Problem 57 - Chemistry The Molecular Nature of Matter

Question

At \(773^{\circ} \mathrm{C}\), a mixture of \(\mathrm{CO}(g), \mathrm{H}_{2}(g)\), and \(\mathrm{CH}_{3} \mathrm{OH}(g)\) was allowed to come to equilibrium. The concentrations were determined to be: \([\mathrm{CO}]=0.105 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.250 \mathrm{M}\) \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.00261 M .\) Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$

Step-by-Step Solution

Answer

The equilibrium constant \(K_{c}\) for the reaction at \(773^{\text{\textdegree}} C\) is 0.3978.

Step 1: Write the expression for the equilibrium constant

For the given reaction \text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g), the equilibrium constant expression is written as: \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\).

Step 2: Substitute the equilibrium concentrations into the expression

Substitute the given concentrations into the equilibrium expression: \(K_c = \frac{0.00261}{0.105 \times (0.250)^2}\).

Step 3: Calculate the equilibrium constant

Perform the calculations to find the value of \(K_c\): \(K_c = \frac{0.00261}{0.105 \times 0.0625}\). Simplify this to find the value of \(K_c\).

Step 4: Simplify and find the answer

After the calculation, \(K_c = \frac{0.00261}{0.0065625}\). Therefore, \(K_c = 0.3978\).

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant is fundamental when dealing with chemical reactions that have reached a state of balance. At this point, the rate of the forward reaction equals that of the reverse reaction, and the concentrations of reactants and products remain constant.

Mathematically, we express the equilibrium constant, K_c, for a reaction in terms of the concentrations of the reactants and products, raised to the power of their respective stoichiometric coefficients in the balanced equation. For instance, if our reaction is A(g) + 2B(g) <=> C(g), the equilibrium constant would be K_c = \([\text{C}]\) / (\([\text{A}]\) \([\text{B}]^2\)).

On substituting the equilibrium concentrations into the equilibrium constant expression for our original exercise, the student correctly identified the constants for reactants and products, and by applying the formula derived from the balanced chemical equation, reached to a calculated K_c value, reflecting the ratio of product over reactants concentrations at equilibrium.

Reaction Quotient

The reaction quotient, Q_c, serves as a 'snapshot' of a reaction's position in relation to equilibrium. Unlike K_c, which is constant at a given temperature, Q_c changes as a reaction progresses. To determine Q_c, one uses the same formula as K_c, but with the initial concentrations of reactants and products instead of those at equilibrium.

By comparing Q_c to K_c, we can predict which direction the reaction will proceed to reach equilibrium. If Q_c < K_c, the forward reaction is favored. If Q_c > K_c, the reverse reaction is favored. If Q_c = K_c, the system is at equilibrium, with no net change in the quantities of reactants and products.

Equilibrium Concentrations

The equilibrium concentrations of reactants and products are the concentrations measured when a chemical system is in a state of dynamical equilibrium. These values are crucial as they are plugged into the equilibrium constant expression to calculate K_c.

In our example, the steps meticulously show how the concentrations of CO, H₂, and CH₃OH were used. When dealing with equilibrium problems, it's pivotal for students to correctly interpret and insert these values into the equilibrium expression. Accuracy here is essential for a correct calculation of the equilibrium constant, and thus in understanding the system's condition.

Le Chatelier's Principle

When a system at equilibrium is disrupted, it will adjust to counteract the change and re-establish equilibrium; this is Le Chatelier's Principle. This concept helps predict the effect of a change in conditions (concentration, pressure, temperature) on a system at equilibrium.

If we imagine increasing the concentration of a reactant in our exercise, according to Le Chatelier's Principle, the system would shift to favor the formation of more products, aiming to reduce the concentration of that reactant again. Le Chatelier’s Principle is quintessential in optimizing industrial chemical processes, like the synthesis of methanol in our case, where conditions are carefully controlled to maximize product yield.