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Chapter 13: Problem 62

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of 4.5 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of 9.0 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

Short Answer

Expert verified
a. For the given reaction with initial pressure of $\mathrm{N}_{2}\mathrm{O}_{4}$ as 4.5 atm, the equilibrium pressures are \(\displaystyle P_{N_2O_4} = 3.0 \ atm\) and \(P_{NO_2} = 3.0 \ atm\). b. For the given reaction with initial pressure of $\mathrm{NO}_{2}$ as 9.0 atm, the equilibrium pressures are \(\displaystyle P_{N_2O_4} = 3.0 \ atm\) and \(P_{NO_2} = 3.0 \ atm\). c. The equilibrium partial pressures are the same in both scenarios, so it does not matter from which direction the equilibrium is reached.

Step by step solution

01

Part a - Initial situation with only N2O4#g

We are given that the initial pressure of N2O4 is 4.5 atm. Since there is no NO2 initially, the reaction can only proceed in the forward direction. Set up the ICE table: | | N2O4 | 2NO2 | |------|--------------|-----------| | I | 4.5 atm | 0 atm | | C | -x atm | +2x atm | | E | 4.5 - x atm | 2x atm | The equilibrium constant Kp is given by: \[K_p = \frac{[NO{_2}]^2}{[N_2O_4]}\] Substitute the equilibrium values: \[\begin{aligned} 0.25 &= \frac{(2x)^2}{4.5 - x} \end{aligned}\] Solve for x to find the change in pressure.
02

Part a - Calculating the change and equilibrium pressures#g

Once the equation for x is solved, we can find the equilibrium pressures for both N2O4 and NO2 by substituting the value of x back into the equilibrium row of the ICE table.
03

Part b - Initial situation with only NO2#g

We are given that the initial pressure of NO2 is 9.0 atm. Since there is no N2O4 initially, the reaction can only proceed in the reverse direction. Set up the ICE table: | | N2O4 | 2NO2 | |------|-------------|-----------| | I | 0 atm | 9.0 atm | | C | +x atm | -2x atm | | E | x atm | 9 - 2x atm| Use Kp for the reverse reaction, which equals 1/Kp for the forward reaction, and substitute the equilibrium values in the equilibrium expression: \[\begin{aligned} \frac{1}{0.25} &= \frac{x}{(9 - 2x)^2} \end{aligned}\] Solve for x to find the change in pressure.
04

Part b - Calculating the change and equilibrium pressures#g

Once the equation for x is solved, we can find the equilibrium pressures for both N2O4 and NO2 by substituting the value of x back into the equilibrium row of the ICE table.
05

Part c - Analyzing the effect of starting conditions on equilibrium#g

Compare the equilibrium partial pressures obtained in parts a and b. If they are the same, then the direction in which the equilibrium is reached does not matter.

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Most popular questions from this chapter

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is 93.71% carbon by mass, and a 0.256-mole sample of naphthalene has a mass of 32.8 g. What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene \((s) \rightleftharpoons\) naphthalene \((g)\) $K=4.29 \times 10^{-6}(\( at \)298 \mathrm{~K})\( If \)3.00 \mathrm{~g}$ solid naphthalene is placed into an en with a volume of \(5.00 \mathrm{~L}\) at $25^{\circ} \mathrm{C},$ what percentage thalene will have sublimed once equilibriur estahlished?

Consider the following reaction at \(725^{\circ} \mathrm{C} :\) $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ At equilibrium, a \(4.50-\mathrm{L}\) container has 2.6 \(\mathrm{g}\) of carbon, \(\mathrm{CO}_{2}\) at a partial pressure of \(0.0020 \mathrm{atm},\) and a total pressure of 0.572 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(725^{\circ} \mathrm{C}\)

An 8.00 -g sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ At equilibrium the total pressure and the density of the gaseous mixture were 1.80 \(\mathrm{atm}\) and 1.60 \(\mathrm{g} / \mathrm{L}\) , respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

Calculate a value for the equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ given $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) \quad K=6.8 \times 10^{-49}$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad K=5.8 \times 10^{-34}$$ (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

The reaction to prepare methanol from carbon monoxide and hydrogen $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ is exothermic. If you wanted to use this reaction to produce methanol commercially, would high or low temperatures favor a maximum yield? Explain.
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