/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 For the reaction \(\mathrm{SO}_{... [FREE SOLUTION] | 91Ó°ÊÓ

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For the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\), if \(K_{\mathrm{p}}=K_{\mathrm{c}}(R T)^{x}\). where the symbols have usual meaning then the value of \(x\) is (assuming ideality): (a) \(-1\) (b) \(-\frac{1}{2}\) (c) \(\frac{1}{2}\) (d) 1

Short Answer

Expert verified
The value of \( x \) is \( -\frac{1}{2} \), option (b).

Step by step solution

01

Understand the Reaction Equation

The given reaction is \( \mathrm{SO}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) \). This is a gas-phase reaction where stoichiometric coefficients are not integers. Recognize the standard equation connecting \( K_p \) and \( K_c \).
02

Recall the Formula Connecting Kp and Kc

For a gaseous reaction, the relation between \( K_p \) and \( K_c \) is given by: \[ K_p = K_c (RT)^{\Delta n} \] where \( \Delta n \) is the change in moles of gas.
03

Calculate Change in Moles (Δn)

Calculate \( \Delta n \) for the reaction:- Products: \(1\) mole of \( \mathrm{SO}_3 \)- Reactants: \(1 + \frac{1}{2} = \frac{3}{2} \) moles of gas (\( \mathrm{SO}_2 \) and \( \mathrm{O}_2 \))Now calculate \( \Delta n = 1 - \frac{3}{2} = -\frac{1}{2} \).
04

Substitute Δn into the Formula

Substitute \( \Delta n = -\frac{1}{2} \) into the formula: \[ K_p = K_c (RT)^{-\frac{1}{2}} \] This shows that \( x = -\frac{1}{2} \).
05

Select the Correct Answer Option

Match your calculated value of \( x \), which is \( -\frac{1}{2} \), with the given options. The correct answer is option (b).

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Key Concepts

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

Mole Concept
The mole concept is an essential foundation in chemistry. It allows chemists to count atoms and molecules in a given sample by relating mass to the number of particles. A mole is defined as the amount of substance containing the same number of entities as there are atoms in 12 grams of carbon-12. This number, known as Avogadro's number, is approximately \(6.022 \times 10^{23}\).Understanding the mole concept is crucial for solving chemical equations, especially in reactions. It provides a bridge between the atomic scale and the macroscopic quantities we measure. In the context of gas-phase reactions, such as the one given in this exercise, the mole concept helps determine the number of moles involved in the reactants and products.
  • It allows for the conversion of masses of substances to the number of particles.
  • Helps in quantifying reactants and products in a chemical reaction.
  • Allows for the prediction of yields and stoichiometry in reactions.
By applying the mole concept to reactions, chemists can ascertain the exact proportions of substances required or produced.
Gas-Phase Reactions
Gas-phase reactions involve substances in the gaseous state. These reactions are often studied due to their relevance in industrial processes and atmospheric chemistry. One of the main characteristics of gas-phase reactions is their dependency on pressure and volume, making them unique compared to reactions in other phases.The stoichiometry of gas-phase reactions is essential, as it directly impacts the calculations of constants like \( K_p \) and \( K_c \). For example, in the sulfur dioxide and oxygen reaction to form sulfur trioxide, the balanced equation is crucial to understand the number of moles of gases involved.
  • In gas-phase reactions, it's important to account for pressure, as it affects the rate and extent of reaction.
  • Stoichiometry requires careful accounting of gaseous moles to correctly apply the relation between \( K_p \) and \( K_c \).
  • Temperature also plays a significant role, impacting the ideal gas law applicability and the calculation of equilibrium constants.
Understanding these properties is vital, as it ensures accurate predictions and interpretations of reaction behavior in gaseous systems.
Relation Between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is a fundamental concept in the study of chemical equilibrium for gas-phase reactions. \( K_c \) is the equilibrium constant expressed in terms of concentrations, while \( K_p \) is expressed in terms of partial pressures.This relationship is defined by the equation:\[ K_p = K_c (RT)^{\Delta n} \]Where:
  • \( R \) is the ideal gas constant (0.0821 L·atm/mol·K).
  • \( T \) is the temperature in Kelvin.
  • \( \Delta n \) is the change in moles of gas (moles of products - moles of reactants).
For gas-phase reactions, it is crucial to calculate \( \Delta n \) accurately, as it affects the exponent in the equation. In the example given, \( \Delta n \) was calculated to be \(-\frac{1}{2}\), thus showing that \( K_p = K_c (RT)^{-\frac{1}{2}} \).Understanding this relation:- Affords insights into how pressure and concentration change as a system approaches equilibrium.- Clarifies how reaction conditions and stoichiometry impact the equilibrium constants.Mastery of this concept allows for predicting and manipulating the conditions necessary for achieving the desired state in chemical reactions.

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Most popular questions from this chapter

The \(\mathrm{p} K_{a}\) of acetylsalicyclic and (aspirin) is \(3.5\). The \(\mathrm{pH}\) of gastric juice in human stomach is about \(2-3\) and the \(\mathrm{pH}\) in the small intestine is about 8. Aspirin will be (a) unionised in the small intestine and in the stomach (b) completely ionised in the small intestine and in the stomach (c) ionised in the stomach and almost unionised in the small intestine (d) ionised in the small intestine and almost unionised in the stomach.

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Consider the following equilibrium in a closed container $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant \(\left(K_{p}\right)\) and degree of dissociation \((\alpha)\) ? (a) neither \(K_{p}\) nor \(\alpha\) changes (b) both \(K_{p}\) and \(\alpha\) change (c) \(K_{p}\) changes, but \(\alpha\) does not change (d) \(K_{p}\) does not change, but \(\alpha\) changes

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