We saw, in the previous post, how the **reaction quotient** is used to **predict the direction of a reaction **before an equilibrium is reached.

To do so, we plug the values of initial concentrations in the expression of the equilibrium constant and, depending on how the values of the quotient and equilibrium constant relate, we predict the direction of the reaction.

Remember, if;

*Q < K**Reaction tends to***form more products**.*Q > K**Reaction tends to***form more reactants**.*Q = K**Reaction is already at***equilibrium**.

Also, if any **reactant or product is missing** in the initial mixture, the **reaction will shift in the direction forming some of that component**.

So, let’s see how the reaction quotient is used to determine the equilibrium concentrations.

**For example**, let’s consider the decomposition reaction of POCl_{3} to POCl and Cl_{2} gases.

POCl_{3}(*g*) ⇌ POCl(*g*) + Cl_{2}(*g*) *K*_{c} = 0.650

If the following amounts of reactants and products were mixed, what will the equilibrium concentration of all components be?

[POCl_{3}] = 0.650 *M*, [POCl] = 0.450 *M*, and [Cl_{2}] = 0.250 *M*

**Solution**

These are the initial concentrations of the components and, most likely, they are different than what we’ll have at equilibrium. They would be **equal to equilibrium concentrations if Q = K** and this is what we need to find out.

So, the first step is to determine the reaction quotient and thus the direction of the reaction.

\[Q\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;\]

\[Q\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.450 }} \times \;{\rm{0}}{\rm{.250}}}}{{0.650}}\; = 0.173\]

Because* Q *

**<**. Remember, the tendency is to reach equilibrium, and therefore, Q “wants to” increase and become equal to the

*K,*the reaction will proceed to the right*K*. Q will increase if more products are formed since their concentrations appear in the numerator.

This means, the concentrations of POCl and Cl_{2 }are going to increase, while the concentration of POCl_{3} will decrease.

Once we know the direction of the reaction, we are going to set up what is called an **ICE table**. ICE is an abbreviation for **I**nitial, **C**hange, **E**quilibrium concentrations of the reactants and products.

This is how we label the rows and columns in the table:

POCl3(g) ⇌ POCl(g) + Cl_{2}(g)

[POCl_{3}] | [POCl] | [Cl_{2}] | |

Initial | |||

Change | |||

Equil |

We know the **initial concentrations** ([POCl_{3}] = 0.650 M, [POCl] = 0.450 M, and [Cl_{2}] = 0.250 M) as they are given in the problem. The “change” is the amounts of the components that react before the equilibrium is established.

Since we don’t know this, we assign x mol/L of POCl_{3 }as the reacted amount. Next, we determine how much of each product will be formed if x mol/L of POCl_{3} decomposes (reacts). This is done based on the stoichiometric ratio of the components, and because it is 1 : 1 mole ratio between all the components, there is going to be x mol/L of POCl(*g*) and x mol/L Cl_{2}(*g*) formed.

We can put x on top/bottom of each component in the chemical equation:

x x x

POCl_{3}(*g*) ⇌ POCl(*g*) + Cl_{2}(*g*)

Now, instead of simply using x, we also add a sign to it. In this case, for the reactant(s), it is going to “-x” because its concertation is decreasing, and for the products, it is “+x” since their concertation is increasing:

-x +x +x

POCl_{3}(g) ⇌ POCl(g) + Cl_{2}(g)

At his point, we know the initial concentrations (I in the table), and the change in concentrations (c in the table) which is how much the concentration of the reactant has decreased, and the ones of the products have increased.

So, let’s write these in the table:

POCl3(g) ⇌ POCl(g) + Cl_{2}(g)

[POCl_{3}] | [POCl] | [Cl_{2}] | |

Initial | 0.650 | 0.45 | 0.250 |

Change | -x | +x | +x |

Equil |

The last part to fill up the table is determining the equilibrium concentrations. And, since we have already set the sign for x, all you need to do here is add the values in the “initial” and “change” cells:

POCl3(g) ⇌ POCl(g) + Cl_{2}(g)

[POCl_{3}] | [POCl] | [Cl_{2}] | |

Initial | 0.650 | 0.45 | 0.250 |

Change | -x | +x | +x |

Equil | 0.650-x | 0.450+x | 0.250+x |

Now that we have the expressions for equilibrium concentrations, we can plug them in the equation for the equilibrium constant and find the concentrations by solving for the x:

\[K\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\;\]

\[K\;{\rm{ = }}\;\frac{{\left( {{\rm{0}}{\rm{.450 + x}}} \right)\left( {{\rm{0}}{\rm{.250 + x}}} \right)}}{{0.650 – x}}\; = 0.650\]

We have a quadratic equation which is what you are going to work with for most problems on equilibrium concentrations. First, we simplify it to be able to use the formula for quadratic equations:

0.1125 + 0.450x + 0.250x + x^{2} = 0.4225 – 0.650x

0.1125 + 0.700x + x^{2 }= 0.4225 – 0.650x

x^{2 }+ 1.35x – 0.310 = 0

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\]

a = 1, b = 1.35, c = -0.310

Therefore,

\[{\rm{x}}\;{\rm{ = }}\;\frac{{ – 1.35\, \pm \,\sqrt {1.8225\; – \,4\; \cdot \;1\; \cdot \;( – 0.310)} }}{{2 \cdot \;1}}\]

**x = 0.200 ***or x= -1.55*

*x=-1.55 doesn’t work because that would indicate increasing the concentration of *POCl_{3, }and therefore,

