Problem 31 - 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: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\). An equilibrium mixture in a \(2.00\) - L vessel is found to contain \(0.0406 \mathrm{~mol}\) \(\mathrm{CH}_{3} \mathrm{OH}, 0.170 \mathrm{~mol} \mathrm{CO}\), and \(0.302 \mathrm{~mol} \mathrm{H}\) at \(500 \mathrm{~K}\). Calculate \(K_{c}\) at this temperature.

Step-by-Step Solution

Answer

The equilibrium constant Kc for the given reaction at 500 K, CO(g) + 2 H₂(g) ⇌ CH₃OH(g), is 5.04 when the equilibrium concentrations are [CH₃OH]=0.0203 M, [CO]=0.0850 M, and [H₂]=0.151 M.

Step 1: Write the expression for Kc using the balanced chemical equation

For the reaction: CO(g) + 2 H₂(g) ⇌ CH₃OH(g) The equilibrium constant expression, Kc, is: \[K_c = \frac{[\mathrm{CH}_{3}\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_{2}]^2}\]

Step 2: Calculate the equilibrium concentrations

Divide the number of moles of each substance by the volume of the container (2.00 L) to obtain the equilibrium concentrations: [CH₃OH] = \(\frac{0.0406 \mathrm{~mol}}{2.00 \mathrm{~L}}\) = 0.0203 M [CO] = \(\frac{0.170 \mathrm{~mol}}{2.00 \mathrm{~L}}\) = 0.0850 M [H₂] = \(\frac{0.302 \mathrm{~mol}}{2.00 \mathrm{~L}}\) = 0.151 M

Step 3: Substitute the equilibrium concentrations into the expression for Kc and solve for Kc

Now, we will plug the equilibrium concentrations into the Kc expression: \[K_c = \frac{[0.0203]}{[0.0850][0.151]^2}\] Now solve for Kc: \(K_c = \(5.04) So, the equilibrium constant Kc for the given reaction at 500 K is 5.04.

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for any budding chemist. But don't worry, the concept is quite straightforward once you get to know it. In essence, chemical equilibrium occurs when the rates of the forward and reverse reactions in a closed system become equal. At this point, the concentrations of reactants and products remain constant over time, but it's important to note that the reactions are still happening; they've just achieved a state of balance. Imagine a see-saw that has two children of equal weight on each end, perfectly balanced. They're still there, and they can still move, but the see-saw stays level. That's a lot like chemical equilibrium.

When we talk about the equilibrium in the context of the methanol production reaction, it means that the rate at which carbon monoxide and hydrogen react to form methanol is the same as the rate at which methanol decomposes back into these gases. Once this state is reached, even though both reactions continue to occur, the amount of methanol, carbon monoxide, and hydrogen in the vessel doesn't change.

Reaction Quotient

The reaction quotient, denoted as Q, is like a snapshot of a reaction at any point in time. It has the same form as the equilibrium constant expression, but it can be calculated before the system has reached equilibrium. By comparing Q to the equilibrium constant, Kc, you can predict which direction the reaction will proceed to reach equilibrium.

• If Q < Kc, the forward reaction is favored, and more products will be formed.
• If Q > Kc, the reverse reaction is favored, and more reactants will be formed.
• If Q = Kc, the system is at equilibrium, and no net change will occur.

When working with the methanol reaction, if we were to calculate Q at any moment before reaching equilibrium using the same expression for Kc, we could determine the direction in which the reaction is likely to proceed.

Le Chatelier's Principle

Le Chatelier's principle is a handy guideline for predicting how a system at equilibrium responds to changes in concentration, temperature, or pressure. According to this principle, if you tweak one aspect of a system in equilibrium, the system will adjust itself to counteract that change and restore a new equilibrium.

For example, if we increase the concentration of CO in our methanol reaction, the principle suggests that the reaction will adjust by making more methanol to reduce the additional CO. Similarly, if we were to decrease the temperature of a system where the forward reaction is exothermic, the system would shift to produce more products to generate heat and counterbalance the change.

This insight helps chemists control the yield of a reaction by altering conditions in a way that favors either the forward or reverse reaction, based on what they're trying to achieve.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of reactants and products present when the reaction has reached equilibrium. These concentrations are used to calculate the equilibrium constant, Kc, which is a ratio that reflects the proportion of products to reactants at equilibrium. Understanding how to find these values is a critical skill.

In our methanol problem, we calculated equilibrium concentrations by dividing the moles of each substance by the volume of the vessel. These concentrations were then plugged into the Kc expression to solve for the equilibrium constant. It’s crucial for students to grasp this calculation process, as knowing the equilibrium concentrations can also provide insights into the position of equilibrium and the extent to which a reaction will proceed under certain conditions.