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Consider the following reaction at \(100^{\circ} \mathrm{C}\) : $$ \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{NOCl}(g) $$ (a) Write an equilibrium constant expression for the reaction and call it \(K^{\prime}\). (b) Write an equilibrium constant expression for the decomposition of \(\mathrm{NOCl}\) to produce one mole of chlorine gas. Call the constant \(K^{\prime \prime}\). (c) Relate \(K^{\prime}\) and \(K^{\prime \prime}\).

Short Answer

Expert verified
Question: Write equilibrium constant expressions for the following reaction and then relate the two constants: (a) $\mathrm{NO}(g) + \frac{1}{2}\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{NOCl}(g)$ (b) Decomposition of $\mathrm{NOCl}$ to produce one mole of chlorine gas. Answer: (a) We can write the equilibrium constant expression for the given reaction as: $$K' = \frac{[\mathrm{NOCl}]}{[\mathrm{NO}] \times [\mathrm{Cl}_2]^{1/2}}$$ (b) The equilibrium constant expression for the decomposition of $\mathrm{NOCl}$ is: $$K'' = \frac{[\mathrm{NO}] \times [\mathrm{Cl}_2]^{1/2}}{[\mathrm{NOCl}]}$$ (c) Based on the expressions, we can relate the two constants as follows: $$K' = \frac{1}{K''}$$

Step by step solution

01

(a) Writing an equilibrium constant expression for the given reaction

To write the equilibrium constant expression, we will use the formula: $$K' = \frac{[\mathrm{NOCl}]}{[\mathrm{NO}] \times [\mathrm{Cl}_2]^{1/2}}$$
02

(b) Writing an equilibrium constant expression for the decomposition of NOCl

First, we need to write the balanced equation for the decomposition of NOCl: $$\mathrm{NOCl}(g) \rightleftharpoons \mathrm{NO}(g) + \frac{1}{2}\mathrm{Cl}_{2}(g) $$ Now let's write the equilibrium constant expression for this reaction: $$K'' = \frac{[\mathrm{NO}] \times [\mathrm{Cl}_2]^{1/2}}{[\mathrm{NOCl}]}$$
03

(c) Relating K' and K''

From part (a) and (b), we can relate \(K'\) and \(K''\) as follows: \(K' = \frac{[\mathrm{NOCl}]}{[\mathrm{NO}] \times [\mathrm{Cl}_2]^{1/2}}\) and \(K'' = \frac{[\mathrm{NO}] \times [\mathrm{Cl}_2]^{1/2}}{[\mathrm{NOCl}]}\) Since the two expressions are inverse of each other, we can write: $$K' = \frac{1}{K''}$$

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Key Concepts

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

Equilibrium Constant Expression
When we speak of chemical reactions, a fascinating concept to explore is the equilibrium constant expression. This refers to a special formula that denotes the ratio of the concentration of the products to the reactants, raised to the power of their coefficients, at equilibrium. In our example, the reaction involving nitrogen monoxide (NO) and chlorine (Clâ‚‚) gases forming nitrosyl chloride (NOCl), the equilibrium constant expression (denoted as \(K'\)) is crucial for understanding the reaction's behavior at equilibrium conditions at \(100^\circ \mathrm{C}\).

The expression \(K' = \frac{[\mathrm{NOCl}]}{[\mathrm{NO}] \times [\mathrm{Cl}_2]^{1/2}}\) illustrates that the concentration of the products and reactants are in a delicate balance. This balance is not a static one but denotes a dynamic state where both the forward and reverse reactions continue to occur at equal rates, maintaining a constant ratio of concentrations. It is important to note that only gaseous and aqueous species are included in the expression, while pure solids and liquids are omitted due to their constant concentration values.
Reaction Quotient
Alongside the equilibrium constant, we have the reaction quotient (\(Q\)), a measure that serves a similar purpose but under non-equilibrium conditions. It is calculated using the same formula as the equilibrium constant expression but with the current concentrations of reactants and products, not necessarily those at equilibrium.

