Complete Guide To Chemical Equilibrium (Expression, Constant and ICE Chart)

Chemical Equilibrium Definition

In the previous chapters, we learned about reactions going to completion and calculations associated with them, such as limiting reagent calculation. However, many reactions do not go to completion. Instead, as reactants produce products, products react back producing reactants. These reactions are called reversible. In these reactions, both forward and reverse reactions occur simultaneously. At some point, equilibrium is reached. At equilibrium, the rate of forward reaction is equal to the rate of reverse reaction and the concentrations of reactants and products remain constant. The arrow that is used for these reaction is a double arrow, ⇌. For example, A+B ⇌ C

Chemical Equilibrium Constant And Equilibrium Constant Expression

Equilibrium constant, K_c, will be used in the majority of problems in this chapter. K_c is equal to the concentration of products raised to the power of their coefficients over the concentration of reactants raised to the power of their coefficients. For example, consider a hypothetical reaction

aA + bB ⇌  cC + dD

Please note, that pure solids and liquids do NOT go into the equilibrium constant expression. If you see (s) or (l) next to a molecule in the equation, do NOT put this molecule into the expression.

Now, let’s practice writing equilibrium constant expression.

Practice: Write the equilibrium-constant expression Kc for the following equations:

Chemical Equilibrium Constant in Terms of Pressure

When an equation has gaseous reactants and/or products, we can write the equilibrium expression in terms of partial pressures. The equilibrium constant for this will be K_p.

For example, for the equation N₂(g)+3H₂(g) ⇌ 2NH₃(g), the equilibrium constant expression using partial pressures, will be

How To Convert Kc to Kp And Vice Versa

We can convert Kc to Kp and vice versa using an equation Kp = Kc(RT)^Δn, where R is the ideal gas constant = 0.08206, T is temperature in Kelvin, and Δn is equal to moles of gaseous product (use coefficients) – moles of gaseous reactant(use coefficients).

Practice: For the Haber process, N₂(g)+3H₂(g) ⇌ 2NH₃(g), Kc = 9.60 at 300°C. Calculate Kp for this reaction at the same temperature.

Solution: Kp = Kc(RT)^Δn
Kc = 9.60, R = 0.08206, T = 300+273 = 573K
Δn = moles of gas of products (use coefficients) – moles of gas of reactants (use coefficients)= 2-(1+3)=-2
Kp = 9.60*(0.08206*573K)^-2 = 4.34*10^-3

What Does Equilibrium Constant Tell Us?

Equilibrium constant gives us valuable information about the composition of the equilibrium mixture. Remember that equilibrium constant is equal to the concentration of products over reactants.
This means that if K is large, there are a lot more products than reactants at equilibrium. If K is smaller than 1, there are more reactants than products.

K>>1, equilibrium mixture consists of mostly products.
K<<1, equilibrium mixture consists of mostly reactants.
K=1 , equilibrium mixture has the same amount of products and reactants

Chemical Equation Stoichiometry and Equilibrium Constants

We can use known equilibrium constants for the reactions to calculate an unknown equilibrium constant for another reaction, similarly to the way we used Hess’s Law in the Thermochemistry chapter. First, let’s go over three rules for K’s.
1. Reverse an equation = Take reciprocal of K (1/K)
2. Multiply an equation by a constant = Raise K to the power of that constant
3. Add equations = Multiply their K’s

The best way to explain this is by doing a problem.

Given the reactions:
1. HF(aq) ⇌ H⁺(aq)+F^–(aq), K = 6.8*10^-4
2. H₂C₂O₄(aq)⇌2H+(aq)+C₂O₄^2-(aq), K = 3.8*10^-6
Determine the value of K for the following reaction:
2HF(aq)+C₂O₄^2-(aq)⇌2F^–(aq)+ H₂C₂O₄(aq), K =???

