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Consider the equilibrium at \(25^{\circ} \mathrm{C}\) $$ 2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \quad K_{\mathrm{c}}=3.58 \times 10^{-3} $$ Suppose that \(0.15 \mathrm{~mol} \mathrm{SO}_{3}(\mathrm{~g}), 0.015 \mathrm{~mol} \mathrm{SO}_{2}(\mathrm{~g}),\) and \(0.0075 \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\) are placed into a \(10.0-\mathrm{L}\) flask at \(25^{\circ} \mathrm{C}\) and the flask is sealed. (a) Is the system at equilibrium? (b) If the system is not at equilibrium, in which direction must the reaction proceed to reach equilibrium? Explain your answer.

Short Answer

Expert verified
(a) No, the system is not at equilibrium. (b) The reaction must proceed towards the products to reach equilibrium.

Step by step solution

01

Find Initial Concentrations

To determine if the system is at equilibrium, we start by calculating the initial concentrations of the gases in the flask. Given the moles and volume, the concentration \( [C] \) is \( \frac{\text{moles}}{\text{volume}} \). For \(\text{SO}_3\), the concentration \([\text{SO}_3] = \frac{0.15 \, \text{mol}}{10.0 \, \text{L}} = 0.015 \, \text{M}\). For \(\text{SO}_2\), \([\text{SO}_2] = \frac{0.015 \, \text{mol}}{10.0 \, \text{L}} = 0.0015 \, \text{M}\). For \(\text{O}_2\), \([\text{O}_2] = \frac{0.0075 \, \text{mol}}{10.0 \, \text{L}} = 0.00075 \, \text{M}\).
02

Write the Reaction Quotient Expression

The reaction quotient \( Q_c \) is similar to the equilibrium constant \( K_c \) but uses the initial concentrations. For the given reaction, \[ Q_c = \frac{[\text{SO}_2]^2 \times [\text{O}_2]}{[\text{SO}_3]^2} \].
03

Calculate the Reaction Quotient \(Q_c\)

Plug in the initial concentrations into the \( Q_c \) expression: \[ Q_c = \frac{(0.0015)^2 \times (0.00075)}{(0.015)^2} \]. \( Q_c = \frac{2.25 \times 10^{-6} \times 0.00075}{0.000225} = \frac{1.6875 \times 10^{-9}}{0.000225} \approx 7.50 \times 10^{-6} \).
04

Compare \( Q_c \) to \( K_c \)

Now compare \( Q_c \) with the given \( K_c \): \( Q_c = 7.50 \times 10^{-6} \) and \( K_c = 3.58 \times 10^{-3} \). Since \( Q_c < K_c \), the system is not at equilibrium.
05

Determine Reaction Direction

Since \( Q_c < K_c \), the reaction must proceed in the forward direction (towards the products), increasing the concentration of \( \text{SO}_2 \) and \( \text{O}_2 \), and decreasing the concentration of \( \text{SO}_3 \) to reach equilibrium.

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Key Concepts

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

Reaction Quotient
The reaction quotient, commonly denoted as \( Q_c \), is a measure that helps determine the current status of a chemical reaction in relation to its equilibrium. It essentially uses the same formula as the equilibrium constant \( K_c \), but with the concentrations of the reactants and products at any given moment in time, instead of those at equilibrium. Understanding the value of \( Q_c \) can give us insight into whether a reaction is at equilibrium or, if not, which direction it must shift to reach equilibrium.

When calculating \( Q_c \), for any reaction of the type \( aA + bB \rightleftharpoons cC + dD \), the expression for \( Q_c \) will be:
  • \( Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
This formula is similar to the one used for \( K_c \), with the crucial difference that \( Q_c \) uses existing concentrations instead of equilibrium concentrations.

By comparing \( Q_c \) to \( K_c \):
  • If \( Q_c = K_c \), the system is at equilibrium.
  • If \( Q_c < K_c \), the reaction will proceed in the forward direction to increase the concentration of products.
  • If \( Q_c > K_c \), the reaction will shift in the reverse direction to increase the concentration of reactants.
Thus, calculating \( Q_c \) helps determine whether a mixture of reactants and products will move forward or backward to achieve equilibrium.
Equilibrium Constant
The equilibrium constant \( K_c \) is a fundamental concept in understanding chemical equilibrium. It quantifies the relative concentrations of reactants and products at the time when a chemical reaction is at equilibrium. For the reaction \( 2 \text{SO}_3(\text{g}) \rightleftharpoons 2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \), the given \( K_c \) value is \( 3.58 \times 10^{-3} \) at 25°C.

