Question

The equilibrium constant for the reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}(g) $$ is \(1.6 \times 10^{-2}\) at a certain temperature. If there are \(1.17 \times 10^{-2}\) mole of \(\mathrm{H}_{2}\) and \(3.46 \times 10^{-3}\) mole of \(\mathrm{CH}_{3} \mathrm{OH}\) at equilibrium in a \(5.60-\mathrm{L}\) flask, what is the concentration of \(\mathrm{CO} ?\)

Short answer

The concentration of CO at equilibrium is 8.85 M.

Step-by-step solution

Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the given reaction is\[K_c = \frac{[\mathrm{CH}_3\mathrm{OH}]}{[\mathrm{H}_2]^2[\mathrm{CO}]}\]where \([\mathrm{CH}_3\mathrm{OH}]\), \([\mathrm{H}_2]\), and \([\mathrm{CO}]\) are the equilibrium concentrations of methanol, hydrogen, and carbon monoxide, respectively.

Calculate the Concentrations at Equilibrium

The concentration of any substance in a solution is given by the formula \[\text{Concentration} = \frac{\text{moles}}{\text{volume (L)}}\]Substitute the given values to find\[[\mathrm{H}_2] = \frac{1.17 \times 10^{-2}}{5.60} = 2.0893 \times 10^{-3} \, M\]\[[\mathrm{CH}_3\mathrm{OH}] = \frac{3.46 \times 10^{-3}}{5.60} = 6.1786 \times 10^{-4} \, M\]

Substitute Values into the Equilibrium Expression

Plug the known concentrations into the equilibrium expression and solve for \([\mathrm{CO}]\): \[1.6 \times 10^{-2} = \frac{6.1786 \times 10^{-4}}{(2.0893 \times 10^{-3})^2[\mathrm{CO}]}\]

Solve for the Concentration of CO

Re-arrange the equation to solve for \([\mathrm{CO}]\):\[[\mathrm{CO}] = \frac{6.1786 \times 10^{-4}}{(1.6 \times 10^{-2})(4.3633 \times 10^{-6})}\]This evaluates to\[[\mathrm{CO}] = \frac{6.1786 \times 10^{-4}}{6.9813 \times 10^{-8}} = 8.85 \, M\]

Key concepts

Chemical Equilibrium

In the context of chemistry, equilibrium refers to the point in a reversible chemical reaction where the rates of the forward and reverse reactions are equal. This creates a state where the concentrations of all reactants and products remain constant over time, even though they are not equal. It’s like a balanced seesaw where the weight on both sides is equal but not necessarily the same type or amount.
For the reaction between hydrogen gas and carbon monoxide to produce methanol, represented as \(2 \mathrm{H}_2(g) + \mathrm{CO}(g) \rightleftarrows \mathrm{CH}_3\mathrm{OH}(g)\), the equilibrium constant \(K_c\) provides insight into the concentration ratios of products and reactants at equilibrium. Knowing and understanding how to use this constant is crucial for predicting how reaction conditions will affect product formation.

Concentration Calculation

To understand and predict the behavior of a chemical reaction at equilibrium, it is important to calculate the concentration of the substances involved. Concentration is a measure of how much solute is present in a given volume of solution and is usually expressed in molarity, denoted as \(M\) which is moles per liter.
In this example, we calculate the concentrations of hydrogen and methanol using the formula:

• Concentration \(= \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)

For instance, given \(1.17 \times 10^{-2}\) moles of \(\mathrm{H}_2\) in a 5.60 L flask, the concentration of \(\mathrm{H}_2\) is calculated to be \(2.0893 \times 10^{-3}\, M\). Similarly, the concentration of \(\mathrm{CH}_3\mathrm{OH}\) is found to be \(6.1786 \times 10^{-4}\, M\). These calculations play an essential role in determining the unknown concentrations involved in the equilibrium reaction.

Reaction Kinetics

Reaction kinetics is the study of the speed at which chemical reactions occur and the factors that affect this speed. In the context of equilibrium, these factors include concentration, temperature, surface area, and the presence of catalysts.
For the synthesis of methanol, understanding the kinetics helps chemists optimize the conditions under which the reaction proceeds most efficiently towards methanol formation. The role of catalysts, often a metal in industrial methanol synthesis, is particularly crucial. They help lower the activation energy, speeding up reactions without being consumed.

Methanol Synthesis

Methanol is synthesized industrially through the reaction of hydrogen and carbon monoxide in the presence of a catalyst. This is a classic example of a chemical reaction that reaches equilibrium, meaning not all the reactants are converted to products. Methanol synthesis is key in the production of important chemical feedstocks, fuel and as a potential clean energy carrier.
The equilibrium constant in this synthesis reflects how much methanol can be theoretically produced under a specific set of conditions. It is also important for estimating the reactions’ efficiency in larger scales, guiding adjustments in the industrial process for optimal yield. Understanding this balance between the forward and reverse reactions is essential for improving methanol production methods and meeting global demands.