ PCl_{5}(g)
If a reaction mixture initially contains 0.3500 M PCl_{3 }and 0.375 M Cl_{2} at 800 K, what are the equilibrium concentrations of all the species in the mixture?
[PCl_{3}] = 0.0253 mol/L
[Cl_{2}] = 0.0503 mol/L
[PCl_{5}] = 0.3247 mol/L
The equilibrium constant for this reaction is:
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\left[ {{\rm{PC}}{{\rm{l}}_{\rm{5}}}} \right]}}{{\left[ {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}\;{\rm{ = }}\,{\rm{255}}\]
To set up the ICE table, we need to determine which gas will increase and which one will decrease. Because no PCl_{5 }is present in the system, the reaction will proceed to the right, thus decreasing the concentration of PCl_{3 }and Cl_{2.}
Let’s suppose x mol/L of PCl_{3 }and Cl_{2} reacts to form x mol/L PCl_{5}_{. }This is based on the stoichiometric ratio of the gases – every one mole of PCl_{3 }and Cl_{2} produces one mol PCl_{5}.
PCl_{3}(g) + Cl_{2}(g) ⇆ PCl_{5}(g)
[PCl_{3}] | [Cl_{2}] | [PCl_{5}] | |
Initial | 0.3500 | 0.375 | 0 |
Change | -x | -x | +x |
Equil | 0.3500-x | 0.375-x | x |
Plugging the numbers, we get an equation based on the equilibrium constant:
\[{K_{\rm{c}}}\;{\rm{ = }}\;\frac{{\rm{x}}}{{\left( {{\rm{0}}{\rm{.3500 – x}}} \right)\left( {{\rm{0}}{\rm{.375 – x}}} \right)}}\;{\rm{ = }}\,{\rm{255}}\]
255x^{2 }– 185.875x + 33.46875 = 0
\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]
a = -255, b = 185.9, c = -33.47
x=0.324691 or x=0.404231
x=0.404231 doesn’t work becasue 0.3500-0.404231 gives a negative value for the concentration.
x=0.3247
Therefore, the equilibrium concentrations are:
[PCl_{3}] = 0.3500 – 0.3247 = 0.0253 mol/L
[Cl_{2}] = 0.375 – 0.3247 = 0.0503 mol/L
[PCl_{5}] = x=0.3247 mol/L
Consider the following reaction characterized with K_{p} = 2.34 x 10^{-4} at 250 K:
I_{2}(g) + Cl_{2}(g) ⇆ 2ICl(g)
A reaction mixture initially contains I_{2} with partial pressure of 655 torr and Cl_{2} with partial pressure of 864 torr at 250 K. Calculate the equilibrium partial pressure of ICl.
0.0141 atm = 11.4 torr
Remember, when converting Kc and Kp, the value for R is most often used as 0.08206 L atm/ mol K, and therefore, the units for pressure should also be in atm.
\[{P_{({{\rm{I}}_{\rm{2}}})}}\; = \;{\rm{655}}\;{\rm{torr\;}} \times {\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}{\rm{\; = }}\,{\rm{0}}{\rm{.862}}\;{\rm{atm}}\]
\[{P_{({\rm{C}}{{\rm{l}}_{\rm{2}}})}}\; = \;{\rm{864}}\;{\rm{torr\;}} \times {\rm{\;}}\frac{{{\rm{1}}\;{\rm{atm}}}}{{{\rm{760}}\;{\rm{torr}}}}{\rm{\; = }}\,{\rm{1}}{\rm{.14}}\;{\rm{atm}}\]
The equilibrium constant is:
\[{K_p}\; = \;\frac{{{P^2}_{{\rm{ICl}}}}}{{{P_{{{\rm{I}}_{\rm{2}}}}}{P_{{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\; = \;2.34{\rm{ }} \times {\rm{ }}{10^{ – 4}}\]
However, remember that these are the partial pressures at equilibrium so, we cannot use the numbers above.
