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Chapter 12: Problem 96

For the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g), K_{\mathrm{p}}=0.25\) at a certain temperature. If 0.040 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is reacted initially, calculate the equilibrium partial pressures of \(\mathrm{NO}_{2}(g)\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\).

Short Answer

Expert verified
The equilibrium partial pressures of \(\mathrm{N}_{2}\mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) are 0.02 atm and 0.04 atm, respectively.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation given in the problem is \(\mathrm{N}_{2}\mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\).
02

Set up Qp

The reaction quotient, Qp, is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the partial pressures of the reactants raised to their stoichiometric coefficients. In this case, Qp is \(\frac{(\mathrm{NO}_2)^2}{(\mathrm N_2O_4)}\).
03

Find change in pressure at equilibrium

Let x represent the change in pressure at equilibrium. Then, the equilibrium pressure of N2O4 will be (0.040 - x) atm and the pressure of NO2 will be 2x atm.
04

Set up the Kp equation

Given, \(K_{p} = 0.25\). We can set up the Kp equation using the equilibrium pressures of NO2 and N2O4: \[\frac{(2x)^2}{(0.040-x)}=0.25\]
05

Solve for x

Solve the equation for x: \[(2x)^2 = 0.25(0.040-x)\] \[4x^2 = 0.01 - 0.25x\] Rearrange the equation into a quadratic: \[4x^2 + 0.25x - 0.01 = 0\] Now, use the quadratic formula to find the value of x: \[x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\] \[x = \frac{-0.25\pm\sqrt{(0.25)^2 - 4(4)(-0.01)}}{2(4)}\] Find two possible values of x: \[x_1 \approx 0.02 \, \text{atm}\] \[x_2 \approx -0.005 \, \text{atm}\] We can ignore x2 since it gives a negative pressure, which is not possible in this context.
06

Calculate the equilibrium partial pressures

Now that we have the value of x, we can find the equilibrium partial pressures of N2O4 and NO2 as follows: \[\mathrm{N}_{2}\mathrm{O}_{4}: 0.040 - x \approx 0.040 - 0.02 = 0.02 \, \text{atm}\] \[\mathrm{NO}_{2}: 2x \approx 2(0.02) = 0.04 \, \text{atm}\] The equilibrium partial pressures of N2O4 and NO2 are 0.02 atm and 0.04 atm, respectively.

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Most popular questions from this chapter

Consider the following reaction:$$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$.Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C},\) will \(^{14} \mathrm{C}\) be found only in CO molecules for an indefinite period of time? Explain.

For the reaction $$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$,\(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium in a \(2.0-\mathrm{L}\) container it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} \mathrm{M}\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} \mathrm{M} .\) Calculate the moles of \(\mathrm{O}_{2}(g)\) present under these conditions.

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned}2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\\P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm}\end{aligned}$$. Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

Lexan is a plastic used to make compact discs, eyeglass lenses, and bulletproof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right),\) an extremely poisonous gas. Phosgene decomposes by the reaction,$$\operatorname{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$,for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\operatorname{COCl}_{2}(g) \rightleftharpoons \operatorname{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)
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