SCH4U - Chemistry 12 (2024-25) - A

A. Introduction

When a system is at equilibrium, the forward and reverse reactions are occurring at the same rate and the concentration of reactants and products remain constant.  From this knowledge we can piece together a quantitative equation that describes a value known as the equilibrium constant (K_eq).

B. Equilibrium Constant Expression

A general equilibrium reaction can be expressed as follows:

aA + bB ↔ cC +dD

From this reaction equation we can piece together a general formula for the equilibrium constant (K_eq)

K_eq = ([C]^c[D]^d) / ([A]^a[B]^b)

Notice that the coefficient that is in front of each species in the reaction equation becomes an exponent in the equilibrium constant expression (for example, cC becomes C^c ).  Equilibrium constant equations are dependent on the equilibrium concentration of the reactants and products.  When calculating equilibrium constants, we do not include the concentrations of solids and pure liquids because these values are constant.  The concentrations of solids and pure liquids do not change over time as a system heads towards equilibrium.  Therefore in equilibrium constant expressions we only include species that are in their gaseous _(g) or aqueuos _(aq) states. 

For example, for the following reaction:

2H₂O_(l) ↔ 2H_2(g)+ O_2(g)

K_eq = [H_2(g)]²[O_2(g)]

We do not include the concentration of pure water in its liquid state (2H₂O_(l))

When calculating K_eq we need to make sure that our values for concentration are in moles/litre (mol/L), which gives us units of mol²/L^2, but typically equilibrium constants are expressed without units.  Lets try a real example to piece this equation together:

For the following system at equilibrium write the equilibrium constant expression:

2H₂S_(g) + CH_4(g) ↔ 4H_2(g) + CS_2(g)

K_eq = ([H_2(g)]⁴[CS_2(g)]) / ([H₂S_(g)]²[CH_4(g)])

The value of the equilibrium constant can tell us some qualities of a given reaction.  For example, a large K_eq implies that at equilibrium, the concentration of products is higher than the concentration of reactants.  If the the K_eq is small, the reactants are at a higher concentration at equilibrium than the products.

Click here to review equilibrium constants from Khan Academy.

Click here to review equilibrium constants from Chemguide.

C. Solving Problems with K_eq

There are different types of problems that we can solve using the formula for K_eq, involving determining what the value of K_eq is or alternatively determining the concentration of an unknown reactant or product when the K_eq is known.  When solving problems with K_eq you must first ensure that the chemical equation is properly balanced. 

Solving for K_eq with All Equilibrium Concentrations

When 1.0 mol of ammonia gas is injected into a 0.50L flask at 50°C the reaction below proceeds to equilibrium

2NH_3(g) ↔ N_2(g) + 3H_2(g)

Once at equilibrium the flask contains 0.30 mol H_2(g), 0.10 mol N_2(g) and 0.8 mol NH_3(g). Calculate the value of  K_eq.

Step 1. Check if the equation is balanced. 

Step 2. Convert given concentrations into mol/L by dividing the number of moles by the volume

[H₂] = 0.30 mol / 0.50 L = 0.6 mol/L

[N₂] = 0.10 mol / 0.50 L = 0.2 mol/L

[NH₃] = 0.80 / 50 L = 1.6 mol/L

Step 3. Calculate K_eq

K_eq = ([N₂] * [H₂]³) / ([NH₃]²)

K_eq = (0.2 * 0.6³) / (1.6²) = (0.2 * 0.216) / (2.56) = 0.0432 / 2.56 = 0.017 

Therefore the value of K_eq is 0.017.  Because this number is small, we know that at equilibrium the concentration of the products is much higher than the reactants.

