Let's consider the following reaction at equilibrium:
\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2 \text{HI}(g) \]
To solve for the equilibrium constant \(K_c\), we can use an ICE table (Initial, Change, Equilibrium) to keep track of the concentrations of the reactants and products:
\[
\begin{array}{c|ccc}
& \text{H}_2 & \text{I}_2 & \text{HI} \\
\hline
\text{Initial} & a & b & 0 \\
\text{Change} & -x & -x & +2x \\
\text{Equilibrium} & a-x & b-x & 2x \\
\end{array}
\]
In this table:
- \(a\) and \(b\) represent the initial concentrations of \(\text{H}_2\) and \(\text{I}_2\), respectively.
- \(x\) is the change in concentration of \(\text{H}_2\) and \(\text{I}_2\), and \(2x\) is the change in concentration of \(\text{HI}\).
The equilibrium constant \(K_c\) for the reaction is defined as:
\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \]
At equilibrium, the concentrations are:
\[ K_c = \frac{(2x)^2}{(a-x)(b-x)} \]
To find the value of \(K_c\), you need to know the initial concentrations \(a\) and \(b\), and the value of \(x\), which can be determined from experimental data or additional information about the equilibrium system.
For example, if \(a = 0.1 \, \text{M}\), \(b = 0.1 \, \text{M}\), and at equilibrium, the concentration of \(\text{HI}\) is found to be \(0.15 \, \text{M}\) (meaning \(2x = 0.15 \, \text{M}\) and \(x = 0.075 \, \text{M}\)), then:
\[ K_c = \frac{(0.15)^2}{(0.1-0.075)(0.1-0.075)} = \frac{0.0225}{0.000625} = 36 \]
So, the equilibrium constant \(K_c\) for this reaction under these conditions is 36.