First, we need to set up an ICE table to determine the partial pressures at equilibrium. Because the system does not contain any ICl at the beginning, its pressure (concentration) is going to increase, while the pressure of Cl_{2} and I_{2 }is going to decrease. Let’s assume the pressure of Cl_{2} and I_{2 }drops by x atm which means the pressure of ICl will increase by 2x atm:
P(l_{2}) | P(Cl_{2}) | P(lCl) | |
Initial | 0.862 | 1.14 | 0 |
Change | -x | -x | +2x |
Equil | 0.862-x | 1.14-x | 2x |
Plug in the numbers and assign x for the partial pressure of ICl:
\[{K_p}\; = \;\frac{{{{{\rm{(2x)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862 – x}}} \right)\left( {{\rm{1}}{\rm{.14 – x}}} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}\]
\[{K_p}\; = \;\frac{{{\rm{4}}{{\rm{x}}^{\rm{2}}}}}{{{{\rm{x}}^{\rm{2}}}\;{\rm{ – }}\;{\rm{2x}}\;{\rm{ + }}\;{\rm{098268}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}\]
Solving this quadratic equation, we get x=0.00753
Therefore, the partial pressure of ICl at equilibrium is 2 x 0.00753 atm = 0.0141 atm = 11.4 torr
A good approach to avoid solving long quadratic equations is to make an approximation/assumption that x << 0.862, meaning that there is very little pressure drop (reaction).
In this case, we can simplify the quadratic equation as:
\[{K_p}\; = \;\frac{{{{{\rm{(2x)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862}}} \right)\left( {{\rm{1}}{\rm{.14}}} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.34 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}\]
Solving for x, we find that x = 0.007582 which is very close to 0.00753 (error < 1%).
Let’s assume though that we did not know the actual value for x and only solve for it by the approximation. To test whether the assumption was applicable to this reaction, plug the values in the equilibrium expression and compare the result to the actual Kc given in the problem.
\[{K_p}\; = \;\frac{{{{{\rm{(2}} \times 0.007582{\rm{)}}}^{\rm{2}}}}}{{\left( {{\rm{0}}{\rm{.862 – }}0.007582} \right)\left( {{\rm{1}}{\rm{.14 – }}0.007582} \right)}}\;{\rm{ = }}\;{\rm{2}}{\rm{.38 \times 1}}{{\rm{0}}^{{\rm{ – 4}}}}\]
And this confirms that the approximation was suitable for this problem. If it wasn’t, and the difference between the Kc values was more than 5%, then the equation must be solved without the approximation.
Consider the following equilibrium:
2NO(g) + 2H_{2}(g) ⇆ N_{2}(g) + 2H_{2}O(g)
Initially, there are 0.15 moles of NO and 0.25 moles of H_{2}, in a 10.0-L container. If there are 0.056 moles of NO at equilibrium, how many moles of N_{2} are present at equilibrium?
0.047 mol
The reaction is going to proceed to the right because there are only reactants present in the mixture initially (Q = 0).
Even though we don’t need an ICE table here, let’s put one to visualize the changes. Here is the date we have:
2NO(g) + 2H_{2}(g) ⇆ N_{2}(g) + 2H_{2}O(g)
[NO] | [H_{2}] | [N_{2}] | [H_{2}O] | |
Initial | 0.15 | 0.25 | 0 | 0 |
Change | ||||
Equil | 0.056 | ? |
Next is to determine how many moles of NO have reacted to establish the equilibrium. For this, subtract the final (equilibrium) and initial concentrations of NO:
Change (NO) = 0.15-0.056=0.094 mol
Add this to the ICE table and determine the moles of NO based on the stoichiometric ratio. Every 2 moles of NO produce one mole of N_{2}, and therefore,
n (N2) = 0.094/2 = 0.047 mol
2NO(g) + 2H_{2}(g) ⇆ N_{2}(g) + 2H_{2}O(g)
[NO] | [H_{2}] | [N_{2}] | [H_{2}O] | |
Initial | 0.15 | 0.25 | 0 | 0 |
Change | 0.094 | 0.094 | 0.047 | 0.094 |
Equil | 0.056 | 0.156 | 0.047 | 0.094 |
And since there is no N_{2 }present in the beginning, this is also the number of moles that are present at equilibrium.
Check Also
- Chemical Equilibrium
- Equilibrium Constant
- K_{p}and Partial Pressure
- K_{p }and K_{c}Relationship
- K Changes with Chemical Equation
- Equilibrium Constant K from Two Reactions
- Reaction Quotient – Q
- ICE Table – Calculating Equilibrium Concentrations
- ICE Table Practice Problems
- Le Châtelier’s principle
- Le Châtelier’s principle Practice Problems
- Chemical Equilibrium Practice Problems