/*! 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 The equilibrium-constant express... [FREE SOLUTION] | 91Ó°ÊÓ

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The equilibrium-constant expression for a gas reaction is $$ K_{c}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{2}\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}\left[\mathrm{O}_{2}\right]^{3}} $$ Write the balanced chemical equation corresponding to this expression.

Short Answer

Expert verified
The balanced chemical equation is \( 2\mathrm{H}_2\mathrm{S} + 3\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + 2\mathrm{SO}_2 \).

Step by step solution

01

Understand the Denominator Expression of Kc

The equilibrium constant expression \( K_c \) has a denominator of \( \left[\mathrm{H}_2\mathrm{~S}\right]^2\left[\mathrm{O}_2\right]^3 \). This indicates that the reactants in the balanced chemical equation include 2 moles of \( \mathrm{H}_2\mathrm{~S} \) and 3 moles of \( \mathrm{O}_2 \). This means that two molecules of hydrogen sulfide and three molecules of oxygen are the reactants.
02

Understand the Numerator Expression of Kc

The expression \( \left[\mathrm{H}_2\mathrm{O}\right]^2\left[\mathrm{SO}_2\right]^2 \) in the numerator denotes that the products are \( \mathrm{H}_2\mathrm{O} \) and \( \mathrm{SO}_2 \), each with a coefficient of 2. This means there are 2 moles of water and 2 moles of sulfur dioxide as products.
03

Use Stoichiometry to Formulate the Balanced Equation

Based on the equilibrium expression \( K_c \), the balanced chemical equation can be written as:\[ 2 \mathrm{H}_2\mathrm{~S} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + 2 \mathrm{SO}_2 \]This equation has equal numbers of atoms of each element on both sides, thus confirming it is balanced.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, often represented as \( K_c \) when dealing with concentrations, is a key concept in chemical equilibrium. It quantifies the ratio of the concentrations of products to the concentrations of reactants for a reaction at equilibrium. This value provides insight into the position of equilibrium:
  • If \( K_c \) is large, it means the equilibrium is weighted towards the products—indicating a tendency to form products.
  • If \( K_c \) is small, it suggests the reactants are favored at equilibrium.
For the given reaction, the expression \( K_{c} = \frac{[\mathrm{H}_2\mathrm{O}]^2[\mathrm{SO}_2]^2}{[\mathrm{H}_2\mathrm{S}]^2[\mathrm{O}_2]^3} \) is used to describe how products and reactants distribute themselves when the reaction reaches equilibrium. Each concentration in the expression is raised to a power that corresponds to its stoichiometric coefficient in the balanced equation.
Balanced Chemical Equation
A balanced chemical equation represents a reaction with equal numbers of each type of atom on both sides. This reflects the law of conservation of mass, meaning matter cannot be created or destroyed in a chemical reaction.
In the step-by-step solution, the balanced chemical equation derived was:\[ 2 \mathrm{H}_2\mathrm{~S} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + 2 \mathrm{SO}_2 \]This tells us not only the reactants and products but also their molar ratios.Ensuring a balanced equation helps in understanding the stoichiometry of a reaction, which is crucial for predicting reaction behavior and calculating reactants or products needed or produced in a given amount.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction, based on the balanced equation. It involves calculating the amounts of reactants needed or products formed during a reaction.
  • This includes mole ratios derived from the coefficients of the balanced equation.
  • Stoichiometry can determine how much of a reactant is necessary to completely react with another reactant or how much product will form from given amounts of reactants.
In the balanced equation \( 2 \mathrm{H}_2\mathrm{~S} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + 2 \mathrm{SO}_2 \), each component's coefficient indicates the mole ratio needed for the reaction to proceed completely while maintaining mass conservation.
Gas Reactions
Gas reactions involve reactants and products in the gaseous state. These reactions are particularly interesting due to the behavior of gases under various conditions. In the context of chemical equilibria, gas reactions follow Le Chatelier's principle where changes in pressure, temperature, or concentration can shift the equilibrium position.
For our equation:\[ 2 \mathrm{H}_2\mathrm{~S} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + 2 \mathrm{SO}_2 \]the reaction occurs in the gaseous phase. This implies:
  • The reaction can be influenced by changes in pressure and temperature, given that gases behave predictably under varying conditions.
  • Understanding gas behavior is important, notably the ideal gas law, which helps in predicting how changes in conditions may affect the reaction outcomes, especially when a system is not at equilibrium.
Gas reactions are often studied for their dynamic equilibria and are critical in industrial applications where controlling reaction conditions is essential to optimize product yields.

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Most popular questions from this chapter

The following reaction has an equilibrium constant \(K_{c}\) equal to \(3.07 \times 10^{-4}\) at \(24^{\circ} \mathrm{C}\) $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ For each of the following compositions, decide whether the reaction mixture is at equilibrium. If it is not, decide which direction the reaction should go. a. \([\mathrm{NOBr}]=0.0720 M,[\mathrm{NO}]=0.0162 M,\left[\mathrm{Br}_{2}\right]=0.0123 M\) b. \([\mathrm{NOBr}]=0.121 M,[\mathrm{NO}]=0.0159 M,\left[\mathrm{Br}_{2}\right]=0.0139 M\) c. \([\mathrm{NOBr}]=0.103 M,[\mathrm{NO}]=0.0134 M,\left[\mathrm{Br}_{2}\right]=0.0181 M\) d. \([\mathrm{NOBr}]=0.0472 M,[\mathrm{NO}]=0.0121 M,\left[\mathrm{Br}_{2}\right]=0.0105 M\)

Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is manufactured industrially by the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ A gaseous mixture at \(500 \mathrm{~K}\) is \(0.020 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}, 0.10 \mathrm{M} \mathrm{CO}\), and \(0.10 M \mathrm{H}_{2} .\) What will be the direction of reaction if this mixture goes to equilibrium? The equilibrium constant \(K_{c}\) equals \(10.5\) at \(500 \mathrm{~K}\).

You place \(0.600\) mol of nitrogen, \(\mathrm{N}_{2}\), and \(1.800 \mathrm{~mol}\) of hydrogen, \(\mathrm{H}_{2}\), into a reaction vessel at \(450^{\circ} \mathrm{C}\) and \(10.0 \mathrm{~atm}\). The reaction is $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ What is the composition of the equilibrium mixture if you obtain \(0.048\) mol of ammonia, \(\mathrm{NH}_{3}\), from it?

Phosgene was used as a poisonous gas in World War I. At high temperatures it decomposes as follows: $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ with \(K_{c}=4.6 \times 10^{-3}\) at \(800 \mathrm{~K}\). a. A sample of \(6.55 \mathrm{~g}\) of \(\mathrm{COCl}_{2}\) is placed in a \(1.00\) - \(\mathrm{L}\) reaction vessel and heated to \(800 \mathrm{~K}\). What are the equilibrium concentrations of all of the species? b. What fraction of \(\mathrm{COCl}_{2}\) has decomposed? C. If \(3 \mathrm{~g}\) of carbon monoxide is inserted into the reaction vessel, what qualitative effect would this have on the fraction of \(\mathrm{COCl}_{2}\) that has decomposed?

What would you expect to be the effect of an increase of pressure on each of the following reactions? Would the pressure change cause the reaction to go to the right or left? a. \(\mathrm{CH}_{4}(g)+2 \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g)\) b. \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) c. \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\)

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