I2 (3.0 x 10-2 moles) and H2 (2.0 x 10-2 moles) were mixed in a closed container and heated to 700K. The equilibrium constant Kc for the reaction is 55. How many moles of HI will be present at equilibrium?
Solution: This requires a number of steps.
Step 1: Set up a "balance sheet" using symbols for the amounts of the reactants and products:
| I2 | H2 | HI | |
| Amount (moles) at the start | m | n | 0 |
| Amount (moles) used up | x | x | 0 |
| Amount (moles) formed | 0 | 0 | 2x |
| Amount at equilibrium | m - x | n - x | 2x |
| Concentration at equilibrium (mol.dm-3) | (m - x)/V | (n - x)/V | 2x/V |
where V is the unknown volume of the container.
Step 2: Write down the equation linking the equilibrium constant Kc with the equilibrium concentrations:
Note that the volume V cancels out, so its value is irrelevant. This may not always be true for other reactions!
Step 3: Rearrange the equation to give a quadratic equation in the standard form ax2 + bx + c = 0:
Step 4: Solve for x, using the formula used to solve quadratic equations:
Therefore, x = 3.6 x 10-2 or x = 1.8 x 10-2. The first value must be rejected, since it is greater than the starting amount of either H2 or I2. The amount of HI formed is 2x, i.e. 3.6 x 10-2 mol.
NOTE: By approaching the problem in an algebraic manner, one can easily keep track of the various steps, and the arithmetic can be done right at the end. This makes it easy to go over the calculations in order to check the answer.
For a general quadratic equation y = x2 + bx + c,
the roots of the equation (i.e., the values of x when the equation is zero, y = 0), are given by
