Problem 42 - Chemistry The Science in Context

Question

If the equilibrium constant \(K_{\mathrm{c}}\) for the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) is \(5 \times 10^{12}\) at a given temperature, what is the value of the equilibrium constant \(K_{\mathrm{c}}\) for each of the following reactions at the same temperature? a. \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g)\) b. \(2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\)

Step-by-Step Solution

Answer

The equilibrium constants for the related reactions are: Reaction a: \(K_{c_{a}} \approx 2.236 \times 10^6\) Reaction b: \(K_{c_{b}} = 2 \times 10^{-13}\) Reaction c: \(K_{c_{c}} \approx 4.472 \times 10^{-7}\)

Step 1: Reaction a

To compare the given reaction with reaction a, we need to divide the entire given reaction by 2: \(NO(g)+\frac{1}{2} O_2(g) \rightleftharpoons NO_2(g)\) When we divide the reaction by the coefficient 2, we must take the square root of the equilibrium constant: \(K_{c_{a}} = \sqrt{K_c} = \sqrt{5 \times 10^{12}}\)

Step 2: Reaction b

Reaction b is the reverse reaction of the given reaction: \(2 NO_2(g) \rightleftharpoons 2 NO(g) + O_2(g)\) The equilibrium constant for the reverse reaction is the reciprocal of the given equilibrium constant: \(K_{c_{b}} = \frac{1}{K_c} = \frac{1}{5 \times 10^{12}}\)

Step 3: Reaction c

To compare the given reaction with reaction c, we need to divide the given reaction by 2 and reverse it: \(NO_2(g) \rightleftharpoons NO(g) + \frac{1}{2} O_2(g)\) As in Reaction a, we will take the square root of the equilibrium constant since we are dividing the reaction by 2. Then, we will take the reciprocal of the result as for the reverse reaction: \(K_{c_{c}} = \frac{1}{\sqrt{K_c}} = \frac{1}{\sqrt{5 \times 10^{12}}}\) Now we can calculate the equilibrium constants for reactions a, b, and c: \(K_{c_{a}} = \sqrt{5 \times 10^{12}} \approx 2.236 \times 10^6\) \(K_{c_{b}} = \frac{1}{5 \times 10^{12}} = 2 \times 10^{-13}\) \(K_{c_{c}} = \frac{1}{\sqrt{5 \times 10^{12}}} \approx 4.472 \times 10^{-7}\)

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible chemical reaction where the rates of forward and reverse reactions are equal. At equilibrium, the concentrations of reactants and products remain constant over time, even though reactions continue to occur at the molecular level. This dynamic process indicates that the system has reached a balance, but it is important to note that equilibrium does not imply that the reactants and products are in equal amounts.

Reaction Coefficients

Reaction coefficients are essential in chemical equations as they denote the number of moles of each reactant and product involved in a reaction. These coefficients directly influence the equilibrium constant, represented as \(K_c\) for reactions at a particular temperature. For a general reaction:

• \(aA + bB \rightleftharpoons cC + dD\)

The equilibrium constant \(K_c\) can be expressed as:\[K_c = \frac{{[C]^c [D]^d}}{{[A]^a [B]^b}}\]

This formula shows how the reaction coefficients play a pivotal role in determining \(K_c\). In practice, when the coefficients of a balanced chemical equation change, it impacts the calculation of \(K_c\). For instance, if you divide the entire equation by a number, like 2, you would need to apply a corresponding operation to the equilibrium constant, such as taking the square root.

Reverse Reactions

In chemistry, a reverse reaction is one where the products can react to form the original reactants. The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. This concept is crucial when examining chemical equilibria, as it affects how we understand and calculate equilibrium constants for reactions that have been reversed.

• If a forward reaction has an equilibrium constant \(K_c\), its reverse reaction will have \(K_{c_{reverse}} = \frac{1}{K_c}\).
• This relationship helps in predicting how concentrations of substances will shift when conditions in a reaction change, such as pressure or temperature.

Understanding the concept of reverse reactions is important for accurately analyzing reaction systems and predicting their behavior under various circumstances.