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Chapter 15: Problem 18

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) at \(1300 \mathrm{~K} K_{c}=0.57\) (b) \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) at \(900 \mathrm{~K} K_{p}=0.0572\)

Short Answer

Expert verified
For reaction (a) with \(K_c=0.57\) at \(1300 \mathrm{~K}\), and reaction (b) with \(K_p=0.0572\) at \(900 \mathrm{~K}\), both equilibrium constants are less than \(1\), indicating that both reactions favor the formation of reactants and lie to the left.

Step by step solution

01

Reaction (a) analysis

We are given the reaction: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) at \(1300 \mathrm{~K}\) with \(K_c=0.57\). Since the given equilibrium constant, \(K_c\), is less than \(1\), the reaction favors the formation of reactants. Therefore, the reaction lies to the left.
02

Reaction (b) analysis

We are given the reaction: \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) at \(900 \mathrm{~K}\) with \(K_p=0.0572\). Since the given equilibrium constant, \(K_p\), is less than \(1\), the reaction favors the formation of reactants. Therefore, the reaction lies to the left. Based on the analysis of the given equilibrium constants, both reactions (a) and (b) lie to the left, favoring formation of reactants.

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Key Concepts

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

Equilibrium Constant
In chemistry, the equilibrium constant is a numerical value that helps us understand the position of a chemical reaction at equilibrium. Represented as either \(K_c\) or \(K_p\), depending on whether the concentration or pressure is used, this constant informs us about the relative amounts of products and reactants at equilibrium.

An equilibrium constant greater than 1 suggests that, at equilibrium, products are favored over reactants. This means the reaction lies to the right. Conversely, if the equilibrium constant is less than 1, reactants are favored, and the reaction lies to the left. This indicates more reactants are present than products when the reaction reaches equilibrium.

In the exercises given, both reactions have equilibrium constants less than 1, implying they lie to the left. This means they favor the formation of reactants over products.
Reaction Direction
For any given reaction, knowing whether it lies to the right or left is essential in predicting the dominant species at equilibrium. The direction of a reaction is determined by its equilibrium constant:
  • If \( K > 1 \), the reaction proceeds to the right, favoring product formation.
  • If \( K < 1 \), the reaction proceeds to the left, favoring reactant formation.
These guidelines help chemists understand the distribution of chemicals at equilibrium. In our case, both exercise reactions have equilibrium constants less than 1, signaling a leftward direction, or a preference for reactants. This insight helps in practical settings where maximizing product yield is crucial.
Gaseous Reactions
Gaseous reactions involve reactants and products in the gas phase. Their behavior is often described by partial pressures, especially when using the equilibrium constant \(K_p\). The exercises given are excellent examples of gaseous reactions and demonstrate how equilibrium concepts are applied to them.

In such reactions, changes in conditions like temperature and pressure can significantly affect the position of equilibrium. For instance, increasing pressure may favor the side of the equation with fewer moles of gas. However, the equilibrium constant itself is only temperature-dependent and remains unchanged by shifts in pressure or concentration.

Gaseous reactions require a careful balance, making knowledge of equilibrium principles vital for effectively managing and predicting their outcomes. Understanding these principles allows chemists and students alike to manipulate conditions favorably for desired results.

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Most popular questions from this chapter

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C}\). The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium, the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) taken together) is \(62.21 \mathrm{kPa}\). What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

If \(K_{c}=0.013 \mathrm{~L} / \mathrm{mol}\) for \(2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\) at \(1000 \mathrm{~K}\), what is the value of \(K_{p}\) for this reaction at this temperature?

Consider the hypothetical reaction $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)+2 \mathrm{C}(g)$$ A flask is charged with \(100 \mathrm{kPa}\) of pure \(\mathrm{A}\), after which it is allowed to reach equilibrium at \(25^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{B}\) is \(25 \mathrm{kPa}\). (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?(\mathbf{c})\) What could we do to maximize the yield of \(\mathrm{B}\) ?

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) in a fetus, compared to \(K_{c}\) for the same reaction in an adult.

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15 ), \(K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).
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