Calculations When the Value of K is Known

Most of the time, you actually will know the equilibrium constant for a reaction, since someone else has already measured it.  This means that you can calculate the amount of products you expect a certain reaction to give, without having to do the reaction at all.

This is an incredibly useful thing to be able to do.  If you are trying to design a Sabatier reactor to generate methane fuel from the Maratian atmosphere, you want to know quantitatively the kinds of conditions that will cause high product yield before you start designing the system.  Using known values of Keq for the reaction you can predict whether it is even feasible to build such a system.  Spending millions of dollars to build a system that will hardly make any useful product is hardly sensible.  You need to check out the conditions before you start the design.

With the value of Keq, the balanced equation, and some of the concentrations you can work out the calculation of each substance at equilibrium.

Example 1: K for the reaction H2 (g) + I2 (g) 2HI (g) at 450 șC is 22.5.   If at equilibrium, the [H2] = 3.4 x 10-2 mol/L, [I2] = 9.0 x 10-3 mol/L, then what is the [HI]?

This is an easy problem to solve because there is only one unknown – the concentration of [HI].  We will represent this unknown by the algebraic variable x.

Step 1: Write out the K expression for the balanced equation = 22.5
Step 2: Let the unknown concentration of HI be x.  Substitute this and the known concentrations into the K expression

Step 3: Isolate the unknown value.

Step 4: Solve for x (in this case take the square root of both sides)

calcsamp4_05.gif (276 bytes)

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Step 5: Substitute x back for its actual value

[HI] = x = 8.30 x 10-2

 

Example 2:At 4000 oC Keq = 3.94 x 102 for the reaction  2H2 (g) + O2 (g) 2H2O (g).   If at equilibrium, the [H2] = 1.60 x 10-2 mol/L, and [H2O] = 3.02 x 10-2 mol/L, then what is the [O2]?

There is only one unknown – the concentration of [O2].  We will represent this unknown by the algebraic variable x.

Step 1: Write out the K expression for the balanced equation = 3.94 x 102
Step 2: Let the unknown concentration of O2 be x.  Substitute this and the known concentrations into the K expression

Step 3: Isolate the unknown value.

Step 4: Solve for x.

x = 9.04 x 10-3

Step 5: Substitute x back for its actual value

[O2] = x = 9.04 x 10-3

Try the following problems:

To enter numbers in scientific notation on the computer, use "E" notation.  For example: 2.34 x 105 is 2.34E5, 2.34 x 10-5 is 2.34E-5

 

At 986 șC, the water gas shift equilibrium can be represented by this equation: H2 (g) + CO2 (g) H2O (g) + CO (g).  The concentrations (in mol/L) at equilibrium of [H2] = 6.90 x 10-4, [CO2] = 4.60 x 10-3, and [H2O] = 2.40 x 10-3  and K = 1.57.  What is the concentration of CO (g) (to three significant digits)?

[CO] =  

 

At 500 K, nitrogen dioxide gas will be in equilibrium with nitrogen monoxide and oxygen gases:

2NO (g) + O2 (g) 2NO2 (g)

If K = 7.90 x 105, [O2] = 5.70 x 10-4, and [NO2] = 2.40 x 10-2 mol/L, then what is [NO]?

[NO] =  

Click here for more example problems.

expt.gif (1651 bytes) Measuring K for the solubility of calcium hydroxide