Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at 298 \(\mathrm{K} )\) : \begin{equation} \begin{array}{l}{2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) \quad K_{c}=2.0} \\ {2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30}}\end{array} \end{equation}

Short Answer

Expert verified
The equilibrium constant \(K_{p}\) for the reaction \(N_{2}(g)+O_{2}(g)+Br_{2}(g) \rightleftharpoons 2\, NOBr(g)\) can be found by manipulating the given reactions and their equilibrium constants. Reverse the second reaction and divide it by 2. Then, add this manipulated reaction to the first reaction. Calculate the overall equilibrium constant by multiplying the equilibrium constants of these individual reactions. The resulting equilibrium constant, \(K_{p}\), is approximately \(1.38 \times 10^{-15}\).

Step by step solution

01

Analyze the given reactions

We are given the following two reactions with their equilibrium constants: 1. \(2\, NO(g) + Br_2(g) \rightleftharpoons 2\, NOBr(g)\) with \(K_{c1} = 2.0\) 2. \(2\, NO(g) \rightleftharpoons N_2(g) + O_2(g)\) with \(K_{c2} = 2.1 \times 10^{30}\) Our goal is to find the equilibrium constant for the following reaction: \(N_{2}(g)+O_{2}(g)+Br_{2}(g) \rightleftharpoons 2\, NOBr(g)\)
02

Add reactions to create the overall reaction

Manipulate the given reactions so that when they are combined, the net reaction is equal to the overall reaction. First, reverse the second reaction and divide it by 2: \(\dfrac{1}{2}\left(N_{2}(g) + O_{2}(g) \rightleftharpoons NO(g)\) with \(K^\prime_{c2} = \dfrac{1}{\sqrt{K_{c2}}}\) Now, add this manipulated second reaction to the first reaction: \(\begin{aligned} &2\, NO(g) + Br_2(g) \rightleftharpoons 2\, NOBr(g) \\ + \hspace{25pt} &\dfrac{1}{2}\left(N_{2}(g) + O_{2}(g) \rightleftharpoons NO(g) \end{aligned}\) The overall reaction is now: \(N_{2}(g)+O_{2}(g)+Br_{2}(g) \rightleftharpoons 2\, NOBr(g)\)
03

Calculate the equilibrium constant

We added the manipulated second reaction to the first reaction, so now we multiply their equilibrium constants to obtain the equilibrium constant for the overall reaction: \(K_{p} = K_{c1} \times K^\prime_{c2}\) \(K_{p} = 2.0 \times \frac{1}{\sqrt{2.1 \times 10^{30}}}\) \(K_{p} = \frac{2.0}{\sqrt{2.1 \times 10^{30}}}\) After calculating the result, we get: \(K_{p} \approx 1.38 \times 10^{-15}\) Therefore, the equilibrium constant \(K_{p}\) for the reaction \(N_{2}(g)+O_{2}(g)+Br_{2}(g) \rightleftharpoons 2\, NOBr(g)\) is approximately \(1.38 \times 10^{-15}\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a dynamic state where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the amounts of substances involved. At this point, the concentrations of reactants and products remain constant over time, although they are not necessarily equal. Equilibrium can be achieved from any direction of the reaction and is characterized by specific conditions such as temperature, pressure, and concentration.

Understanding the state of equilibrium is crucial for predicting the outcome of chemical reactions and is particularly important in industrial processes where yields must be maximized. A key aspect of equilibrium lies in its quantification through what's known as the equilibrium constant, which offers a measure of the extent of the reaction and the relative concentrations of the reactants and products at equilibrium.
Le Chatelier's Principle
Le Chatelier's principle provides insight into how a system at equilibrium responds to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. This is important for predicting the behavior of a system when it faces perturbations such as changes in concentration, pressure, or temperature.

Here are some common scenarios where Le Chatelier's principle applies:
  • If the concentration of a reactant is increased, the system shifts to consume the added reactants and produce more products.
  • When pressure is increased for a gaseous reaction, the equilibrium shifts towards the side with fewer gas molecules.
  • An increase in temperature for an exothermic reaction will shift the equilibrium towards the reactants because heat is a product of the reaction.
These adjustments are the system's natural way to achieve a new equilibrium state according to the changes imposed on it.
Equilibrium Expressions
Equilibrium expressions are mathematical formulas that allow us to calculate the equilibrium constant for a chemical reaction. For a generic reaction given by:

\(aA + bB \rightleftharpoons cC + dD\)

the equilibrium constant (\(K_c\)) is defined as the ratio of the concentration of the products raised to their coefficients to the concentration of the reactants raised to their coefficients at equilibrium:

\(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)

For gases, a related expression, the equilibrium constant in terms of partial pressure (\(K_p\)), is used:

\(K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\)

It is the equilibrium constants that quantify the position of equilibrium, and they remain the same at a given temperature, regardless of the initial concentrations of reactants and products.
Reaction Quotient
The reaction quotient (\(Q\)) is a measure that tells us how far a system is from reaching equilibrium at any point during a reaction. It has the same form as the equilibrium constant expression but is calculated using the initial concentrations or partial pressures of reactants and products before the system reaches equilibrium.

By comparing \(Q\) to the equilibrium constant \(K\), we can predict which way a reaction will shift to reach equilibrium:
  • If \(Q < K\), the reaction will proceed in the forward direction to produce more products.
  • If \(Q > K\), the reaction will move in the reverse direction to produce more reactants.
  • If \(Q = K\), the reaction is already at equilibrium.
To find the direction of the shift or to predict what will happen if conditions change, we can use \(Q\) along with Le Chatelier's principle for a comprehensive understanding of the behavior of the reaction under new circumstances.

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Most popular questions from this chapter

Water molecules in the atmosphere can form hydrogen-bonded 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?

Methane, \(\mathrm{CH}_{4},\) reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) .\) At \(630 \mathrm{K}, K_{p}\) for this reaction is \(2.26 \times 10^{-4} .\) A reaction was set up at 630 \(\mathrm{K}\) with initial partial pressures of methane of 105.1 torr and of 7.96 torr for \(\mathrm{I}_{2}\) . Calculate the pressures, in torr, of all reactants and products at equilibrium.

At \(373 \mathrm{K}, K_{p}=0.416\) for the equilibrium $$2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g) ?\)

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\operatorname{HF}(a q) \Longrightarrow \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\) (f) \(\mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (g) \(2 \mathrm{H}_{2} \mathrm{O}(I) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}^{-}(a q)\)

Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{B}(g) K_{c}\) and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

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