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The relation between \(K_{n}\) and \(K_{c}\) for the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (c) \(K_{p}^{p}=K_{c} /(R T)^{2}\) (d) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} / \mathrm{RT}\)

Short Answer

Expert verified
Option (a): \(K_p = K_c (RT)^{-1}\).

Step by step solution

01

Understand the Reaction

The balanced chemical equation given is \(2 \text{NO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons 2 \text{NOCl(g)}\). This reaction consists of gases and involves a change in the number of gas molecules, going from 3 molecules on the reactant side to 2 on the product side.
02

Recall the Relationship Between \(K_p\) and \(K_c\)

The relationship between \(K_p\) and \(K_c\) for a gaseous equilibrium is given by the formula: \(K_p = K_c (RT)^{\Delta n}\), where \(\Delta n\) is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants), \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
03

Calculate \(\Delta n\)

For the given reaction, calculate \(\Delta n\): there are 2 moles of gaseous products (\(2 \text{NOCl}\)) and 3 moles of gaseous reactants (\(2 \text{NO} + \text{Cl}_2\)). Thus, \(\Delta n = 2 - 3 = -1\).
04

Apply the \(K_p\) \(K_c\) Equation

Substitute \(\Delta n = -1\) into the equation \(K_p = K_c (RT)^{\Delta n}\). This yields \(K_p = K_c (RT)^{-1}\).
05

Identify the Correct Option

Compare the derived expression \(K_p = K_c (RT)^{-1}\) with the given options. The correct relation matches option (a): \(K_p = K_c (RT)^{-1}\).

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Key Concepts

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

Gaseous Reactions
In chemistry, gaseous reactions involve the transformation of substances where gases are either reactants or products. These reactions are central to many processes and can happen under a variety of conditions, like in closed containers or open systems. Understanding these reactions is crucial because the behavior of gases can significantly influence the reaction dynamics. Gaseous reactions often involve a shift in the number or types of gas molecules, which can change the equilibrium states. For example, consider the reaction given: \(2 \text{NO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons 2 \text{NOCl(g)}\). Here, we start with three gaseous molecules on the reactant side and end with two gaseous molecules on the product side. Such transformations affect properties like pressure and volume, highlighting why it's essential to analyze these factors during equilibrium studies.
Equilibrium Constant
The equilibrium constant, denoted as \(K\), is a vital concept in chemical equilibrium, representing the ratio of the concentrations of products to reactants at equilibrium. It provides insight into the position of equilibrium and helps predict the direction of a reaction. The specific form of equilibrium constant depends on the phases of the substances involved. For gaseous reactions, two forms are predominantly discussed: \(K_c\) and \(K_p\).
  • \(K_c\): The equilibrium constant when concentrations are used. Calculated using molarity or moles per liter.
  • \(K_p\): The equilibrium constant for gases, calculated using partial pressures.
Both constants tie deeply into the laws governing chemical equilibria and offer a quantitative measure for gauging equilibrium under different conditions. While \(K_p\) is usually more convenient for gaseous reactions due to the direct involvement of pressure, \(K_c\) applies generally across states. What's important to remember is that these constants remain unchanged at a given temperature, signifying the consistency of equilibrium in a given reaction system.
Relationship Between Kp and Kc
The relationship between \(K_p\) and \(K_c\) is crucial in understanding how equilibrium constants adapt based on the nature of gases versus concentration. This link is described by the equation: \[K_p = K_c (RT)^{\Delta n}\]where:
  • \(R\): the universal gas constant (8.314 J/mol·K)
  • \(T\): temperature in Kelvin
  • \(\Delta n\): change in the moles of gas (calculated by subtracting the moles of gaseous reactants from the moles of gaseous products)
The equation highlights how the conditions of temperature and changes in moles of gas, represented by \(\Delta n\), shift the relation between \(K_p\) and \(K_c\). In simpler terms, when there is no change in moles of gas (\(\Delta n = 0\)), \(K_p\) equals \(K_c\). However, when there is a change, the relationship modifies proportionally to \(RT\) raised to the power of \(\Delta n\), like in our exercise where \(\Delta n = -1\). This alteration directly influences the reaction's dynamics at equilibrium.
Reaction Stoichiometry
Reaction stoichiometry centers on understanding the quantitative relationships between reactants and products in a chemical reaction. It involves using the balanced equation to calculate the proportions of elements and compounds. This concept is especially useful in predicting the amounts needed or produced during a chemical reaction. Let's break it down with the provided equation: For the reaction \(2 \text{NO(g)} + \text{Cl}_2\text{(g)} \rightleftharpoons 2 \text{NOCl(g)}\), stoichiometry tells us that:
  • 2 molecules of NO react with 1 molecule of \(\text{Cl}_2\)
  • 2 molecules of \(\text{NOCl}\) are formed
This 2:1:2 ratio is crucial for calculating the exact amounts of each substance participating in the reaction.
Stoichiometry allows one to transform theoretical equations into practical measures, such as determining how much reactant is required to achieve a desired product yield. It lays the foundation for deeper analysis of reactions, such as determining the limiting reactant, yield calculations, and understanding the overall efficiency of a reaction system.

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Most popular questions from this chapter

For the reaction: \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) at a given temperature, the equilibrium amount of \(\mathrm{CO}_{2}(\mathrm{~g})\) can be increased by (a) Adding a suitable catalyst (b) Adding an inert gas (c) Decreasing the volume of the container (d) Increasing the amount of \(\mathrm{CO}(\mathrm{g})\)

The rate constants for the forward and backward reactions of hydrolysis of ester are \(1.1 \times 10^{-2}\) and \(1.5 \times\) \(10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) respectively. The equilibrium constant of the reaction, \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is: (a) \(6.53\) (b) \(7.34\) (c) \(7.75\) (d) \(8.33\)

For equilibrium reaction: \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+14.6 \mathrm{~J}\), increase in tempera- ture would (a) Favour the formation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (b) Stop reaction (c) Favour the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) (d) No alter the equilibrium

In the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\), when 100 \(\mathrm{mL}\) of \(\mathrm{N}_{2}\) has reacted, the volumes of \(\mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are: (a) \(300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(300 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (b) \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (c) \(300 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(200 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\) (d) \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) and \(100 \mathrm{~mL}\) of \(\mathrm{NH}_{3}\)

For the hypothetical reactions, the equilibrium constant \((\mathrm{K})\) values are given: \(\mathrm{A} \rightleftharpoons \mathrm{B} \mathrm{K}_{1}=2\) \(\mathrm{B} \rightleftharpoons \mathrm{C} \mathrm{K}_{2}^{1}=4\) \(\mathrm{C} \rightleftharpoons \mathrm{D} \mathrm{K}_{3}^{2}=3\) The equilibrium constant \((\mathrm{K})\) for the reaction: \(\mathrm{A} \rightleftharpoons \mathrm{D}\) is (a) 3 (b) 6 (c) 12 (d) 24

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