Solution: My trick is to find one substance in equation 1 that you also see in the final equation but not in equal 2. This could be HF. I see HF in equation 1 and the final equation but not in equation 2. Next, I ask what happened to HF as we go from equation 1 to the final equation. I see that in the final equation there is a 2 in front of HF. This mean we must multiply equation 1 by 2 and take K to the power. We get:
2(HF(aq) ⇌ H⁺(aq)+F^–(aq)), K = (6.8*10^-4)²
New equation 1: 2HF(aq) ⇌ 2H⁺(aq)+2F^–(aq)), K =4.6*10^-7

Next, I need to do the same thing to equation 2. H₂C₂O₄ is a common molecule with the final equation that equation 1 does not have. What happened to it? In the equation 2, it is on the left (reactant) side and in the final equation it is on the right (product side). I have to reverse equation 2 and take reciprocal of its K.
Reverse(H₂C₂O₄(aq)⇌2H+(aq)+C₂O₄^2-(aq)), K = 1/(3.8*10^-6)
New Equation 2: 2H+(aq)+C₂O₄^2-(aq)⇌ H₂C₂O₄(aq), K = 2.6*10⁵
Notice, that if we now add the two equation, we will get the final equation.
2HF(aq) ⇌ 2H⁺(aq)+2F^–(aq))
2H+(aq)+C₂O₄^2-(aq)⇌ H₂C₂O₄(aq)
To find the new K, we have to multiply the new K values. K = 4.6*10^-7 * 2.6*10⁵= 0.12

How To Predict The Direction Of Reaction

In order to predict the direction of a reaction, we must calculate Reaction Quotient, Q. Q will have the same expression as K, EXCEPT that the values for concentration or pressure you plug in, will NOT be equilibrium values.
aA + bB ⇌  cC + dD

Memorize:
If Q>K, the reaction will go to the left
If Q<K, the reaction will go to the right
If Q=K, the reaction is at equilibrium

Practice: A 50.0-L reaction vessel contains 1.00 mol N₂, 3.00 mol H₂, and 0.500 mol NH₃. Will more ammonia, NH₃, be formed or will it dissociate when the mixture goes to equilibrium at 400°C? The equation is
N₂(g) + 3H2(g) ⇌  2NH₃(g).  K is 0.500 at 400°C.

Solution: Let’s write the expression for Q, which would be the same as the expression for K (products over reactants to their coefficients).

We need molarity of each component. M= moles/ Liters. [H₂]=3.00 mol/50.0-L=0.0600M, [N₂] = 1.00 mol/50.0-L=0.0200M, [NH₃]=0.500 mol/50.0-L= 0.0100
Q= (0.0100)²/(0.0200*0.0600³) = 23.1
23.1>0.500, Q>K, reaction will go to the left, dissociating more ammonia.

How To Predict Chemical Equilibrium Concentrations When K Is Given

When K is given, we can calculate the equilibrium concentrations of reactants and products by making an ICE CHART.

What is an ICE Chart and how do we use one?
I: Initial= We put any initial concentrations given in this row. If we have Kp, we can put initial pressures.
C: Change= We must know in which direction the reaction is going. Look at the Initial row. If there is a zero on either a product or reactant side, that is the side whose concentration should increase (since we can’t have less than a zero concentration) and will have plus signs. The other side will decrease and will have minus signs. We use X times the coefficient from the equation, to signify by how much the concentration is increasing or decreasing. If you find this confusing, we will go through a problem to make it easier.
E: Equilibrium = Add the first two rows to get to the equilibrium: I+C

Let’s do an example and see how we can use an ICE chart

Problem: H₂(g)+ I₂(g) ⇌ 2HI(g)
A 1.000L flask is filled with 1.000mol H₂ and 2 mol of I₂ at 275°C. The value of the equilibrium constant, Kc, for the reaction at this temperature is 50.5. What are the equilibrium concentrations of H₂, I₂, and HI?

Solution: The steps are: 1. Determine molarity and write out the equation 2. Write out the ICE chart and fill it in 3. Write the Kc expression 4. Plug in equilibrium values from the ICE chart in the Kc expression and solve for X.

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References:
General Chemistry by Ebbing and Gammon
Chemistry The Central Science by Brown, LeMay, Bursten, Murphy, Woodward, Stoltzfus