/*! 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 37 Consider the following reaction ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following reaction at a certain temperature: $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ An equilibrium mixture contains \(1.0 \mathrm{~mol} \mathrm{Fe}, 1.0 \times 10^{-3} \mathrm{~mol} \mathrm{O}_{2}\), and \(2.0 \mathrm{~mol} \mathrm{Fe}_{2} \mathrm{O}_{3}\) all in a 2.0-L container. Calculate the value of \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant, K, for the given reaction is approximately \(1.6 \times 10^{10}\).

Step by step solution

01

Calculate the initial concentrations

In order to calculate the initial concentrations, we need to divide the number of moles of each substance by the volume of the container. We are given: - 1.0 mol of Fe in a 2.0 L container. - \(1.0\times10^{-3}\) mol of \(O_2\) in a 2.0 L container. - 2.0 mol of Fe\(_2\)O\(_3\) in a 2.0 L container. Calculate the initial concentrations by dividing moles by the volume (2 L): - \([Fe] = \frac{1.0}{2.0} = 0.5\: M\) - \([O_2] = \frac{1.0\times10^{-3}}{2.0} = 5.0\times10^{-4}\: M\) - \([Fe_2O_3] = \frac{2.0}{2.0} = 1.0\: M\)
02

Write the expression for the equilibrium constant, K

Based on the balanced chemical equation \(4\: Fe(s) + 3\: O_2(g) \rightleftharpoons 2\: Fe_2O_3(s)\), the expression for the equilibrium constant K is given by: \[K = \frac{[Fe_2O_3]^2}{[Fe]^4[O_2]^3}\]
03

Substitute the initial concentrations into the K expression and solve for K

Now that we have the initial concentrations of each substance, we can substitute these values into the K expression: \[K = \frac{[1.0]^2}{[0.5]^4[5.0\times10^{-4}]^3}\] Next, calculate the value of K: \[K = \frac{1}{(0.5)^4(5.0\times10^{-4})^3} \approx 1.6 \times 10^{10}\] The equilibrium constant, K, for the given reaction is approximately \(1.6 \times 10^{10}\).

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Key Concepts

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

Equilibrium Constant Expression
Understanding the equilibrium constant expression is essential when exploring chemical reactions that reach a state of balance, where the rate of the forward reaction equals the rate of the reverse reaction. For the given balanced chemical equation, \[4 \text{Fe}(s) + 3 \text{O}_2(g) \rightleftharpoons 2 \text{Fe}_2\text{O}_3(s)\],the equilibrium constant expression, denoted as \(K\), encapsulates the ratio of the concentration of the products to the reactants, each raised to the power of their stoichiometric coefficients in the balanced equation. Solid substances, like iron (\text{Fe}) and iron(III) oxide (\text{Fe}_2\text{O}_3), are omitted from the expression because their concentrations do not change in a closed system, leaving only the gaseous constituents.When writing the expression: \[K = \frac{[\text{Fe}_2\text{O}_3]^2}{[\text{Fe}]^4[\text{O}_2]^3}\],it's crucial to note that \([X]\) represents the molar concentration of a substance. For gases, partial pressures may be used instead of concentrations. To strengthen comprehension, take a moment to recognize that exponents reflect the stoichiometry: for every two moles of iron(III) oxide produced, four moles of iron and three moles of oxygen are consumed. Calculating the equilibrium constant provides insights into the position of equilibrium, and whether products or reactants are favored under certain conditions.
Initial Concentration Calculation
A pivotal step in understanding chemical equilibrium is calculating the initial concentrations of reactants and products. These concentrations are essential inputs when determining the equilibrium constant \(K\). To find them, divide the number of moles of each substance by the volume of the reaction container.
In our example, the volumes were all the same, simplifying our task. Quick calculations yield the concentrations:
  • \([Fe] = 0.5 M\)
  • \([O_2] = 5.0\times10^{-4} M\)
  • \([Fe_2O_3] = 1.0 M\)
Remember that solutes dissolved in liquid and gases have units of molarity (M), which is moles per liter (mol/L). The accuracy of these initial concentration calculations is paramount as it directly affects the accuracy of the equilibrium constant determined later. As a tip for improvement, always double-check your units during calculation and ensure coherence throughout the process. It's also worth noting that these 'initial' concentrations refer to the moment right before the system begins to approach equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is an invaluable tool for predicting the behavior of a system at equilibrium when subjected to an external change such as concentration, pressure, or temperature modulation. It posits that if a dynamic equilibrium is disturbed, the system will adjust to minimize that disturbance and re-establish equilibrium. In simpler terms, the system 'fights back' against the imposed change.
For example, increasing the concentration of reactants will shift the equilibrium towards the formation of more products to lower the reactants' concentration, while a decrease leads to the opposite effect. Similarly, changing the volume or pressure of a gaseous system will shift the equilibrium towards the side with fewer gas molecules if pressure is increased, or towards the side with more gas molecules if pressure is decreased. Temperature changes will drive the equilibrium in the direction that either absorbs or releases heat, depending on whether the reaction is endothermic or exothermic.Understanding Le Chatelier's Principle can also aid in optimizing industrial processes by predicting system responses to controlled alterations, which could improve yield and efficiency. It's a key concept for students to master, not only for solving textbook exercises but also for its practical applications in various scientific fields.

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Most popular questions from this chapter

Consider the following reactions. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) and \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.

Explain the difference between \(K, K_{\mathrm{p}}\), and \(Q\).

The equilibrium constant is \(0.0900\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift? a. \(P_{\mathrm{H}_{2} \mathrm{O}}=1.00 \mathrm{~atm}, P_{\mathrm{Cl}_{2} \mathrm{O}}=1.00 \mathrm{~atm}, P_{\mathrm{HOCl}}=1.00 \mathrm{~atm}\) b. \(P_{\mathrm{H}_{2} \mathrm{O}}=200 .\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=49.8\) torr, \(P_{\mathrm{HOCl}}=21.0\) torr c. \(P_{\mathrm{H}_{2} \mathrm{O}}=296\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=15.0\) torr, \(P_{\mathrm{HOCl}}=20.0\) torr

At \(327^{\circ} \mathrm{C}\), the equilibrium concentrations are \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.15 \mathrm{M}\), \([\mathrm{CO}]=0.24 M\), and \(\left[\mathrm{H}_{2}\right]=1.1 M\) for the reaction $$ \mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) $$ Calculate \(K_{\mathrm{p}}\) at this temperature.

An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ At equilibrium the total pressure and the density of the gaseous mixture were \(1.80 \mathrm{~atm}\) and \(1.60 \mathrm{~g} / \mathrm{L}\), respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

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