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In the relation, \(K_{p}=K_{c}(R T)^{\mathrm{a} n}\) the value of \(\Delta n\) is (a) number of moles of gaseous reactants - number of moles of gaseous products in a balanced equation (b) number of moles of gaseous products - number of moles of gaseous reactants in a balanced equation (c) number of moles of gaseous products \(\times\) number of moles of gaseous reactants in a balanced equation (d) number of moles of gaseous reactants + number of moles of gaseous products in balanced equation.

Short Answer

Expert verified
The value of \(\Delta n\) is the number of moles of gaseous products minus the number of moles of gaseous reactants in a balanced equation (option b).

Step by step solution

01

Understand the Relation between Kp and Kc

The equation given relates the equilibrium constants in terms of partial pressures (Kp) and concentrations (Kc). Here, R is the gas constant, T is the temperature in Kelvin, and \(\Delta n\) represents the change in the number of moles of gas going from reactants to products.
02

Define \(\Delta n\)

The term \(\Delta n\) in the equation is defined as the difference in the number of moles of gaseous products and the number of moles of gaseous reactants. It represents the net change in the number of moles of gas after a reaction has reached equilibrium.
03

Identify the Correct Option

Using the definition of \(\Delta n\) from Step 2, we can look at the options provided to determine which one correctly expresses \(\Delta n\).
04

Eliminate Incorrect Options

Options (c) and (d) can be quickly eliminated because \(\Delta n\) is a difference, not a product or sum. Option (a) suggests that \(\Delta n\) is the number of moles of gaseous reactants minus the number of moles of gaseous products, which is the opposite of the correct definition. Therefore, option (b) is correct.

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Key Concepts

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

Equilibrium Constant (Kp)
In chemistry, understanding how reactions reach an equilibrium state is crucial. The equilibrium constant (Kp) is one such measure that indicates the extent of a reaction at equilibrium when dealing with gases. It is expressed in terms of partial pressures. Kp is determined by taking the partial pressures of the gaseous products, raising each to the power of its stoichiometric coefficient, and dividing by the partial pressure of each reactant also raised to its stoichiometric coefficient.
When a chemical reaction achieves equilibrium in a closed system, the ratio of the products to reactants remains constant at a given temperature. This ratio is what we call Kp, signifying that despite the ongoing forward and reverse reactions, the system's observable properties remain unchanged. As Kp is dimensionless, it provides a straightforward way to predict the position of equilibrium; higher values suggest a greater concentration of products relative to reactants.
Equilibrium Constant (Kc)
Moving beyond gaseous systems, we encounter the equilibrium constant (Kc). This version of the equilibrium constant is based on the molar concentrations (also known as molarity) of reactants and products in solution. In a balanced chemical equation, the concentration of each product (in moles per liter) is raised to the power of its coefficient, and the result is divided by the product of the concentrations of each reactant raised to the power of its respective coefficient.
Since Kc directly deals with concentrations, it is temperature dependent—like Kp—but it is more appropriate to use in reactions where the substances are in a liquid solution or any phase other than gas. A higher Kc similarly suggests a reaction that favors product formation, and it is particularly useful when comparing different reactions or predicting the direction of equilibrium shifts.
Relation between Kp and Kc
The relationship between Kp and Kc is pivotal for chemists to convert between different expressions of equilibrium constants. This relationship can be described by the equation

\(K_{p}=K_{c}(R T)^{abla n}\)


where R is the ideal gas constant, T is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas. This equation signifies that if there is a change in the number of moles of gases during the reaction, the two constants are related by the factor \((R T)^{\Delta n}\). When \(\Delta n\) is zero, meaning there's no change in the number of moles of gases between reactants and products, Kp and Kc become numerically equal. However, if \(\Delta n\) is not zero, this factor has to be considered to accurately convert between Kp and Kc.
Change in Number of Moles of Gas (Δn)
The term \(\Delta n\) is significant in the context of the reaction's stoichiometry as it represents the change in the number of moles of gas when a reaction proceeds to equilibrium. To find \(\Delta n\), we subtract the sum of the stoichiometric coefficients of gaseous reactants from the sum of the coefficients of gaseous products.
For instance, if a reaction has more gaseous products than reactants, \(\Delta n\) will be positive, indicating an increase in the number of gaseous molecules, and vice versa. This change impacts the volumes of gases and, consequently, their partial pressures. It’s this alteration that the equation for Kp takes into account by the factor \((R T)^{\Delta n}\), making \(\Delta n\) a crucial factor when relating Kp and Kc.

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Most popular questions from this chapter

For the following reaction: $$ \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} $$ The value of \(K_{c}\) is \(8.2 \times 10^{4} .\) What will be the value of \(K_{c}\) for the reverse reaction? (a) \(8.2 \times 10^{4}\) (b) \(\frac{1}{8.2 \times 10^{4}}\) (c) \(\left(8.2 \times 10^{4}\right)^{2}\) (d) \(\sqrt{8.2 \times 10^{4}}\)

\(0.05\) mole of \(\mathrm{NaOH}\) is added to 5 litres of water. What will be the \(\mathrm{pH}\) of the solution? (a) 12 (b) 7 (c) 2 (d) 10

What is the minimum concentration of \(\mathrm{SO}_{4}^{2-}\) required to precipitate \(\mathrm{BaSO}_{4}\) in a solution containing \(1 \times 10^{-4}\) mole of \(\mathrm{Ba}^{2+} ?\left(K_{i p}\right.\) for \(\left.\mathrm{BaSO}_{4}=4 \times 10^{-10}\right)\) (a) \(4 \times 10^{-10} \mathrm{M}\) (b) \(2 \times 10^{-10} \mathrm{M}\) (c) \(4 \times 10^{-6} \mathrm{M}\) (d) \(2 \times 10^{-3} \mathrm{M}\)

Given below are the dissociation constant values of few acids. Arrange them in order of increasing acidic strength. \(\mathrm{H}_{2} \mathrm{SO}_{3}=1.3 \times 10^{-2}, \mathrm{HNO}_{2}=4 \times 10^{-4}\) \(\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}, \mathrm{HCN}=4 \times 10^{-10}\) (a) \(\mathrm{HCN}<\mathrm{CH}_{3} \mathrm{COOH}<\mathrm{HNO}_{2}<\mathrm{H}_{2} \mathrm{SO}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}<\mathrm{HNO}_{2}<\mathrm{HCN}<\mathrm{H}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}<\mathrm{HCN}<\mathrm{H}_{2} \mathrm{SO}_{3}<\mathrm{HNO}_{2}\) (d) \(\mathrm{HNO}_{2}<\mathrm{H}_{2} \mathrm{SO}_{3}<\mathrm{CH}_{3} \mathrm{COOH}<\mathrm{HCN}\)

If the equilibrium constant for the given reaction is \(0.25\) \(\mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2}\), then the equilibriumconstant for the reaction \(\frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) will be (a) 1 (b) 2 (c) 3 (d) 4

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