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Write the equilibrium expression ( \(K\) ) for each of the following gas-phase reactions. a. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) b. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. \(\mathrm{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g)\) d. \(2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)\)

Short Answer

Expert verified
The equilibrium expressions for the given reactions are: a. \( K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}] \times [\mathrm{O}_{2}]} \) b. \( K = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]} \) c. \( K = \frac{[\mathrm{SiCl}_{4}] \times [\mathrm{H}_{2}]^2}{[\mathrm{SiH}_{4}] \times [\mathrm{Cl}_{2}]^2} \) d. \( K = \frac{[\mathrm{PCl}_{3}]^2 \times [\mathrm{Br}_{2}]^3}{[\mathrm{PBr}_{3}]^2 \times [\mathrm{Cl}_{2}]^3} \)

Step by step solution

01

Equilibrium expression for reaction a

For the reaction: \[ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \] The equilibrium expression can be written as: \[ K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}] \times [\mathrm{O}_{2}]} \]
02

Equilibrium expression for reaction b

For the reaction: \[ \mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \] The equilibrium expression can be written as: \[ K = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]} \]
03

Equilibrium expression for reaction c

For the reaction: \[ \mathrm{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g) \] The equilibrium expression can be written as: \[ K = \frac{[\mathrm{SiCl}_{4}] \times [\mathrm{H}_{2}]^2}{[\mathrm{SiH}_{4}] \times [\mathrm{Cl}_{2}]^2} \]
04

Equilibrium expression for reaction d

For the reaction: \[ 2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g) \] The equilibrium expression can be written as: \[ K = \frac{[\mathrm{PCl}_{3}]^2 \times [\mathrm{Br}_{2}]^3}{[\mathrm{PBr}_{3}]^2 \times [\mathrm{Cl}_{2}]^3} \]

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Key Concepts

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

Chemical Equilibrium
Understanding the concept of chemical equilibrium is essential in grasping how reactions work in a closed system. In a chemical reaction, reactants convert to products, but under certain conditions, the products can also revert back to the original reactants. Equilibrium is the state in which the rates of the forward reaction (reactants to products) and the reverse reaction (products to reactants) are equal. This balance results in the concentrations of both reactants and products remaining constant over time, even though the reactions continue to occur.

In an equilibrium state, it's important to note that the system is dynamic, meaning reactants and products are constantly interchanging, but their overall concentrations do not change. This concept is often demonstrated in reversible gas-phase reactions, where gaseous reactants and products are enclosed in a sealed container. Students sometimes assume equilibrium means that the quantities of reactants and products are equal, but this is not necessarily the case; what is equal are the rates at which the products form and reactants are replenished.
Equilibrium Constants
The equilibrium constant, represented as \(K\), is a value that gives us insight into the proportions of reactants and products at equilibrium. It is a snapshot of the system's position at equilibrium and indicates whether reactants or products are favored. For a given balanced chemical equation at a specific temperature, \(K\) remains constant and is calculated using the concentrations (for solutions) or partial pressures (for gases) of the reactants and products.

The expression for the equilibrium constant for a reaction is specific to the form of the reaction equation. You determine \(K\) by dividing the product of the concentrations of products, each raised to the power of its coefficient in the balanced equation, by the product of the concentrations of reactants, also each raised to the power of its coefficient. When writing equilibrium expressions, it's critical to remember that only the concentrations of gases and species dissolved in solution are included in the expression; solids and pure liquids are omitted because their concentrations do not change.
Gas-Phase Reactions
Gas-phase reactions are characterized by the involvement of substances in their gaseous state. These types of reactions are particularly interesting from an equilibrium perspective because the behavior of gases can be explained and predicted using the ideal gas law. Moreover, the concentration of a gas is directly proportional to its partial pressure, a concept that's key when dealing with equilibrium constants for gas-phase reactions.

In the space of gas-phase reactions, the term 'partial pressure' is used and is represented as \( P \) with the formula for partial pressure being \( P = nRT/V \), where \( n \) is the number of moles, \( R \) is the ideal gas constant, \( T \) is the temperature, and \( V \) is the volume. In equilibrium expressions for gas-phase reactions, we often use partial pressures instead of concentrations, especially when dealing with dilute gases. Understanding gas-phase reactions demands not only an awareness of the relevant equations but also the practical conditions under which these reactions occur, such as temperature and pressure, which can affect reaction rates and equilibrium positions.

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Most popular questions from this chapter

For the reaction $$ \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s) $$ \(K=400\). at \(35.0^{\circ} \mathrm{C}\). If \(2.00\) moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{~S}\), and \(\mathrm{NH}_{4} \mathrm{HS}\) are placed in a \(5.00-\mathrm{L}\) vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{~S}\) at equilibrium?

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction: \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \quad K=2.2\) \(\begin{array}{ll}\text { Acetic acid } \text { Ethanol } & \text { Ethyl acetate }\end{array}\) For the following mixtures \((a-d)\), will the concentration of \(\mathrm{H}_{2} \mathrm{O}\) increase, decrease, or remain the same as equilibrium is established? a. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.010 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.010 \mathrm{M}\) b. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}\right]=0.0020 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.0020 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.10 \mathrm{M}\) c. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.88 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.12 \mathrm{M}\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.044 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=6.0 \mathrm{M}\) d. \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=4.4 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=4.4 M\), \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.88 \mathrm{M},\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=10.0 \mathrm{M}\) e. What must the concentration of water be for a mixture with \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=2.0 \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.10 M\), and \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=5.0 M\) to be at equilibrium? f. Why is water included in the equilibrium expression for this reaction?

The following equilibrium pressures were observed at a certain temperature for the reaction $$ \begin{array}{c} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\ P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{~atm} \\ P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{~atm} \end{array} $$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{~atm}\), and \(\underline{P_{\mathrm{H}}}=0.00761\) atm, does this represent a system at equilibrium?

The following equilibrium pressures at a certain temperature were observed for the reaction $$ \begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{~atm} \\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{~atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{~atm} \end{aligned} $$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

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