/*! 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 6 The following reaction is found ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 6

The following reaction is found to be at equilibrium at 25掳C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) What is the expression for the equilibrium constant, \(K_{\mathrm{c}} ?\) (A) \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}\) (B) \(\frac{2\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}\) (C) \(\frac{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{SO}_{3}\right]^{2}}\) (D) \(\frac{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}{2\left[\mathrm{SO}_{3}\right]}\)

Short Answer

Expert verified
Option (C), \(K_{c} = \frac{{[O_{2}] \cdot [SO_{2}]^{2}}}{{[SO_{3}]^{2}}}\), is the expression for the equilibrium constant for the given reaction.

Step by step solution

01

Analyze the Balanced Reaction

We have the balanced reaction: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g) + 2 \mathrm{SO}_{2}(g)\). The reactant is \(SO_{3}\) with stoichiometric coefficient 2 and the products are \(O_{2}\) with stoichiometric coefficient 1 and \(SO_{2}\) with stoichiometric coefficient 2.
02

Apply the Law of Mass Action

According to the law of mass action, the equilibrium constant, \(K_{\mathrm{c}}\), is given by the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. Applying this to our reaction, we get: \(K_{c} = \frac{{([O_{2}]^{1} \cdot [SO_{2}]^{2})}}{{[SO_{3}]^{2}}}\). Simplifying this expression yields: \(K_{c} = \frac{{[O_{2}] \cdot [SO_{2}]^{2}}}{{[SO_{3}]^{2}}}\)
03

Match the Expression with Options

The derived expression for the equilibrium constant \(K_{c} = \frac{{[O_{2}] \cdot [SO_{2}]^{2}}}{{[SO_{3}]^{2}}}\) matches with option (C). So, option (C) is the correct answer.

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Key Concepts

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

Law of Mass Action
The law of mass action forms the foundation for understanding the equilibrium constant. This law states that the rate of any chemical reaction is proportional to the product of the masses (or concentrations) of the reactants. In simpler terms, when a reaction reaches equilibrium, the ratio of the concentrations of products to reactants remains constant. This is what we call the equilibrium constant, represented as \( K_c \) for reactions involving concentrations.
For the given reaction \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g) + 2 \mathrm{SO}_{2}(g)\), the expression for \( K_c \) is derived using the law of mass action. We take the concentrations of products \( \mathrm{O}_{2} \) and \( \mathrm{SO}_{2} \), each raised to the power of their stoichiometric coefficients, and divide by the concentration of the reactant \( \mathrm{SO}_{3} \) raised to its stoichiometric coefficient. This gives \( K_c = \frac{[\mathrm{O}_2] \cdot [\mathrm{SO}_2]^2}{[\mathrm{SO}_3]^2} \). This constant helps to understand how favored a reaction is towards forming products at equilibrium.
Chemical Equilibrium
Chemical equilibrium is a state in which the forward and reverse reactions occur at the same rate. As a result, the concentrations of reactants and products remain constant over time. It's important to note that equilibrium does not mean that the concentrations of reactants and products are equal, but rather that their rates of formation are balanced.
In the context of our reaction \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g) + 2 \mathrm{SO}_{2}(g)\), reaching equilibrium means that the amount of \( \mathrm{SO}_{3} \) being converted to \( \mathrm{O}_{2} \) and \( \mathrm{SO}_{2} \) is equal to the amount of \( \mathrm{O}_{2} \) and \( \mathrm{SO}_{2} \) recombining into \( \mathrm{SO}_{3} \). By using the equilibrium constant expression, we can predict the concentration ratios of reactants to products at equilibrium, and determine how the reaction shifts under different conditions.
Stoichiometric Coefficients
Stoichiometric coefficients are the numbers placed in front of the compounds in a chemical equation to balance it. These coefficients indicate the proportional amounts of each compound involved in the reaction. Understanding these coefficients is crucial because they inform how the concentrations in the equilibrium expression are handled.
In the balanced equation for our reaction, the stoichiometric coefficient for \( \mathrm{SO}_{3} \) is 2, for \( \mathrm{O}_{2} \) it is 1, and for \( \mathrm{SO}_{2} \) it is 2. These coefficients are used as exponents in the expression for the equilibrium constant. Thus, the concentration of \( \mathrm{SO}_{3} \) is squared, \( [\mathrm{SO}_{3}]^2 \), and so is \( \mathrm{SO}_{2} \) \( [\mathrm{SO}_2]^2 \), while \( \mathrm{O}_{2} \) remains \( [\mathrm{O}_2] \) since its coefficient is 1.
Considering the stoichiometric coefficients is essential for accurately calculating the equilibrium constant and understanding the dynamic balance of the reaction at equilibrium.

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Most popular questions from this chapter

In a voltaic cell with a Cu(s) \(| \mathrm{Cu}^{2+}\) cathode and \(\mathrm{a} \mathrm{Pb}^{2+} | \mathrm{Pb}\) (s) anode, increasing the concentration of \(\mathrm{Pb}^{2+}\) causes the voltage to decrease. What is the reason for this? (A) The value for Q will increase, causing the cell to come closer to equilibrium. (B) The solution at the anode becomes more positively charged, leading to a reduced electron flow. (C) The reaction will shift to the right, causing a decrease in favor ability. (D) Cell potential will always decrease anytime the concentration of any aqueous species present increases.

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Silver sulfate, \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) , has a solubility product constant of \(1.0 \times 10^{-5} .\) The below diagram shows the products of a precipitation reaction in which some silver sulfate was formed. (Diagram Can't Copy) Which ion concentrations below would have led the precipitate to form? (A) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (B) \(\left[\mathrm{Ag}^{+}\right]=0.10 M\left[\mathrm{SO}_{4}^{2-}\right]=0.01 M\) (C) \(\left[\mathrm{Ag}^{+}\right]=0.01 M\left[\mathrm{SO}_{4}^{2-}\right]=0.10 M\) (D) This is impossible to determine without knowing the total volume of the solution.

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