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Chapter 15: Problem 22

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for $2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\( if \)K_{p}=0.0572$ at this temperature.

Short Answer

Expert verified
The equilibrium constant, \(K_c\), for the reaction \(2 CO(g) \rightleftharpoons CO_2(g) + C(s)\) at 900 K is 4.227.

Step by step solution

01

Calculate \(\Delta n\) for the reaction

To determine the change in moles of gas, subtract the moles of gaseous reactants from the moles of gaseous products: \[\Delta n = (\text{moles of } CO_2) - (\text{moles of } CO)\] In this reaction, there is 1 mole of \(CO_2\) and 2 moles of \(CO\). So, the change in moles is: \[\Delta n = 1 - 2 = -1\]
02

Calculate \(K_c\) using the relationship between \(K_p\) and \(K_c\)

Now that we know the \(\Delta n\) value, we can use the formula to find \(K_c\): \[K_p = K_c (RT)^{\Delta n}\] We are given that \(K_p = 0.0572\), \(R = 0.0821 \, L \, atm/mol \, K\), and \(T = 900 \, K\). Plug these values into the equation: \[0.0572 = K_c \left( 0.0821 \times 900 \right)^{-1}\] Now, calculate the value inside the parentheses and raise it to the power of \(\Delta n\): \(0.0572 = K_c \times (73.89)^{-1}\) Divide both sides of the equation by \((73.89)^{-1}\) to solve for \(K_c\): \[K_c = 0.0572 \times 73.89 = 4.227\] So, the equilibrium constant \(K_c\) at 900 K for the reaction \(2 CO(g) \rightleftharpoons CO_2(g) + C(s)\) is 4.227.

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Most popular questions from this chapter

Methane, \(\mathrm{CH}_{4}\), reacts with \(I_{2}\) according to the reaction $\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g)\(. At \)600 \mathrm{~K}, K_{p}$ for this reaction is \(1.95 \times 10^{-4}\). A reaction was set up at 600 \(\mathrm{K}\) with initial partial pressures of methane of \(13.3 \mathrm{kPa}\) and of $6.67 \mathrm{kPa}\( for \)\mathrm{I}_{2}$. Calculate the pressures, in \(\mathrm{kPa}\), of all reactants and products at equilibrium.

At \(2000^{\circ} \mathrm{C}\), the equilibrium constant for the reaction $2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$ is \(K_{c}=2.4 \times 10^{3} .\) If the initial concentration of NO is \(0.250 \mathrm{M}\), what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) ?

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$. A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up $15.0 \mathrm{~L}\( of a solution that is initially \)0.275 \mathrm{M}$ in acetic acid and \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

The following equilibria were attained at \(298 \mathrm{~K}:\) $$\begin{array}{c} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) \quad K_{c}=5.6 \times 10^{9} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \\ K_{c}=1.6 \times 10^{7}\end{array}$$ Based on these equilibria, calculate the equilibrium constant $\text { for } \mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)$ at \(298 \mathrm{~K}\).

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?
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