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At \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathbf{b})\) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Short Answer

Expert verified
(a) For the reverse reaction, \(K_{p\,reverse} = 0.54\) (b) For the multiplied reaction, \(K_{p\,multiplied} = 3.42\) (c) For the multiplied reaction, \(K_c = 280.79\)

Step by step solution

01

Calculate Kp for the reverse reaction

To obtain Kp for the reverse reaction, simply take the reciprocal of the given Kp: \( K_{p\,reverse}=\frac{1}{K_{p\,forward}} \) Reciprocal of the given Kp: \( K_{p\,reverse} = \frac{1}{1.85} = 0.54 \)
02

Calculate Kp for the multiplied reaction

To find Kp for the multiplied reaction, raise the Kp of the forward reaction to the power which we multiplied the reaction by: $$ K_{p\,multiplied} = (K_{p\,forward})^{n} $$ In this case, we multiplied the forward reaction by 2, so we get: $$ K_{p\,multiplied} = (1.85)^2 = 3.42 $$
03

Convert Kp to Kc

To convert from Kp to Kc, use the relationship: $$ K_p = K_c (RT)^{\Delta{n}} $$ Where R is the gas constant, T is the temperature in Kelvin, and ∆n is the difference of stoichiometric coefficients of products and reactants. In this case, we have: Temperature: T = 1000 K Gas constant: R = 0.0821 \( \frac{L\,atm}{mol\,K} \) Δn = (2 moles of SO3) - (2 moles of SO2 + 1 mole of O2) = -1 Now, solving for Kc: $$ K_c = \frac{K_p}{(RT)^{-1}} = 3.42 \times \left(\frac{0.0821\times1000}{1}\right)^{1} $$ $$ K_c = 3.42 \times 82.1 = 280.79 $$ Final answers: (a) Kp (reverse reaction) = 0.54 (b) Kp (multiplied reaction) = 3.42 (c) Kc (multiplied reaction) = 280.79

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Key Concepts

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

Kp and Kc relationship
Understanding the relationship between the equilibrium constants expressed in terms of pressure, denoted as \( K_p \), and concentration, denoted as \( K_c \), is crucial in the study of chemical equilibrium. These constants offer insights into the ratio of the products to the reactants at equilibrium under different measurement units—partial pressures for \( K_p \) and molar concentrations for \( K_c \).

The general relationship between \( K_p \) and \( K_c \) is governed by the formula:\[ K_p = K_c(RT)^{\Delta n} \], where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) represents the change in moles of gas particles when going from reactants to products (products minus reactants).

When dealing with exercises that include this relationship, always check if the temperature is given in Kelvin and verify that you're considering the proper units for \( R \). Additionally, the calculation of \( \Delta n \) must accurately reflect the stoichiometry of the balanced equation.
Equilibrium constant for reverse reactions
The direction of a chemical equation is essential when determining the equilibrium constant. For a given chemical equilibrium, if the reaction is reversed, the equilibrium constant for the reverse reaction, \( K_{p\_reverse} \), is the reciprocal of the equilibrium constant for the forward reaction, \( K_{p\_forward} \). Thus, the mathematical representation is:\[ K_{p\_reverse} = \frac{1}{K_{p\_forward}} \].

This principle was used in the textbook exercise, where we found the reverse \( K_p \) of \( SO_3 \) decomposing into \( SO_2 \) and \( O_2 \) by using the reciprocal of the given \( K_p \) for the forward reaction. This concept emphasizes that equilibrium is a dynamic balance where the forward and reverse reactions are inversely related. When assisting students with such problems, highlight the importance of understanding reaction directions and how they affect equilibrium constants.
Determining equilibrium constants
Determining the equilibrium constant for a chemical reaction is a fundamental skill in chemistry that describes the extent of the reaction at equilibrium. For any balanced chemical equation, the equilibrium constant can be calculated if the concentrations or pressures of the reactants and products at equilibrium are known.

In some cases, you may need to adjust the equilibrium constant for a reaction that has been multiplied by a factor. For instance, if you double the coefficients in the balanced equation, the new equilibrium constant, \( K_{p\_multiplied} \), will be the square of the original \( K_p \), as seen in the exercise with \[ K_{p\_multiplied} = (K_{p\_forward})^{n} \], where \( n \) is the factor by which the reaction is multiplied.

Furthermore, through precise calculation and understanding of the relation between different forms \( K_p \) and \( K_c \), and the stoichiometry of the reaction, students can confidently determine the equilibrium constants required for various reactions.

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Most popular questions from this chapter

Consider \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g),\) \(\Delta H=-904.4 \mathrm{~kJ} .\) How does each of the following changes affect the yield of \(\mathrm{NO}\) at equilibrium? Answer increase, decrease, or no change: (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathrm{c})\) decrease \(\left[\mathrm{O}_{2}\right]\) (d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

(a) How does a reaction quotient differ from an equilibrium constant? (b) If \(Q_{c}

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ An equilibrium mixture in a \(5.00-\mathrm{L}\) vessel at \(100{ }^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of \(\mathrm{NOBr}, 3.08 \mathrm{~g}\) of \(\mathrm{NO},\) and \(4.19 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) (a) Calculate \(K_{c}\) (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of NOBr?

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

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