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Chapter 15: Problem 11

Determine \(K_{c}\) for the reaction $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NOBr}(\mathrm{g})$$ from the following information (at \(298 \mathrm{K}\) ). $$\begin{aligned} 2 \mathrm{NO}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) K_{\mathrm{c}}=2.1 \times 10^{30} \\ \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NOBr}(\mathrm{g}) \quad K_{\mathrm{c}}=1.4 \end{aligned}$$

Short Answer

Expert verified
Using the provided equilibrium constants and corresponding reactions, the equilibrium constant \(K_c\) for the given reaction is \(K_c = K_{c1} * K_{c2} = \sqrt{2.1 * 10^{30}} * 1.4\).

Step by step solution

01

Analysis of given reactions

First, let's examine the given reactions. The first reaction produces N2 and O2 from 2NO, and the second reaction produces NOBr from NO and Br2. Both reactions have known \(K_c\) values. Our goal is to manipulate these two reactions in a way that it will resemble our target reaction: \(\frac{1}{2}N2(g) + \frac{1}{2}O2(g) + \frac{1}{2}Br2(g) \rightleftharpoons NOBr(g)\).
02

Manipulate the first reaction

In the first reaction, we the N2 and O2 being produced from 2NO. However, in the our target reaction, we want only half of these amounts. To match the desired reaction, therefore, we need to divide the first reaction by 2, which gives: \(NO(g) \rightleftharpoons \frac{1}{2}N2(g) + \frac{1}{2}O2(g)\). According to the rules of chemical equilibrium, when the reaction is divided by a number (in this case, 2), we should also take the square root of the equilibrium constant to get the new \(K_c\). Therefore, the new \(K_c\) for this reaction is \(K_{c1} = \sqrt{2.1 * 10^{30}}\).
03

Combine the first reaction with the second reaction

The second reaction: \(NO(g) + \frac{1}{2}Br2(g) \rightleftharpoons NOBr(g)\) already matches the remainder of our target reaction, and it does not require any manipulation. So, now we can add the manipulated first reaction with the second reaction to achieve our target equation. When adding the two reactions, we also multiply the corresponding equilibrium constants. Therefore, the \(K_c\) for the desired reaction will be \(K_{c1} * K_{c2}\), which is \(\sqrt{2.1 * 10^{30}} * 1.4\).

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Most popular questions from this chapter

Show that in terms of mole fractions of gases and total gas pressure the equilibrium constant expression for $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$$ is $$K_{\mathrm{p}}=\frac{\left(x_{\mathrm{NH}_{3}}\right)^{2}}{\left(x_{\mathrm{N}_{2}}\right)\left(x_{\mathrm{H}_{2}}\right)^{2}} \times \frac{1}{\left(P_{\mathrm{tot}}\right)^{2}}$$

The decomposition of \(\mathrm{HI}(\mathrm{g})\) is represented by the equation $$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})$$ \(\mathrm{HI}(\mathrm{g})\) is introduced into five identical \(400 \mathrm{cm}^{3}\) glass bulbs, and the five bulbs are maintained at \(623 \mathrm{K}\) Each bulb is opened after a period of time and analyzed for \(I_{2}\) by titration with \(0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq})\) $$\begin{array}{l} \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \end{array}$$ Data for this experiment are provided in the table below. What is the value of \(K_{\mathrm{c}}\) at \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

In the Ostwald process for oxidizing ammonia, a variety of products is possible- \(\mathrm{N}_{2}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO},\) and \(\mathrm{NO}_{2}-\) depending on the conditions. One possibility is $$\begin{aligned} \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) &+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}} &=2.11 \times 10^{19} \mathrm{at} 700 \mathrm{K} \end{aligned}$$ For the decomposition of \(\mathrm{NO}_{2}\) at \(700 \mathrm{K}\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.524$$ (a) Write a chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\) (b) Determine \(K_{\mathrm{p}}\) for the chemical equation you have written.

In the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K=31.4\) at \(588 \mathrm{K} .\) Equal masses of each reactant and product are brought together in a reaction vessel at \(588 \mathrm{K}\). (a) Can this mixture be at equilibrium? (b) If not, in which direction will a net change occur?

The reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}), \quad \Delta H^{\circ}=\) \(+181 \mathrm{kJ},\) occurs in high-temperature combustion processes carried out in air. Oxides of nitrogen produced from the nitrogen and oxygen in air are intimately involved in the production of photochemical smog. What effect does increasing the temperature have on (a) the equilibrium production of \(\mathrm{NO}(\mathrm{g})\) (b) the rate of this reaction?
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