# Determining the Equilibrium Constant 2

• Page ID
1374
• The Equilibrium Constant is unitless. It simply signifies the ratio between the forward and reversal rates of a chemical reaction at equilibrium.

### Introduction

There are a few different Equilibrium Constants such as:

• $$K_c$$ (dealing with concentration)
• $$K_p$$ (dealing with partial pressure)

For the sake of brevity, let us stick to the former, $$K_c$$ to follow this Module.

Generally, we follow this equation when dealing with the calculation of the equilibrium constant, $$K_c$$:

$$K_c = \dfrac{Products}{Reactants}$$

• Where the concentrations of the Products are multiplied in the numerator (top) and the concentrations of the Reactants are multiplied in the denominator (bottom).
• We separate and denote each chemicals' concentration inside of closed brackets.

Important: We only include the concentrations of aqueous and gaseous substances. Do not include those of solids or liquids!

### The Calculation of K

1. Determine whether the corresponding chemical (i.e. HCl, NaCl, etc…) is on the products or reactants side of the chemical equation. Be sure to only include those in the (aq) or (g) states.
2. Observe your ICE table and look at the values in the E row of the table.
3. Based on your observations from step 2, place those values in the general equation stated above (products divided by reactants).
4. Multiply, then divide your answers in order to obtain the equilbrium constant.

This guide assumes you know how to setup and ICE table. If you do not, please refer to this excellent guide: Using ICE Tables. Caution: Remember that we are obtaining the equilibrium constant K, NOT the Q. Q deals with using the initial values from the I row of the ICE table.

### That's All? Be Careful…

• If your balanced chemical equation contains a coefficient (i.e., 2, 3, 4, ect…) for a specific chemical (i.e. 2H3O+) then you MUST raise it's concentration to the power of that coefficient. (i.e. [H3O+]2)

### Example

Take this general, easy to follow example, where A/B/C are different chemicals and M is molarity (mols/liter).

Question: Calculate the equilibrium constant for a reaction between .100 M A and .300 M B, given that the product, C has a concentration of .200 M and that these values are those when the reaction is in a state of equilbrium. Here is the chemical equation and assume it is balanced.

$$A_{(aq)} + B_{(aq)} \rightleftharpoons 3C_{(aq)}$$

Solution:

Given that the concentrations at equilibrium, we can assume that the ICE table (omit IC as they are not needed here) ends up looking like:

 A(aq) + B(aq) $$\rightleftharpoons$$ 3C(aq) 0.100 0.300 0.200

Based on the steps mentioned above:

1. A and B are reactants, while C is a product. They are all in the aqueous (aq) state, so their concentrations are important to us.
2. According to the E row of the ICE table (shown), we see the concentrations at equilibrium of A, B and C.
3. Recall:

$$K_c = \dfrac{Products}{Reactants}$$

Thus:

$$K_c = \dfrac{(0.200)^3}{(0.100)(0.300)}$$

Notice that the concentration of C (0.200) is raised to the power of 3, since the number 3 is the coefficient of C.

4. Following basic order of operations of mathematics, first calculating exponential values then multiplying out the products and the reactants, and finally dividing them, we obtain our answer as follows:

$$K_c = 0.267$$

Therefore, we can state that the ratio of the forward and reversal rates of this reaction is 0.267

### Extra Info: When Balancing a Chemical Equation, K is Manipulated…

• If you reverse the equation, then invert K. In other words, 1/K.
• If you multiply the entire equation by a number, then raise K to the power of that number you multiplied by.
• If you divide the entire equation by a number, then root K according to that number (i.e. dividing by 2 will cause a square root of K, or dividing by 3 would cause a cube root of K).

### References

1.  Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications; Ninth Ed. Upper Saddle River, NJ: Prentice Hall 2007.