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Chapter 12: Problem 6

WEB Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Short Answer

Expert verified
Question: Write the equilibrium constant expressions for the following reactions: (a) \(\mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g)\) (b) \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_{2}(g)\) (d) \(\mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\) Answer: (a) \(K_{a}=\frac{[\mathrm{CO}]([H_{2}]^3)}{[\mathrm{CH}_{4}]}\) (b) \(K_{b}=\frac{([\mathrm{NO}]^4)([\mathrm{H}_{2}\mathrm{O}]^6)}{([\mathrm{NH}_{3}]^4)([\mathrm{O}_{2}]^5)}\) (c) \(K_{c}=[\mathrm{CO}_{2}]\) (d) \(K_{d}=\frac{1}{[\mathrm{NH}_{3}][\mathrm{HCl}]}\)

Step by step solution

01

Write an equilibrium constant expression for reaction (a)

For reaction (a), the given equation is: \(\mathrm{CH}_{4}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CO}(g) + 3 \mathrm{H}_{2}(g)\) To write the expression, we'll use the law of mass action. However, since \(\mathrm{H}_{2} \mathrm{O}(l)\) is a pure liquid, it will not be included in the equilibrium constant expression. So, the equilibrium constant expression is: \(K_{a}=\frac{[\mathrm{CO}]([H_{2}]^3)}{[\mathrm{CH}_{4}]}\)
02

Write an equilibrium constant expression for reaction (b)

For reaction (b), the given equation is: \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\) Now, apply the law of mass action to write the expression: \(K_{b}=\frac{([\mathrm{NO}]^4)([\mathrm{H}_{2}\mathrm{O}]^6)}{([\mathrm{NH}_{3}]^4)([\mathrm{O}_{2}]^5)}\)
03

Write an equilibrium constant expression for reaction (c)

For reaction (c), the given equation is: \(\mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_{2}(g)\) The two solids, \(\mathrm{BaCO}_{3}\) and \(\mathrm{BaO}\), will not be included in the equilibrium constant expression, so we have: \(K_{c}=[\mathrm{CO}_{2}]\)
04

Write an equilibrium constant expression for reaction (d)

For reaction (d), the given equation is: \(\mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\) As \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) is a solid, it will not be included in the equilibrium constant expression. Thus, we have: \(K_{d}=\frac{1}{[\mathrm{NH}_{3}][\mathrm{HCl}]}\) Now all equilibrium constant expressions for the given reactions have been found.

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Most popular questions from this chapter

When carbon monoxide reacts with hydrogen gas, methane and steam are formed. $$\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ At \(1127^{\circ} \mathrm{C}\), analysis at equilibrium shows that \(P_{\mathrm{CO}}=0.921 \mathrm{~atm}\), \(P_{\mathrm{H}_{2}}=1.21 \mathrm{~atm}, P_{\mathrm{CH}_{4}}=0.0391 \mathrm{~atm}\), and \(P_{\mathrm{H}_{2} \mathrm{O}}=0.0124 \mathrm{~atm} .\) What is the equilibrium constant, \(K\), for the reaction at \(1127^{\circ} \mathrm{C}\) ?

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{ccccccc}\hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}} \text { (atm) } & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{\mathrm{B}} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\ \hline\end{array}$$ Prepare a graph of \(P_{\Lambda}\) and \(P_{\mathrm{B}}\) versus time and use it to answer the following questions: (a) Estimate \(P_{\mathrm{A}}\) and \(P_{\mathrm{g}}\) after \(30 \mathrm{~s}\). (b) Estimate \(P_{\mathrm{A}}\) after \(150 \mathrm{~s}\). (c) Estimate \(P_{\mathrm{B}}\) when \(P_{\mathrm{A}}=0.700 \mathrm{~atm}\).

Write equilibrium constant expressions ( \(K\) ) for the following reactions: (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons\) \(2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(g)$$ When equilibrium is established, the partial pressures of the gases are: \(P_{\mathrm{N}_{2}}=\) \(1.2 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.80 \mathrm{~atm}, P_{\mathrm{NO}}=0.022 \mathrm{~atm} .\) (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure \(1.2\) atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?
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