Problem 22 - Chemistry and Chemical Reactivity

Question

The equilibrium constant \(K\) for the reaction $$ \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is \(6.66 \times 10^{-12}\) at \(1000 \mathrm{K}\). Calculate \(K\) for the reaction $$ 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

The equilibrium constant is \( 2.25 \times 10^{22} \).

Step 1: Write the given reaction and its equilibrium constant

The given reaction is \( \mathrm{CO}_2(\mathrm{g}) \rightleftarrows \mathrm{CO}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \) with \( K_1 = 6.66 \times 10^{-12} \) at \( 1000 \mathrm{K} \).

Step 2: Write the target reaction and its relation to the given reaction

The target reaction is \( 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_2(\mathrm{g}) \). This reaction is the reverse of the given reaction, multiplied by 2.

Step 3: Calculate the equilibrium constant for the target reaction

Since the target reaction is the reverse of the given reaction and is multiplied by 2, the equilibrium constant \( K_2 \) can be calculated as the inverse of the square of \( K_1 \). Thus, \[ K_2 = \frac{1}{(K_1)^2} = \frac{1}{(6.66 \times 10^{-12})^2} = 2.25 \times 10^{22}. \]

Step 4: Conclusion

The equilibrium constant for the reaction \( 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_2(\mathrm{g}) \) is \( 2.25 \times 10^{22} \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is an essential concept in chemistry, where the rate of the forward reaction equals the rate of the backward reaction, resulting in the concentrations of reactants and products remaining constant over time. This dynamic state does not mean that the reactions stop but that both the forward and reverse reactions proceed at the same pace.
At equilibrium, a specific ratio of the concentrations of products to reactants is observed, which is described by the equilibrium constant, denoted as \( K \). For a reaction such as \( aA + bB \rightleftarrows cC + dD \), the equilibrium constant is expressed as:

• \( K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)

Here, the square brackets indicate the concentration of the respective species, and \( a, b, c, \) and \( d \) are the stoichiometric coefficients. When calculating \( K \), only species in the gaseous or aqueous state are included. Equilibrium constants provide insight into the position of equilibrium — whether the products or reactants are favored at a given temperature.

Reversible Reactions

Reversible reactions are reactions that can proceed in both the forward and reverse directions. They are depicted by a double arrow (\( \rightleftarrows \)) in a chemical equation. This type of reaction is noteworthy because it can reach a state of equilibrium as we discussed earlier.
The concept is crucial due to its application in understanding chemical processes that approach a balance of reactants and products over time. For example, in the Haber process used for ammonia production, both the forward and reverse reactions play a significant role in determining the yield of ammonia under specific conditions.
Consider the following reversible reaction:

• \( \mathrm{A} + \mathrm{B} \rightleftarrows \mathrm{C} + \mathrm{D} \)

This shows that as products are formed from reactants in the forward reaction, they can also convert back to reactants in the reverse reaction. The equilibrium constant helps us quantify which direction is favored under equilibrium conditions.

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative balance of reactants and products in a chemical reaction. It involves using the coefficients from a balanced chemical equation to determine the proportions of substances involved in the reaction.
For reversible reactions and equilibrium calculations, stoichiometry is crucial since it influences the form of the equilibrium expression. Let's consider:

• The balanced reaction is: \( aA + bB \rightleftarrows cC + dD \).

The equilibrium constant expression will directly depend on these coefficients. For example, if the reaction is multiplied by a factor, the equilibrium constant expression changes. If the reaction is reversed, the equilibrium constant is inverted.

In the given exercise, we saw this in action with the reaction \( 2 \mathrm{CO} + \mathrm{O}_2 \rightleftarrows 2 \mathrm{CO}_2 \), which was derived from the given reaction by reversing and multiplying by 2. This modification required us to adjust the equilibrium constant by the reciprocal of its square, emphasizing the significant role stoichiometry plays in equilibrium calculations.