Consider our textbook example: If we wanted to assess the system's status at a given moment before reaching equilibrium, we would calculate the reaction quotient with the current concentrations. The reaction quotient permits us to predict which direction the reaction will shift to achieve equilibrium - if \(Q\) is less than \(K'\), the forward reaction is favored to produce more products; if higher, the system will shift to favor the reverse reaction. This concept becomes a vital tool in predicting the outcome of reactions when changes occur in a chemical system.
Le Chatelier's Principle
Le Chatelier's Principle is the compass guiding us through the changes in reaction conditions. It indicates that when a dynamic equilibrium is disturbed, the system will adjust itself to counteract the effect of the disturbance and restore equilibrium. To visualize this concept, let's dissect the reaction we've been examining.

If, for example, we add additional NO or Clâ‚‚ to the system, Le Chatelier's Principle suggests that the equilibrium will shift to the right to form more NOCl, minimizing the effects of the change. Conversely, if NOCl is added, the system will shift to the left to reduce the concentration of NOCl by producing more NO and Clâ‚‚. Parameters such as pressure, volume, temperature, and the presence of a catalyst also play significant roles and can cause shifts in the equilibrium position, all of which are encompassed by this principle.
Equilibrium Systems in Chemistry
Finally, the broader context of equilibrium systems in chemistry revolves around the remarkable ability of certain reactions to reach a state of balance, where the rate of the forward reaction equals that of the reverse one. Equilibrium systems are pervasive in both academic studies and real-world applications. From synthesizing chemicals to biological systems and industrial processes, understanding how equilibrium can be achieved and manipulated is foundational.

Our textbook reaction is a classic example of an equilibrium system, where nitrogen monoxide reacts with chlorine to form nitrosyl chloride in a reversible process. It beautifully exhibits the dynamic nature of chemical equilibrium and the applicability of concepts such as the equilibrium constant expression, reaction quotient, and Le Chatelier's Principle. These tools enable chemists to predict and control the behavior of reactions, allowing us to harness chemistry for a plethora of innovative and essential applications.

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Most popular questions from this chapter

Consider the decomposition at \(25^{\circ} \mathrm{C}\) of one mole of NOBr gas into \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\) gases. At equilibrium, the concentrations of \(\mathrm{NOBr}, \mathrm{NO},\) and \(\mathrm{Br}_{2}\) gases are \(0.0162 \mathrm{M}\), \(0.0011 \mathrm{M},\) and \(0.072 \mathrm{M},\) respectively. (a) Write a balanced equation for the reaction. (b) Calculate \(K\) for the reaction at \(25^{\circ} \mathrm{C}\). (Note that the gases need to be expressed as pressure in atm.)

At \(627^{\circ} \mathrm{C}, K=0.76\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ Calculate \(K\) at \(627^{\circ} \mathrm{C}\) for (a) the synthesis of one mole of sulfur trioxide gas. (b) the decomposition of two moles of \(\mathrm{SO}_{3}\).

Consider the following decomposition at \(80^{\circ} \mathrm{C}\). \(\mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g) \quad K=0.054\) Twenty grams of \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) are sealed in a \(5.0-\mathrm{L}\) flask and heated to \(80^{\circ} \mathrm{C}\) (a) What is the total pressure in the flask at equilibrium? (b) How many grams of \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) are left in the flask at equilibrium?

At a certain temperature, the equilibrium constant for the following reaction is 0.0472 . $$ \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) $$ All gases are at an initial pressure of \(0.862 \mathrm{~atm} .\) (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the initial total pressure with the total pressure of the gases at equilibrium. Would that relation be true of all gaseous systems?

Consider the following reaction at \(75^{\circ} \mathrm{C}\) : $$ 3 \mathrm{R}(s)+2 \mathrm{Q}(g) \rightleftharpoons \mathrm{A}(g)+5 \mathrm{~B}(l) \quad K=9.4 $$ A \(10.0-\mathrm{L}\) sample contains \(0.30 \mathrm{~mol}\) of \(\mathrm{R}\) and \(\mathrm{Q}\) and \(0.50 \mathrm{~mol}\) of \(\mathrm{A}\) and \(\mathrm{B}\). In which direction will the reaction proceed?

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