Solving for K_eq with All Initial Concentrations and 1 Equilibrium Concentration

To determine K_eq experimentally, occasionally you will know the initial concentration of all your reactants and the the equilibrium concentration of one of the products.  From this information we can piece together the equilibrium concentrations of each product and reactant as well as the K_eq

Example:

Consider the following reaction PCl_5(g) ↔ PCl_3(g) + Cl_2(g)

A 2.0L flask at 30°C initially contains 0.0087 mol of PCl_5(g) and 0.298 mol PCl_3(g) and zero mol Cl_2(g).  The reaction is allowed to proceed until equilibrium at which point it is determined that the flask contains 0.002 mol of Cl_2(g).  Calculate the concentration of each molecule at equilibrium and the Keq for the reaction.

Step 1. Calculate the concentrations of all the known varibables in terms of mol/L

[PCl₅]_initial = 0.0087 mol / 2.0 L = 0.00435 mol / L

[PCl₃]_initial = 0.298 mol / 2.0 L = 0.149 mol / L

[Cl₂]_initial = 0 mol / L

[Cl₂]_equilibrium = 0.002 mol / 2.0 L = 0.001 mol / L

Step 2. Calculate the concentrations of each variable at equilibrium, using an ICE table

To calculate the concentrations of each variable at equilibrium we use a tool called an ICE table.  ICE is an acronym for Initial, Change, Equilibrium and is set up as follows:

Initial row: we insert all our known initial concentrations

Change row: For the Change row, we insert the variable of the change that occurs from initial to equilibrium.  From the reaction equation we see that the molar ratio of PCl₅ : PCl₃ : Cl₂ is 1 : 1 : 1.  Therefore for every increase in Cl₂ of x mol/L there will be the same increase of x mol/L of PCl₃ and a decrease of x mol/L of PCl₅.    For this example, we know that the concentration of Cl₂ increased by 0.001 mol/L so this will be our value of x.

• • • For the Change in PCl₅ and PCl₃ we need to go back to the reaction equation and use their molar ratios to calculate the change. To do this we look at the value of the coefficients in front of each reactant and product.  For the reaction: PCl_5(g) ↔ PCl_3(g) + Cl_2(g), the molar ratio is 1 : 1 : 1, which tells us that for each mole of Cl₂ formed, one mole of PCl₃ is also formed and 1 mole of PCl₅ is used up.  Therefore if the change in Cl₂ is + x, the change in PCl₃ will also be +x and the change in PCl₅ will be -x.   In step 1 we just calculated that the change in [Cl₂] was 0.001 mol/L therefore this is our x-value.

• • • If the molar ratio is not 1:1 for any variable, you simply multiply or divide the known value by that coefficient.  For example if you have a reaction equation 2A ↔ B + C and you knew the change in C, you would simply multiply the value for the change in C by 2 to get your value for the change in A

Equilibrium Row: calculate the concentration at equilibrium by adding the change to the initial concentration.

• • • [PCl₅]_equilibrium =  0.00435 mol / L - 0.001 mol/L = 0.00335 mol/L
    • [PCl₃]_equilibrium =  0.149 mol / L + 0.001 mol/L = 0.150 mol/L
    • [Cl₂]_equilibrium = 0.001 mol

Step 3. Calculate K_eq

K_eq= [PCl₃][Cl₂] / [PCl₃] = (0.150 * 0.001) / 0.00335

K_eq = 0.045

Check your understanding

A sealed 1.00 L flask was filled with 2.00 mol CO_(g) and 2.00 mol H₂O_(g).  They are allowed to react and form CO_2(g) and H_2(g).  At equilibrium it is determined that 1.30 moles of each product are formed.

a) Write the balanced chemical equation

b) Write the equilibrium constant expression

c) Calculate the K_eq.

Answer 4-2-A-1

D. Summary

• The equilibrium constant (Keq) describes a system at equilibrium based on the concentration of products and reactants at equilibrium.
• K_eq = ([C]^c[D]^d) / ([A]^a[B]^b)
• Solids and pure liquids are not included in the equilibrium constant equation as their concentrations are constant.
• A large K_eq means that the products are favoured (products have a higher concentration than the reactants) at equilibrium while a small K_eq means the reactants are favoured at equilibrium.
• To calculate Keq from initial concentrations, an ICE table can be used.