**x=0.200**

Now, this is how much POCl_{3} has reacted, and to find the equilibrium concatenations, we go based on the expressions in the ICE table:

[POCl_{3}] at equilibrium is 0.650 – 0.200 = **0.450 mol/L**

[POCl] at equilibrium is 0.450 + 0.200 = **0.650 mol/L**

[Cl_{2}] at equilibrium is 0.250 + 0.200 = **0.450 mol/L**

** **** **

**It is ****always a good idea to**** plug this numbers in the expression for K_{c} and **

**check if they are correct:**

** **** **

\[K\;{\rm{ = }}\;\frac{{\left[ {{\rm{POCl}}} \right]\left[ {{\rm{C}}{{\rm{l}}_{\rm{2}}}} \right]}}{{\left[ {{\rm{POC}}{{\rm{l}}_3}} \right]}}\; = \;\frac{{(0.650)(0.450)}}{{(0.450)}}\; = \;0.650\;\;\]

**Can We Do It Without Solving Quadratic Equations?**

The good news is that yes, most often you can determine the equilibrium concentrations without solving a quadratic equation.

This is possible for reactions with a small equilibrium constant (in the 10^{-3} and below range).

Let’s see how it works in the following example.

Consider the following equilibrium:

2NOCl(*g*) ⇆ 2NO(*g*) + Cl_{2}(*g*)

2.00 mole of pure NOCl and 1.65 mole of pure Cl_{2} are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(*g*) considering that with *K* = 2.4 x 10^{–6}.

Again, the first thing here is to determine the direction of the reaction. There are both reactants and products in the initial mixture, however, there is no NO, and remember, when one of the reactants or products is missing from the system, the equilibrium is going to be established by producing some of that compound. Therefore, the reaction is going to proceed to the right to produce some NO and we don’t need to (we cannot actually) calculate the reaction quotient.

So, let’s set up an ICE table and, for convivence, assign 2x as the depletion in NOCl concentration:

2NOCl(*g*) ⇆ 2NO(*g*) + Cl_{2}(*g*)

[NOCl] | [NO] | [Cl_{2}] | |

Initial | 2.00 | 0 | 1.65 |

Change | -2x | +2x | +x |

Equil | 2.00-2x | 2x | 1.65+x |

The equilibrium constant is:

\[K = \frac{{{{[{\rm{NO}}]}^{\rm{2}}}[{\rm{C}}{{\rm{l}}_{\rm{2}}}]}}{{{{[{\rm{NOCl}}]}^{\rm{2}}}}}\]

\[K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + x}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2x}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}\]

This is a cubic equation, and what you can do it these situations, is make an approximation that 2x<<2.00 or x<<1.00 because the equilibrium constant a very small and the amount of the reactant that reacts is going to be insignificant compared to its initial quantity.

This allows to simplify the equation as follows:

\[K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}\]

\[K = \;\frac{{{\rm{4}}{{\rm{x}}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65}}} \right)}}{{\rm{4}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}\]

6.6x^{2} = 9.6×10^{-6}

x = 0.00121

The equilibrium concentration of NO is 2x, so that is **0.00242 mol/L**.

Now, we need to plug the numbers and see if the value for the equilibrium constant is close enough to the given number (*K* = 2.4 x 10^{–6}).

\[K = \;\frac{{{{\left( {{\rm{2x}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + x}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2x}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{–6}}}}\]

\[K = \;\frac{{{{\left( {{\rm{2}} \times {\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}\left( {{\rm{1}}{\rm{.65 + 0}}{\rm{.00121}}} \right)}}{{{{\left( {{\rm{2}}{\rm{.00 – 2}} \times {\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.42 \times 1}}{{\rm{0}}^{{\rm{–6}}}}\]

And that is very close to the value of *K*, and therefore, out approximation was correct, and the equilibrium concentration of NO is 0.00242 mol/L.

Usually, the approximation is considered to be valid if the **x is less than 5% **of the reactant’s initial

concentration (less than 5% of it reacts).

So, to find this percentage, we divide the reactant amount over the initial concentration of NOCl:

0.00242/2.00 x 100% = 0.121%

Therefore, the approximation was valid. If you determine the parentage and find out that it was not, the equation must be solved without the shortcut.

In the end, let’s enter the values for the equilibrium concentrations to make sure everything is correct:

\[K\; = \;\frac{{{{\left( {{\rm{2}}\, \times \;{\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}\left( {{\rm{1}}.{\rm{65}}\,{\rm{ + }}\;{\rm{0}}{\rm{.00121}}} \right)}}{{{{\left( {{\rm{2}}.{\rm{00}}\;{\rm{ – }}\;{\rm{2}}\, \times \;{\rm{0}}{\rm{.00121}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{2}}.{\rm{4\;}} \times \;{\rm{1}}{{\rm{0}}^{{\rm{ – 6}}}}\]

#### Practice

The equilibrium constant for the following reaction at 600 ^{o}C is determined to be *K*c = 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 *K*c 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 *K*p = 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 *K*p.

\[{K_p}\; = \;\frac{{0.598}}{{{{0.248}^2}0.085}}\; = \;114\; > \;6.7{\rm{ }}x{\rm{ }}{10^{ – 3}}\]

This is greater than *K*p (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, *K*c = 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 *K*c and *K*p, 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 *K*c 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*and_{p }*K*Relationship_{c}*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**