Problem 48 - Chemistry: An Atoms-Focused Approach

Question

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

Step-by-Step Solution

Answer

Question: Calculate the equilibrium constants for the following reactions based on the given balanced chemical equation: 2 NO(g) + O2(g) ↔ 2 NO2(g), Kc = 5 x 10^12. a. NO(g) + 0.5 O2(g) ↔ NO2(g) b. 2 NO2(g) ↔ 2 NO(g) + O2(g) c. NO2(g) ↔ NO(g) + 0.5 O2(g) Answer: The equilibrium constants for the given reactions are approximately: a. Kc_a ≈ 2.236 x 10^6 b. Kc_b ≈ 2 x 10^(-13) c. Kc_c ≈ 4.47 x 10^(-7)

Step 1: Reaction a: NO(g) + 0.5 O2(g) ↔ NO2(g)

To determine the equilibrium constant for this reaction, we can observe that it is obtained by dividing the original reaction by a factor of 2. Whenever we divide a reaction by a factor, we raise the equilibrium constant to the power of (1/factor). In this case, the factor is 2. Therefore, we'll raise Kc to the power of 1/2: Kc_a = (Kc)^(1/2) = (5 x 10^12)^(1/2)

Step 2: Calculating Kc_a

Use a calculator to compute the value of Kc_a: Kc_a = √(5 x 10^12) ≈ 2.236 x 10^6

Step 3: Reaction b: 2 NO2(g) ↔ 2 NO(g) + O2(g)

For this reaction, we can see that it is the reverse of the original reaction. Whenever we reverse a reaction, we invert the equilibrium constant, taking its reciprocal. Thus, Kc_b = 1/Kc: Kc_b = 1/(5 x 10^12)

Step 4: Calculating Kc_b

Use a calculator to compute the value of Kc_b: Kc_b ≈ 2 x 10^(-13)

Step 5: Reaction c: NO2(g) ↔ NO(g) + 0.5 O2(g)

For this reaction, we can observe that it is the reverse of reaction a. As we did for reaction b, we need to invert the equilibrium constant Kc_a for reaction c. Thus, Kc_c = 1/Kc_a: Kc_c = 1/(2.236 x 10^6)

Step 6: Calculating Kc_c

Use a calculator to compute the value of Kc_c: Kc_c ≈ 4.47 x 10^(-7) The equilibrium constants for the given reactions are approximately: a. Kc_a ≈ 2.236 x 10^6 b. Kc_b ≈ 2 x 10^(-13) c. Kc_c ≈ 4.47 x 10^(-7)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry where the rate of the forward reaction equals the rate of the reverse reaction. At this stage, the concentrations of reactants and products remain constant over time.

However, it's important to note that these concentrations are not necessarily equal. Instead, the balance is reached when both reactions occur at equal rates, not equal concentrations.

The state of chemical equilibrium can be influenced by various factors such as temperature, pressure, and concentration. By understanding this, students can predict the behavior and composition of a chemical system when it is at equilibrium.

Reversible Reactions

In chemistry, many reactions do not go to completion but instead proceed simultaneously in both directions. These are known as reversible reactions. In a reversible reaction, products can revert back to reactants when the conditions favor this reversal.

An equilibrium position is established where the concentration of reactants and products remains stable. In the exercise, the reaction is shown to be reversible, symbolized by the double arrow (↔). This means that at some point, the rate of the forward reaction equals the rate of the reverse reaction, leading to equilibrium.

• Forward reaction: Converts reactants to products.
• Reverse reaction: Converts products back to reactants.

A deeper understanding of reversible reactions allows us to manipulate reaction conditions to favor the production of desired products.

Reaction Quotient

The reaction quotient, represented as Q, is similar to the equilibrium constant, K, except it applies when a reaction is not at equilibrium.

The formula to calculate Q for a particular reaction involves the initial concentrations or partial pressures of the reactants and products. By comparing Q with K, you can determine the direction in which the reaction will proceed to reach equilibrium:

• If Q < K, the reaction will proceed forward, forming more products.
• If Q > K, the reaction will proceed in reverse, forming more reactants.
• If Q = K, the reaction is at equilibrium.

Understanding the reaction quotient helps predict how changes in conditions affect the reaction's direction.

Stoichiometry

Stoichiometry is the bread and butter of understanding chemical reactions. It deals with the quantitative relationships between reactants and products in a chemical reaction. In simple terms, it helps us know how much of each substance is involved and produced in a chemical reaction.

Stoichiometry makes use of balanced chemical equations, which show the number of moles of each reactant and product involved in the reaction.

In our exercise, stoichiometry becomes important where reactions are compared by transforming coefficients, for example:

• If a reaction is halved, like reaction a in the exercise, the stoichiometric coefficients in the balanced equation are divided by 2.
• In consequence, when stoichiometry changes, we adjust the derived equilibrium constant accordingly, as seen by raising or reciprocating values.

This constant consideration of stoichiometry ensures that we accurately reflect real-life chemical processes.