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For each of these chemical reactions, predict whether the equilibrium constant at \(25^{\circ} \mathrm{C}\) is greater than 1 or less than \(1,\) or state that insufficient information is available. Also indicate whether each reaction is product-favored or reactant-favored. $$ \begin{array}{l} \text { (a) } 2 \mathrm{NaCl}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=823 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=-566 \mathrm{~kJ} / \mathrm{mol} \\ \text { (c) } 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \\ \Delta_{\mathrm{r}} H^{\circ}=2045 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Short Answer

Expert verified
(a) Less than 1, reactant-favored. (b) Greater than 1, product-favored. (c) Less than 1, reactant-favored.

Step by step solution

01

Analyzing Reaction (a)

The reaction \(2 \text{NaCl(s)} \leftrightharpoons 2 \text{Na(s)} + \text{Cl}_2(\text{g})\) has a \( \Delta_r H^\circ = 823 \text{ kJ/mol} \). This is a positive \( \Delta_r H^\circ \), indicating an endothermic reaction. Generally, endothermic reactions at low temperatures are reactant-favored; thus, the equilibrium constant \( K \) is likely less than 1, and the reaction is reactant-favored.
02

Analyzing Reaction (b)

The reaction \(2 \text{CO(g)} + \text{O}_2(\text{g)} \leftrightharpoons 2 \text{CO}_2(\text{g)}\) has a \( \Delta_r H^\circ = -566 \text{ kJ/mol} \). The negative \( \Delta_r H^\circ \) signifies an exothermic reaction. Typically, exothermic reactions are product-favored at low temperatures, so the equilibrium constant \( K \) is expected to be greater than 1, and the reaction is product-favored.
03

Analyzing Reaction (c)

For the reaction \(3 \text{CO}_2(\text{g)} + 4 \text{H}_2\text{O(g)} \leftrightharpoons \text{C}_3\text{H}_8(\text{g)} + 5 \text{O}_2(\text{g)}\), \( \Delta_r H^\circ = 2045 \text{ kJ/mol} \) is highly positive, indicating a very endothermic reaction. Such reactions are usually reactant-favored at low temperatures, implying that \( K \) is likely less than 1.

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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, denoted as \( K \), is a crucial concept in understanding chemical equilibrium. It provides a measure of the extent to which a chemical reaction proceeds. The value of \( K \) is determined by the concentrations of the reactants and products at equilibrium. For a generic reaction \( aA + bB \rightleftharpoons cC + dD \), it is defined as: \[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]- If \( K > 1 \): The reaction is product-favored, meaning that at equilibrium, the concentration of the products is higher than that of the reactants.- If \( K < 1 \): The reaction is reactant-favored, indicating that reactants are more prevalent at equilibrium.- If \( K = 1 \): Neither products nor reactants are favored, and their concentrations are essentially equal.Understanding \( K \) helps predict the direction in which a reaction will proceed and how changes in conditions can affect it.
Endothermic Reaction
In an endothermic reaction, heat is absorbed from the surroundings, resulting in an overall positive change in enthalpy, denoted by \( \Delta H > 0 \). This absorption of energy means that the products have more energy than the reactants. As a result, such reactions are usually less favorable at lower temperatures.Let's break this down: - The positive \( \Delta H \) tells us that energy is required for the reaction to occur.- At low temperatures, endothermic reactions typically remain reactant-favored, meaning most of the substance stays in its original form because the system lacks energy to move forward.Examples of endothermic reactions include: - Photosynthesis - Melting ice cubesUnderstanding endothermic reactions helps predict how temperature changes might shift the equilibrium.
Exothermic Reaction
Exothermic reactions release energy, usually in the form of heat, leading to a negative enthalpy change \( \Delta H < 0 \). This release often makes the reaction more favorable or spontaneous.Here's how it works:- The negative \( \Delta H \) indicates that the products are lower in energy than the reactants.- At low temperatures, exothermic reactions are generally product-favored since the release of heat makes it easier for the reaction to proceed.Common examples of exothermic reactions include: - Combustion of fuels such as gasoline - Mixing acids and basesHaving a basic grasp of exothermic reactions can help foresee which direction a chemical process will naturally take when conditions change.
Product-Favored Reaction
In a product-favored reaction, the equilibrium lies towards the products, meaning more products than reactants are present at equilibrium. This generally results in a high equilibrium constant (\( K > 1 \)).Characteristics include:- An abundance of product molecules in the equilibrium mixture.- Often associated with exothermic reactions, where energy is released, driving the reaction to form more products.The product-favored nature of a reaction is commonly a result of conditions such as:- High \( K \)- Lower temperature favoring exothermicityRecognizing a product-favored reaction can be useful in industrial applications where maximizing the yield is important.
Reactant-Favored Reaction
A reactant-favored reaction occurs when the equilibrium lies toward the reactants, making them more prevalent in the mixture. Here, the equilibrium constant is low, specifically \( K < 1 \).Key points:- A low concentration of products relative to reactants.- Such reactions are generally associated with endothermic processes.The reactant-favored characteristic often results from:- High \( \Delta H \)- Low temperatures insufficient to overcome energy barriersUnderstanding the factors that make a reaction reactant-favored helps in modifying conditions to push the reaction toward products, especially in synthetic chemistry where desirable products are the ultimate goal.

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Most popular questions from this chapter

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 $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$

The atmosphere consists of about \(80 \% \mathrm{~N}_{2}\) and \(20 \% \mathrm{O}_{2}\), yet there are many oxides of nitrogen that are stable and can be isolated in the laboratory. (a) Is the atmosphere at chemical equilibrium with respect to forming NO? (b) If not, why doesn't NO form? If so, how is it that \(\mathrm{NO}\) can be made and kept in the laboratory for long periods?

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ has the value \(5.97 \times 10^{-2}\) at \(500 .{ }^{\circ} \mathrm{C}\). If \(1.00 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{gas}\) and \(1.00 \mathrm{~mol} \mathrm{H}_{2}\) gas are heated to \(500 .{ }^{\circ} \mathrm{C}\) in a \(10.00-\mathrm{L}\) sealed flask together with a catalyst, calculate the percentage of \(\mathrm{N}_{2}\) converted to \(\mathrm{NH}_{3}\). (Hint: Assume that only a very small fraction of the reactants is converted to products. Obtain an approximate answer and use it to obtain a more accurate result.)

Carbon dioxide reacts with carbon to give carbon monoxide according to the equation $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ At \(700 .{ }^{\circ} \mathrm{C},\) a \(2.0-\mathrm{L}\) sealed flask at equilibrium contains $$ 0.10 \mathrm{~mol} \mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}, \text { and } 0.40 \mathrm{~mol} \mathrm{C} . \text { Calculate } $$ the equilibrium constant \(K_{\mathrm{P}}\) for this reaction at the specified temperature.

Write the expression for \(K_{\mathrm{c}}\) for each reaction. (a) \(\mathrm{PCl}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons\) \(\mathrm{CoCl}_{4}^{2-}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(2 \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{OF}_{2}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\)

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