Problem 28 - Chemistry: The Central Science

Question

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\). An equilibrium mixture in a 2.00-L vessel is found to contain \(0.0406\) mol \(\mathrm{CH}_{3} \mathrm{OH}, 0.170 \mathrm{~mol} \mathrm{CO}\), and \(0.302 \mathrm{~mol} \mathrm{H}_{2}\) at \(500 \mathrm{~K}\). Cal- culate \(K_{c}\) at this temperature.

Step-by-Step Solution

Answer

The equilibrium constant Kc for the reaction CO(g) + 2H2(g) ⇌ CH3OH(g) at 500 K is approximately 1.21, considering the given concentrations of CH3OH, CO, and H2 in the 2.00 L vessel.

Step 1: Write the balanced chemical equation

The balanced equation for the reaction is already given: CO(g) + 2H2(g) ⇌ CH3OH(g)

Step 2: Calculate the concentrations of all species at equilibrium

The given amounts of CH3OH, CO, and H2 are in moles, so to find their concentrations in the 2.00-L vessel, we need to divide the moles by the volume of the reaction: [CH3OH] = 0.0406 mol / 2.00 L = 0.0203 M [CO] = 0.170 mol / 2.00 L = 0.0850 M [H2] = 0.302 mol / 2.00 L = 0.151 M

Step 3: Use the equilibrium constant expression to find Kc

The equilibrium constant expression for this reaction is: \(K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\) Now plug in the equilibrium concentrations: \(K_c = \frac{0.0203}{(0.0850)(0.151)^2}\) Now, perform the calculation: \(K_c \approx 1.21\) The equilibrium constant Kc for this reaction at 500 K is approximately 1.21.

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant (K_c) is a crucial figure that helps us understand the balance between reactants and products at equilibrium. It is derived from the concentrations of the products and reactants, raised to the power of their coefficients in the balanced equation.

For the methanol production reaction:
CO(g) + 2H₂(g) ⇌ CH₃OH(g)
The expression for the equilibrium constant, K_c, is given by:

\[K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\]
This formula shows the concentrations of methanol ([CH₃OH]), carbon monoxide ([CO]), and hydrogen ([H₂]) at equilibrium. To find K_c, the concentrations must be plugged into this expression.

Knowing K_c gives insights into the extent of a reaction. A larger K_c suggests that the position of equilibrium lies towards the products, meaning more methanol is formed at equilibrium. It's important to note that K_c is specific to a given temperature and will change if the temperature changes.

Catalyzed Reaction

A catalyzed reaction involves the use of a catalyst to speed up the rate of a chemical reaction without being consumed in the process. In the methanol production process, this catalyst often involves metals such as copper or zinc, which facilitate the interaction between carbon monoxide (CO) and hydrogen (H₂) molecules.

Here’s how the catalysis works:

• The catalyst provides an alternate pathway for the reaction with a lower activation energy, which means the reaction can proceed more rapidly at a given temperature.
• It doesn't alter the position of equilibrium or the equilibrium constant ( K_c ); it only speeds up how quickly that equilibrium is reached.

The employment of catalysts is crucial in industrial processes as it reduces energy consumption and increases efficiency, making processes like methanol production economically viable.

Methanol Production

Methanol ( CH₃OH ) is a highly versatile chemical used in a variety of applications, including as an industrial solvent, antifreeze, and more recently, as a biofuel. The commercial production of methanol is done through the catalyzed reaction of carbon monoxide (CO) and hydrogen (H₂).

The process can be summarized as follows:

• Carbon monoxide and hydrogen are combined in the presence of a catalyst under high pressure and temperature.
• The balanced chemical reaction is: CO(g) + 2H₂(g) ⇌ CH₃OH(g).
• This reaction system is governed by equilibrium principles, where both the rate of forward and reverse reactions are balanced, resulting in the formation of methanol.
• Efficient use of catalysts ensures that the reaction can be economically viable by improving yields and reducing production costs.

Methanol production is a fine example of applying chemical equilibrium and catalysis concepts to achieve practical and industrial uses.