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Calculate a value for the equilibrium constant for the reaction $$\mathbf{O}_{2}(g)+\mathbf{O}(g) \rightleftharpoons \mathbf{O}_{3}(g)$$.given $$\begin{aligned}& \mathrm{NO}_{2}(g) \stackrel{h v}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) & & K=6.8 \times 10^{-49} \\\\\mathrm{O}_{3}(g)+\mathrm{NO}(g) & \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & & K=5.8 \times 10^{-34}\end{aligned}$$.(Hint: When reactions are added together, the equilibrium expressions are multiplied.) (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Short Answer

Expert verified
The equilibrium constant for the reaction O鈧(g) + O(g) 鈬 O鈧(g) is approximately 8.52 脳 10鹿鈦.

Step by step solution

01

Analyze given reactions

We are given the following reactions: 1. NO鈧(g) 鈬 NO(g) + O(g) with K鈧 = 6.8 脳 10鈦烩伌鈦 2. O鈧(g) + NO(g) 鈬 NO鈧(g) + O鈧(g) with K鈧 = 5.8 脳 10鈦宦斥伌 We want to manipulate these reactions in a way that results in the desired reaction: O鈧(g) + O(g) 鈬 O鈧(g).
02

Determine a suitable combination of given reactions

ToAdd the given reactions together in a suitable manner, we can reverse reaction 1 and add it to reaction 2. This will result in the following reaction: 1. -(NO鈧(g) 鈬 NO(g) + O(g)) 2. O鈧(g) + NO(g) 鈬 NO鈧(g) + O鈧(g) -------------------------- Result: O鈧(g) + O(g) 鈬 O鈧(g)
03

Calculate the new equilibrium constants for the manipulated reaction

Since we have reversed reaction 1, the equilibrium constant for this manipulated reaction will be the inverse of the initial equilibrium constant: K'鈧 = 1 / K鈧 = 1 / (6.8 脳 10鈦烩伌鈦)
04

Multiply the equilibrium constants for the combined reactions

Now, we can multiply K'鈧 by K鈧 to obtain the equilibrium constant for the desired reaction: K鈧 = K'鈧 脳 K鈧 = (1 / (6.8 脳 10鈦烩伌鈦)) 脳 (5.8 脳 10鈦宦斥伌)
05

Calculate the value of the equilibrium constant

Finally, we can calculate the value of K鈧: K鈧 = (1 / 6.8 脳 10鈦烩伌鈦) 脳 (5.8 脳 10鈦宦斥伌) K鈧 = (1 / 6.8) 脳 (5.8 脳 10鹿鈦) K鈧 = 0.147 脳 5.8 脳 10鹿鈦 K鈧 鈮 8.52 脳 10鹿鈦 Thus, the equilibrium constant for the reaction O鈧(g) + O(g) 鈬 O鈧(g) is approximately 8.52 脳 10鹿鈦.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is fundamental in chemical reactions, including synthesis and decomposition. It is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. In this balanced state, no net change occurs in the concentrations of reactants and products. However, both processes continue to happen at equal rates, maintaining the system's stability.

When a system reaches equilibrium, it is often represented by an equilibrium constant, denoted as K, which is a numerical value that expresses the ratio of product concentrations to reactant concentrations, each raised to the power of their respective coefficients from the balanced equation. For the reaction \( A + B \rightleftharpoons C + D \), the equilibrium constant K is given by the expression: \[ K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \] where the square brackets indicate concentrations, and a, b, c, d are the stoichiometric coefficients. This constant is crucial for predicting the direction of the reaction and understanding the reaction's extent under given conditions.
Le Chatelier's Principle
Le Chatelier's principle provides insight into the behavior of a system when changes are introduced. It postulates that if a dynamic equilibrium is disturbed by changing conditions, such as concentration, pressure, or temperature, the equilibrium will shift to oppose the change and establish a new equilibrium.

This principle is utilized to predict the effects of such changes. For instance, if the concentration of a reactant is increased, the system will respond by shifting the equilibrium to the right, producing more products to counteract the change. Conversely, decreasing the concentration of a product causes the equilibrium to shift to the left, producing more reactants.

Application in Exercises

Applying Le Chatelier's principle in calculations involves understanding how alterations affect K. When reactions are added together to form a new one, as shown in the step-by-step solution, the principle helps understand why and how the equilibrium constants of individual reactions relate to the new reaction's K value.
Reaction Quotient
The reaction quotient, Q, plays a pivotal role in predicting the direction of equilibrium shifts. It is calculated using the same formula as the equilibrium constant K, but with the initial concentrations instead of the equilibrium concentrations.

The equation for Q is: \[ Q = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \], and it reflects the state of a reaction mixture at any point in time, unlike K, which only represents the mixture at equilibrium. By comparing the values of Q and K, we can determine the reaction's progress:
  • If \( Q < K \), the forward reaction is favored, and more products will be formed.
  • If \( Q > K \), the reverse reaction is favored, and more reactants will be formed.
  • If \( Q = K \), the system is at equilibrium, and no net change occurs.
Through the use of the reaction quotient, students can anticipate how a reaction will proceed and adjust the conditions to achieve a desired outcome or state of equilibrium.

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Most popular questions from this chapter

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$.Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.0 moles of pure NOCl in a 2.0 -L flask b. 1.0 mole of NOCI and 1.0 mole of \(\mathrm{NO}\) in a 1.0 -L flask c. 2.0 moles of NOCl and 1.0 mole of \(\mathrm{Cl}_{2}\) in a 1.0 -L flask

Consider the following exothermic reaction at equilibrium: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$. Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

For the following endothermic reaction at equilibrium: $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T\) ) d. decreasing the volume (constant \(T\) ) e. adding Ne(g) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )

Consider the decomposition of the compound \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) as follows:$$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$.When a 5.63 -g sample of pure \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\) was sealed into an otherwise empty 2.50 -L flask and heated to \(200 .^{\circ} \mathrm{C},\) the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate \(K\) for this reaction.

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