This constant is constant for a given reaction at a specific temperature. It reflects the ratio of product concentrations to reactant concentrations, with each raised to the power of its stoichiometric coefficient as per the balanced chemical equation:
  • \( K_c = \frac{[\text{SO}_2]^2[\text{O}_2]}{[\text{SO}_3]^2} \)
\( K_c \) essentially provides a snapshot of the "equilibrium position" of a reaction, indicating whether products or reactants are favored once equilibrium is achieved.

In practice, \( K_c \) helps us determine:
  • Whether the reaction will proceed forward or backward to reach equilibrium when we measure initial concentrations and calculate \( Q_c \).
  • The extent of a reaction, showing if products or reactants are more prevalent.
It's worth noting that \( K_c \) remains unchanged unless the temperature of the reaction changes, making it a crucial value for predicting reaction behavior under stable conditions.
Le Chatelier's Principle
Le Chatelier's Principle is a powerful concept that helps predict how a chemical system at equilibrium reacts to disturbances or changes. When a system at equilibrium is subjected to a change in concentration, temperature, or pressure, this principle helps us understand the system's response to restore balance.

The principle states that if an external change is applied to a system at equilibrium, the system adapts in a way that partially counteracts that change. Here are some typical scenarios and responses:
  • Concentration Change: If the concentration of a reactant or product is changed, the system will shift to restore equilibrium by consuming the added substance or producing more of the removed substance.
  • Temperature Change: If a reaction is exothermic, increasing temperature shifts equilibrium towards the reactants. For an endothermic reaction, increasing temperature shifts it towards the products.
  • Pressure Change: A change in pressure affects gas equilibria. An increase in pressure shifts the reaction towards the side with fewer gas molecules, whereas a decrease in pressure shifts it towards the side with more gas molecules.
Le Chatelier's Principle is a vital tool in chemical analysis, helping chemists manipulate reaction conditions to favor the formation of desired products. It is crucial in both industrial applications and laboratory settings, allowing precise control over reaction pathways.

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Most popular questions from this chapter

A solid sample of benzoic acid, a carboxylic acid, is in equilibrium with an aqueous solution of benzoic acid. A tiny quantity of \(\mathrm{D}_{2} \mathrm{O},\) water containing the isotope \({ }^{2} \mathrm{H}\), deuterium, is added to the solution. The solution is allowed to stand at constant temperature for several hours, after which some of the solid benzoic acid is removed and analyzed. The benzoic acid is found to contain a tiny quantity of deuterium, D, and the formula of the deuterium-containing molecules is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOD}\). Explain how this can happen.

Write equilibrium constant expressions for these reactions: (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{BaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) (d) \(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\)

The hydrocarbon \(\mathrm{C}_{4} \mathrm{H}_{10}\) can exist in two gaseous forms: butane and 2 -methylpropane. The value of \(K_{\mathrm{c}}\) for conversion of butane to 2 -methylpropane is 2.5 at \(25^{\circ} \mathrm{C}\). CCCCC=CC(C)(C)CO (a) Suppose that the initial concentrations of butane and 2-methylpropane are each \(0.100 \mathrm{~mol} / \mathrm{L}\). Make up a table of initial concentrations, change in concentrations, and equilibrium concentrations for this reaction. (b) Write the equilibrium constant expression in terms of \(x\), the change in the concentration of butane, and then solve for \(x\). (c) If you place 0.017 mol butane in a \(0.50-\mathrm{L}\) sealed flask at \(25^{\circ} \mathrm{C},\) calculate the equilibrium concentration of each isomer.

Chemists carried out a study of the high temperature reaction of sulfur dioxide with oxygen in which a sealed reactor initially contained \(0.0076-\mathrm{M} \mathrm{SO}_{2}, 0.0036-\mathrm{M} \mathrm{O}_{2}\), and no \(\mathrm{SO}_{3}\). After equilibrium was achieved, the \(\mathrm{SO}_{2}\) concentration decreased to \(0.0032 \mathrm{M}\). Calculate \(K_{\mathrm{c}}\) at this temperature for $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$

\(K_{\mathrm{p}}\) for this reaction is 0.16 at \(25^{\circ} \mathrm{C}:\) $$ 2 \operatorname{NOBr}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=16.3 \mathrm{~kJ} / \mathrm{mol} $$ Predict the effect of each change on the position of the equilibrium; that is, state which way the equilibrium shifts (left, right, or no change) when each change is made. Assume constant-volume conditions for parts (a), (b), and (c). (a) Add more \(\mathrm{Br}_{2}(\mathrm{~g})\). (b) Remove some \(\mathrm{NOBr}(\mathrm{g})\). (c) Decrease the temperature. (d) Increase the container volume.

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