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Chapter 13: Problem 33

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
The expressions for \(K\) and \(K_{\mathrm{p}}\) for the given reactions are: a. \(K = \frac{[\mathrm{H}_2 \mathrm{O}]}{[\mathrm{NH}_3]^2 [\mathrm{CO}_2]}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{NH}_3}^2 P_{\mathrm{CO}_2}}\) b. \(K = \frac{[\mathrm{N}_2] [\mathrm{Br}_2]^3}{1}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{N}_2}P_{\mathrm{Br}_2}^3}{1}\) c. \(K = \frac{[\mathrm{O}_2]^3}{1}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{O}_2}^3}{1}\) d. \(K = \frac{[\mathrm{H}_2\mathrm{O}]}{[\mathrm{H}_2]}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{H}_2}}\)

Step by step solution

01

Write the expression for K

The expression for the equilibrium constant for this reaction can be written as: \[K = \frac{[\mathrm{N}_2\mathrm{CH}_4\mathrm{O}] [\mathrm{H}_2\mathrm{O}]^1}{[\mathrm{NH}_3]^2 [\mathrm{CO}_2]^1}\] Since solids and liquids are excluded from the equilibrium constant, the equation becomes: \[K = \frac{1 [\mathrm{H}_2 \mathrm{O}]^1}{[\mathrm{NH}_3]^2 [\mathrm{CO}_2]^1}\]
02

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{NH}_3}^2 P_{\mathrm{CO}_2}}\] b. Reaction: \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\)
03

Write the expression for K

Given that we don't include pure solids in the equilibrium expressions, we can write the equilibrium constant for this reaction as: \[K = \frac{[\mathrm{N}_2]^1 [\mathrm{Br}_2]^3}{1}\]
04

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{N}_2}P_{\mathrm{Br}_2}^3}{1}\] c. Reaction: \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3\mathrm{O}_{2}(g)\)
05

Write the expression for K

Given that we don't include pure solids in the equilibrium expressions, we can write the equilibrium constant for this reaction as: \[K = \frac{[\mathrm{O}_2]^3}{1}\]
06

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{O}_2}^3}{1}\] d. Reaction: \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons\mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)
07

Write the expression for K

Given that we don't include pure solids and liquids in the equilibrium expressions, we can write the equilibrium constant for this reaction as: \[K = \frac{[\mathrm{H}_2\mathrm{O}]}{[\mathrm{H}_2]}\]
08

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{H}_2}}\]

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Most popular questions from this chapter

The reaction $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\). If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is \(0.0159\) atm and the equilibrium partial pressure of \(\mathrm{NOBr}\) is \(0.0768\) atm, calculate the partial pressure of \(\mathrm{NO}\) at equilibrium.

For a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0 \mathrm{~mol} \mathrm{NO}\) and \(1.0 \mathrm{~mol} \mathrm{Cl}_{2}\) are placed into a \(1.0\) - \(\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\ \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\ \mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\ \mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14} \end{array} $$ determine the values for the equilibrium constants for the following reactions. a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .{ }^{\circ} \mathrm{C}\). $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ If a 20.0-g sample of \(\mathrm{CaCO}_{3}\) is put into a \(10.0\) - \(\mathrm{L}\) container and heated to \(800 .{ }^{\circ} \mathrm{C}\), what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?
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