Question

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a 10.00-L vessel is found to contain \(0.050 \mathrm{~mol}\) \(\mathrm{CH}_{3} \mathrm{OH}, 0.850 \mathrm{~mol} \mathrm{CO},\) and \(0.750 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K}\). Calculate \(K_{c}\) at this temperature.

Short answer

The equilibrium constant, \(K_c\), for the given reaction at 500 K can be calculated by first determining the concentrations of each species and then plugging them into the expression \(K_c = \frac{[\mathrm{CH_3OH}]}{[\mathrm{CO}][\mathrm{H_2}]^2}\). The concentrations are found to be: \([\mathrm{CH_3OH}] = 0.005 \text{ M}\), \([\mathrm{CO}] = 0.085 \text{ M}\), and \([\mathrm{H_2}] = 0.075 \text{ M}\). Substituting these values into the equilibrium constant expression, we get \(K_c = \frac{0.005}{(0.085)(0.075)^2} = 12.35\). Thus, the equilibrium constant at this temperature is 12.35.

Step-by-step solution

Write the equilibrium constant expression for the given reaction.

The general expression for an equilibrium constant is given by the formula: \[K_c = \frac{\text{Product Concentrations}^{a}}{\text{Reactant Concentrations}^{b}}\] where 'a' and 'b' are the stoichiometric coefficients of the balanced chemical reaction. For the given reaction, the equilibrium constant expression is: \[K_c = \frac{[\mathrm{CH_3OH}]}{[\mathrm{CO}][\mathrm{H_2}]^2}\]

Calculate the concentration of each species.

To find the concentration, we will use the following formula: \[\text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}}\] Calculate the concentration of each species in the mixture: \[[\mathrm{CH_3OH}] = \frac{0.050 \text{ mol}}{10.00 \text{ L}} = 0.005 \text{ M}\] \[[\mathrm{CO}] = \frac{0.850 \text{ mol}}{10.00 \text{ L}} = 0.085 \text{ M}\] \[[\mathrm{H_2}] = \frac{0.750 \text{ mol}}{10.00 \text{ L}} = 0.075 \text{ M}\]

Substitute the concentrations into the equilibrium constant expression.

Plug the calculated concentrations of the species from Step 2 into the expression for \(K_c\): \[K_c = \frac{0.005}{(0.085)(0.075)^2}\]

Calculate \(K_c\).

Calculate the equilibrium constant \(K_c\) using the values obtained in the previous step: \[K_c = \frac{0.005}{(0.085)(0.075)^2} = 12.35\] The equilibrium constant \(K_c\) for this reaction at 500 K is 12.35.

Key concepts

Chemical Equilibrium

When discussing chemical reactions that can occur in both directions, we talk about chemical equilibrium. In a simple sense, it's a state where the forward reaction (reactants converting into products) and the reverse reaction (products converting back into reactants) happen at the same rate. At this point, concentrations of reactants and products remain unchanged over time. This does not mean that the substances have stopped reacting, but rather that they continue to convert from one side to the other at equal rates.

Equilibrium is described using the equilibrium constant, denoted as \(K_c\) for concentration in moles per liter (M). This constant provides insight into the position of equilibrium. A large \(K_c\) value suggests the reaction favors products, while a small \(K_c\) indicates it favors reactants.

In the reaction to produce methanol, achieving equilibrium means the rates at which carbon monoxide and hydrogen form methanol are identical to the rates at which methanol decomposes back into these reactants. By determining \(K_c\), we can understand whether methanol is predominantly forming or decomposing.

Reaction Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantity relationships of the elements and compounds involved in the reactions. In any chemical equation, stoichiometry is represented by the coefficients in the balanced equation, which tell us the ratio in which reactants combine and products form.

For the production of methanol from carbon monoxide and hydrogen, the reaction is as follows: \(\text{CO} (g) + 2 \text{H}_2 (g) \leftrightharpoons \text{CH}_3\text{OH} (g)\).
Here, the stoichiometry tells us that 1 mole of carbon monoxide reacts with 2 moles of hydrogen to produce 1 mole of methanol. These ratios are essential when writing the expression for \(K_c\), as they ensure the correct powers to which concentrations are raised. For the given reaction, the coefficient of hydrogen is 2, and therefore, its concentration is squared in the \(K_c\) expression.

This accurate balancing ensures all calculations reflect the chemical reality of the situation, providing a reliable basis for predicting how varying the concentrations can shift the equilibrium.

Concentration Calculation

Calculating concentration is pivotal when dealing with chemical equilibria to help understand the quantities present at equilibrium. Concentration, commonly expressed in molarity (M), tells us the amount of a substance in a given volume of solution and is calculated using the formula \(\text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}}\).

In the methanol production equilibrium problem, we have known amounts of reactants and products in moles, and a specified volume of the reaction vessel. The concentration of each species was determined as follows:

• \([\text{CH}_3\text{OH}] = \frac{0.050 \text{ mol}}{10.00 \text{ L}} = 0.005 \text{ M}\)
• \([\text{CO}] = \frac{0.850 \text{ mol}}{10.00 \text{ L}} = 0.085 \text{ M}\)
• \([\text{H}_2] = \frac{0.750 \text{ mol}}{10.00 \text{ L}} = 0.075 \text{ M}\)

These concentrations are key inputs for finding \(K_c\). By substituting them into the equilibrium constant expression, the balance of the reaction can be quantitatively understood, ultimately allowing for precise control of chemical processes.