Question

Consider the reaction. $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ An equilibrium mixture of this reaction at a certain temperature has \([\mathrm{CO}]=0.105 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.114 \mathrm{M}\), and \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.185 \mathrm{M}\). What is the value of the equilibrium constant at this temperature?

Short answer

The equilibrium constant (K) at this temperature is approximately 136.41.

Step-by-step solution

Write the Expression for the Equilibrium Constant

For the reaction \[\begin{equation}\mathrm{CO}(g) + 2\mathrm{H}_2(g) \rightleftharpoons \mathrm{CH}_{3}\mathrm{OH}(g)\end{equation}\]the equilibrium constant expression, K, is given by \[\begin{equation}K = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_2]^2}\end{equation}\]where the concentrations in brackets represent the equilibrium molar concentrations of the products and reactants.

Substitute the Equilibrium Concentrations into the Expression

Substitute the equilibrium concentrations given into the K expression:\[\begin{equation}K = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_2]^2} = \frac{0.185}{0.105 \cdot (0.114)^2}\end{equation}\].

Calculate the Value of K

Carry out the calculation to find the value of K:\[\begin{equation}K = \frac{0.185}{0.105 \cdot (0.114)^2} = \frac{0.185}{0.105 \cdot 0.012996} \approx 136.41\end{equation}\].

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time, but it's important to remember that the reactions are still occurring; they are just balanced. This dynamic balance doesn't mean the reactants and products are present in equal amounts, but rather that their ratios do not change with time.

To understand equilibrium, picture a busy street where the number of people entering a shop is the same as those leaving. Inside, the number of people remains constant, much like the concentrations at equilibrium. Recognizing when a system is at equilibrium is crucial in the calculation of the equilibrium constant, which reveals how far a reaction proceeds before reaching this balanced state.

Improving one's grasp on this concept involves picturing the dynamic nature of equilibrium—it's not static but a continuous, ongoing process where the forward and backward reactions proceed at the same rate.

Equilibrium Constant Expression

The equilibrium constant expression, often represented by the symbol 'K', quantifies the balance between products and reactants in a chemical equilibrium. For a general reaction, where 'aA + bB \rightleftharpoons cC + dD', the equilibrium constant expression is \[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\].

Here, the letters A, B, C, and D represent the chemical species involved in the reaction, the lowercase letters a, b, c, and d denote their respective stoichiometric coefficients, and the square brackets indicate molar concentrations of these species at equilibrium. In the given problem, the equilibrium constant for the formation of \(\text{CH}_3\text{OH}\) from CO and \(\text{H}_2\) is represented in terms of their concentrations. It's essential to raise the concentration of each species to the power of its coefficient in the balanced equation and understand that only gaseous and aqueous species are included in K expressions.

To truly grasp this, students should practice writing K expressions for various reactions and also understand what a high or low K value signifies in terms of product or reactant favorability.

Molar Concentration

Molar concentration, also known as molarity, is the number of moles of a solute divided by the volume of the solution in liters, and it's expressed in moles per liter (\( M \)). Determining the molar concentration of a substance is fundamental in the context of equilibrium calculations. For instance, in the reaction highlighted in our exercise, the molar concentration of CO, \(\text{H}_2\), and \(\text{CH}_3\text{OH}\) forms the crux of finding the equilibrium constant.

Understanding molar concentration requires practice in converting between moles, liters, and molarity. Practicing dilution and mixing problems can also substantially aid in grasping this concept. Accurate measurement and calculation of molar concentration directly affect the accuracy of the equilibrium constant determination.

For students looking to improve, it is beneficial to engage with problems involving varied ways to express concentration (including mass percent, mole fraction, etc.) and to relate them back to molarity, as this reinforces an understanding of the different ways concentration can impact chemical reactions.