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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) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{p}=5.0 \times 10^{12}\) (b) \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) ; K_{c}=5.8 \times 10^{-18}\)

Short Answer

Expert verified
The short answer is: (a) The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) lies to the right, favoring the formation of products, as $$K_p = 5.0 \times 10^{12}$$ is significantly greater than 1. (b) The reaction \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) lies to the left, favoring the formation of reactants, as $$K_c = 5.8 \times 10^{-18}$$ is significantly less than 1.

Step by step solution

01

Reaction (a)

Given the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) with the equilibrium constant $$K_p = 5.0 \times 10^{12}$$. As $$K_p$$ is significantly greater than 1, we can say that this reaction lies to the right, favoring the formation of products \(($$\mathrm{NO}_{2}$$) over reactants (\)\mathrm{NO}\( and \)\mathrm{O}_{2}$).
02

Reaction (b)

For the reaction \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\), the given equilibrium constant is $$K_c = 5.8 \times 10^{-18}$$. Since $$K_c$$ is significantly less than 1, this reaction lies to the left, favoring the formation of reactants (\(\mathrm{HBr}\)) over the products (\(\mathrm{H}_{2}\) and \(\mathrm{Br}_{2}\)). The conclusions are: (a) The reaction lies to the right, favoring the formation of products. (b) The reaction lies to the left, favoring the 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
The equilibrium constant is a critical concept in understanding chemical equilibrium. It tells us the extent to which a reaction proceeds and helps predict the relative concentrations of reactants and products at equilibrium.
In a balanced chemical reaction, the equilibrium constant is represented by either \( K_c \) or \( K_p \). For reactions using concentrations, \( K_c \) is used, while partial pressures use \( K_p \). Each constant offers valuable insights into the reaction's dynamics.
If the equilibrium constant is much larger than 1, it indicates that the reaction favors the formation of products. This means that at equilibrium, the concentration of products is greater than that of the reactants. Conversely, if the constant is much smaller than 1, it implies that the equilibrium favors the reactants.
Key points to remember include:
  • \( K_c \) and \( K_p \) relate to concentrations and pressures, respectively.
  • A large equilibrium constant means more products at equilibrium.
  • A small equilibrium constant signifies a preference for reactants.
  • The equilibrium position represents the ratio of product to reactant concentrations when the reaction is at rest.
Reaction Direction
The direction in which a chemical reaction proceeds towards equilibrium is crucial for understanding the behavior of the reaction under different conditions. By analyzing the equilibrium constant, we can determine whether a reaction moves to the right (favoring products) or to the left (favoring reactants).
In the exercise given, the reaction involving \( 2 \mathrm{NO}(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \) has an equilibrium constant \( K_p = 5.0 \times 10^{12} \). This large value indicates that the reaction proceeds to the right, heavily favoring products. In contrast, the reaction \( 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g) \) with \( K_c = 5.8 \times 10^{-18} \) goes to the left, favoring reactants.
The direction can be easily understood with these pointers:
  • Large \( K \) values indicate a rightward direction with more products formed.
  • Small \( K \) values suggest a leftward direction and a predominance of reactants.
  • Reaction direction is essential for understanding chemical processes and predicting outcomes.
Product Formation
Product formation in a chemical reaction is largely influenced by the reaction's equilibrium state. When a reaction has a large equilibrium constant, it means that, at equilibrium, most of the reactant species have turned into products.
Taking from our exercise, in reaction (a), \( 2 \mathrm{NO}(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \), the high value of \( K_p \) reflects a large tendency to form \( \mathrm{NO}_2 \), the product. This is because the enormous \( K \) value shifts the balance heavily towards products.
Meanwhile, in reaction (b), the formation of products \( \mathrm{H}_2 \) and \( \mathrm{Br}_2 \) is less favored as \( K_c \) is very small. This indicates that under equilibrium, \( \mathrm{HBr} \) remains mostly unreacted, preferring to stay as a reactant.
Important points to recall on product formation:
  • Product formation dominance relies on a high equilibrium constant.
  • Small \( K \) values demonstrate limited product formation.
  • Understanding product formation helps in optimizing reaction conditions for desired outputs.

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Most popular questions from this chapter

Water molecules in the atmosphere can form hydrogenbonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSEPR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at \(300 \mathrm{~K}\) and 0.020 at \(350 \mathrm{~K}\). Is water dimer formation endothermic or exothermic?

Silver chloride, \(\mathrm{AgCl}(s)\), is an "insoluble" strong electrolyte. (a) Write the equation for the dissolution of \(\mathrm{AgCl}(s)\) in \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Write the expression for \(K_{c}\) for the reaction in part (a). (c) Based on the thermochemical data in Appendix \(\mathrm{C}\) and Le Châtelier's principle, predict whether the solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}\) increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of \(\mathrm{AgCl}\) in water is \(1.6 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). In addition, \(\mathrm{Ag}^{+}(a q)\) can react with \(\mathrm{Cl}^{-}(a q)\) according to the reaction $$\mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}_{2}^{-}(a q)$$ where \(K_{c}=1.8 \times 10^{5}\) at \(25^{\circ} \mathrm{C}\). Although \(\mathrm{AgCl}\) is "not soluble" in water, the complex \(\mathrm{AgCl}_{2}^{-}\) is soluble. At \(25^{\circ} \mathrm{C},\) is the solubility of AgCl in a \(0.100 M\) NaCl solution greater than the solubility of AgCl in pure water, due to the formation of soluble \(\mathrm{AgCl}_{2}^{-}\) ions? Or is the \(\mathrm{AgCl}\) solubility in \(0.100 \mathrm{M} \mathrm{NaCl}\) less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations.

The following graph represents the yield of the compound \(\mathrm{AB}\) at equilibrium in the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \longrightarrow \mathrm{AB}(g)\) at two different pressures, \(x\) and \(y\), as a function of temperature. (a) Is this reaction exothermic or endothermic? (b) Is \(P=x\) greater or smaller than $P=y ?

(a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? Explain. (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

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