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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 $$ 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) $$

Short Answer

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

Step by step solution

01

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 \( 1000 \mathrm{K} \).
02

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.
03

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, \[ K_2 = \frac{1}{(K_1)^2} = \frac{1}{(6.66 \times 10^{-12})^2} = 2.25 \times 10^{22}. \]
04

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 \( 2.25 \times 10^{22} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The photographs below show what occurs when a solution of potassium chromate is treated with a few drops of concentrated hydrochloric acid. Some of the bright yellow chromate ion is converted to the orange dichromate ion. \(2 \mathrm{CrO}_{4}^{2-}(\mathrm{aq})+2 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell)\) (a) Explain this experimental observation in terms of Le Chatelier's principle. (b) What would you observe if you treated the orange solution with sodium hydroxide? Explain your observation.

At \(1800 \mathrm{K},\) oxygen dissociates very slightly into its atoms. $$ \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{p}}=1.2 \times 10^{-10} $$ If you place 1.0 mol of \(\mathrm{O}_{2}\) in a \(10 .\).L. vessel and heat it to \(1800 \mathrm{K}\), how many \(\mathrm{O}\) atoms are present in the flask?

Which of the following correctly relates the two equilibrium constants for the two reactions shown? \(\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{Cl}_{2}(\mathrm{g}) \quad K_{1}\) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NOCl}(\mathrm{g}) \quad K_{2}\) (a) \(K_{2}=-K_{1}^{2}\) (c) \(K_{2}=1 / K_{1}^{2}\) (b) \(K_{2}=1 /\left(K_{1}\right)^{1 / 2} \quad\) (d) \(K_{2}=2 K_{1}\)

Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \operatorname{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ \(K_{\mathrm{c}}\) is 0.190 at \(73^{\circ} \mathrm{C} .\) Suppose you place \(0.500 \mathrm{mol}\) of COBr, in a \(2.00-\) I. flask and heat it to \(73^{\circ} \mathrm{C}\) (see Study Question 17 ). After equilibrium has been achieved, you add an additional 2.00 mol of \(\mathrm{CO}\) (a) How is the equilibrium mixture affected by adding more CO? (b) When equilibrium is reestablished, what are the new equilibrium concentrations of \(\mathrm{COBr}_{2}, \mathrm{CO},\) and \(\mathrm{Br}_{2} ?\) (c) How has the addition of CO affected the percentage of COBr \(_{2}\) that decomposed?

Decide whether each of the following statements is true or false. If false, change the wording to make it true. (a) The magnitude of the equilibrium constant is always independent of temperature. (b) When two chemical equations are added to give a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants of the summed equations. (c) The equilibrium constant for a reaction has the same value as \(K\) for the reverse reaction. (d) Only the concentration of \(\mathrm{CO}_{2}\) appears in the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) (e) For the reaction \(\operatorname{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) the value of \(K\) is numerically the same, whether the amount of \(\mathrm{CO}_{2}\) is expressed as moles/liter or as gas pressure.

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