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Chapter 15: Problem 12

Given the equilibrium constant values $$\begin{aligned} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{c}}=2.7 \times 10^{-18} \\ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) & \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) K_{\mathrm{c}}=4.6 \times 10^{-3} \\ \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) & \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.1 \times 10^{-9} \end{aligned}$$ Determine a value of \(K_{\mathrm{c}}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

Short Answer

Expert verified
The equilibrium constant (\(K_c\)) for the reaction 2 N2O(g) + 3 O2(g) ⇌ 2 N2O4(g) is 0.65

Step by step solution

01

Identify and Flip Equations

First, we identify which of the given equations will be useful to get the desired target equation. We can see that the second equation needs to be reversed so as to have N2O4 on the product side, and the first and third equations need to be doubled so as to get 2N2O and NO2, respectively, with the correct stoichiometric coefficients.
02

Find the New Equilibrium Constants

Calculate the new equilibrium constants. Recall the rules that apply to flipping or reversing a reaction (you take the reciprocal of the equilibrium constant, \(K_c\)) and for multiplying the entire equation by a factor (you raise the equilibrium constant to the power equal to the factor). For the first equation, we have \(K_c' = (2.7 \times 10^{-18})^2\), for the second equation, \(K_c' = 1 / (4.6 \times 10^{-3})\), and for the third equation, \(K_c' = (4.1 \times 10^{-9})^2\).
03

Compute \(K_c'C\)'s and \(K_c\) for the Desired Reaction

Determine the product of the altered \(K_c'\) values, which will be the desired equilibrium constant. Hence, \(K_c = K_{c1}' \times K_{c2}' \times K_{c3}' = (2.7 \times 10^{-18})^2 \times 1 / (4.6 \times 10^{-3}) \times (4.1 \times 10^{-9})^2 = 0.65\)

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Most popular questions from this chapter

The following reaction is used in some self-contained breathing devices as a source of \(\mathrm{O}_{2}(\mathrm{g})\) $$\begin{aligned} 4 \mathrm{KO}_{2}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s}) &+3 \mathrm{O}_{2}(\mathrm{g}) \\\ K_{\mathrm{p}} &=28.5 \mathrm{at} 25^{\circ} \mathrm{C} \end{aligned}$$ Suppose that a sample of \(\mathrm{CO}_{2}(\mathrm{g})\) is added to an evacuated flask containing \(\mathrm{KO}_{2}(\mathrm{s})\) and equilibrium is established. If the equilibrium partial pressure of \(\mathrm{CO}_{2}(\mathrm{g})\) is found to be \(0.0721 \mathrm{atm},\) what are the equilibrium partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) and the total gas pressure?

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 \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

For the following reaction, \(K_{\mathrm{c}}=2.00\) at \(1000^{\circ} \mathrm{C}\) $$2 \operatorname{COF}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{CF}_{4}(\mathrm{g})$$ If a \(5.00 \mathrm{L}\) mixture contains \(0.145 \mathrm{mol} \mathrm{COF}_{2}, 0.262 \mathrm{mol}\) \(\mathrm{CO}_{2},\) and \(0.074 \mathrm{mol} \mathrm{CF}_{4}\) at a temperature of \(1000^{\circ} \mathrm{C}\) (a) Will the mixture be at equilibrium? (b) If the gases are not at equilibrium, in what direction will a net change occur? (c) How many moles of each gas will be present at equilibrium?

At \(500 \mathrm{K}\), a 10.0 L equilibrium mixture contains 0.424 \(\mathrm{mol} \mathrm{N}_{2}, 1.272 \mathrm{mol} \mathrm{H}_{2},\) and \(1.152 \mathrm{mol} \mathrm{NH}_{3} .\) The mixture is quickly chilled to a temperature at which the \(\mathrm{NH}_{3}\) liquefies, and the \(\mathrm{NH}_{3}(1)\) is completely removed. The 10.0 L gaseous mixture is then returned to \(500 \mathrm{K}\), and equilibrium is re-established. How many moles of \(\mathrm{NH}_{3}(\mathrm{g})\) will be present in the new equilibrium mixture? $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3} \quad K_{\mathrm{c}}=152 \text { at } 500 \mathrm{K}$$

For the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\) \(\mathrm{H}_{2}(\mathrm{g}), K_{\mathrm{p}}=23.2\) at \(600 \mathrm{K} .\) Explain which of the fol- lowing situations might equilibrium: (a) \(\quad P_{\mathrm{CO}}=P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{CO}_{2}}=P_{\mathrm{H}_{2}} ; \quad\) (b) \(\quad P_{\mathrm{H}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=\) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{CO}} ; \quad(\mathrm{c}) \quad\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2}}\right)=\left(P_{\mathrm{CO}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}^{2}\right)\) (d) \(P_{\mathrm{CO}_{2}} / P_{\mathrm{H}_{2} \mathrm{O}}=P_{\mathrm{H}_{2}} / P_{\mathrm{CO}}\)
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