Problem 27 - Chemistry The Central Science

Question

The following equilibria were attained at \(823 \mathrm{~K}\) : $$\begin{aligned} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c} &=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c} &=490 \end{aligned}$$ Based on these equilibria, calculate the equilibrium constant $$\text { for } \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \text { at } 823 \mathrm{~K} \text { . }$$

Step-by-Step Solution

Answer

The equilibrium constant for the reaction \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at 823 K can be calculated by combining the given equilibria and their constants: \(K_{c} = \frac{K_{c_{2}}}{K_{c_{1}}} = \frac{490}{67} \approx 7.31\).

Step 1: Write down the given equilibria with their equilibrium constants

We are given two equilibria with their equilibrium constants at 823 K: Equilibrium 1: \[\mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c_{1}}=67\] Equilibrium 2: \[\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) \quad K_{c_{2}}=490\] We want to find the equilibrium constant for the following reaction: \[\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\]

Step 2: Combine the given equilibria to obtain the desired reaction

In order to get the desired reaction, we need to subtract the first equilibrium from the second equilibrium: \[(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g)) - (\mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g))\] When we subtract the two equilibria, we get: \[\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\]

Step 3: Combine the equilibrium constants for the combined reaction

Now that we have combined the reactions, we need to combine their equilibrium constants. When we subtract two chemical reactions, we divide their equilibrium constants: \[K_{c} = \frac{K_{c_{2}}}{K_{c_{1}}}\] Using the given equilibrium constants, we calculate the equilibrium constant for the desired reaction: \[K_{c} = \frac{490}{67} \approx 7.31\]

Step 4: Report the answer

The equilibrium constant for the reaction \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at 823 K is approximately 7.31.

Key Concepts

Equilibrium Constants

The equilibrium constant, often denoted as \(K_c\) for reactions in solutions or gases, is a crucial concept in chemical equilibrium. It provides us with a ratio that compares the concentration of the products to the reactants when the reaction has reached equilibrium. These concentrations are raised to the power of their respective coefficients in the balanced chemical equation.

For the reaction \( ext{aA + bB} \rightleftharpoons \text{cC + dD}\), the equilibrium constant expression is: \[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]

• If \(K_c\) is greater than 1, the formation of products is favored at equilibrium.
• If \(K_c\) is less than 1, the formation of reactants is favored.

In this exercise, two given equilibria each had their own equilibrium constants \(K_{c_1} = 67\) and \(K_{c_2} = 490\). When we manipulated the reactions to derive a new one, the new constant was derived by dividing these two constants, yielding \(K_c \approx 7.31\). This value indicates the extent to which the products are formed at equilibrium.

Reversible Reactions

Reversible reactions are at the heart of chemical equilibrium. These are reactions where the reactants form products, which can then transform back into reactants. This cycle continues until a state of balance, or equilibrium, is reached.

• When a reaction is at equilibrium, both forward and reverse reactions occur at the same rate.
• This doesn’t mean the concentrations of reactants and products are equal, but that they remain constant over time.

In the given exercise, each equilibrium involving \(\text{CoO}, \text{H}_{2}, \text{CO}, \text{H}_{2}\text{O}, \text{CO}_{2}\) is reversible. Both reactions reach a point where the rate of the forward reaction equates to the reverse reaction, maintaining the system in equilibrium under the specified conditions. The manipulation of these reactions allowed for the calculation of the unknown equilibrium constant, emphasizing the reversibility's importance in chemical systems.

Reaction Quotients

The reaction quotient, \(Q\), is similar to the equilibrium constant, \(K_c\), but applies to a system that is not at equilibrium. It is useful for predicting the direction in which a reaction will proceed to reach equilibrium. Like \(K_c\), \(Q\) is calculated using the concentrations of the reactants and products.

• If \(Q < K_c\), the reaction will proceed in the forward direction to form more products.
• If \(Q > K_c\), the reaction will proceed in the reverse direction to form more reactants.
• If \(Q = K_c\), the system is at equilibrium.

In the original exercise, the given reactions were already at equilibrium, so \(K_c\) was used directly to find the equilibrium state of the new reaction. However, if the concentrations had been different or if the system was disturbed, the reaction quotient would help in determining how the reaction must proceed to achieve equilibrium once again. Understanding \(Q\) is essential for predicting the effects of changes in concentration or conditions on a chemical equilibrium system.