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Chapter 12: Problem 56

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 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 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
The equilibrium partial pressures of the gases are the same in both cases, with an equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) as 2.25 atm and an equilibrium partial pressure of \(\mathrm{NO}_{2}\) as 4.5 atm. Hence, it does not matter from which direction the equilibrium position is reached; the equilibrium partial pressures will be the same irrespective of the initial conditions.

Step by step solution

01

Write the expression for \(K_p\)

Given the balanced chemical equation, the expression for \(K_p\) is: $$K_p = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]}$$
02

Define the variable for equilibrium pressures

Let the partial pressure of N2O4 at equilibrium be x atm. Since 2 moles of NO2 are produced for each mole of N2O4 consumed, the partial pressure of NO2 at equilibrium will be 2(4.5-x) atm.
03

Write the equation for \(K_p\) at equilibrium

Substitute the partial pressures at equilibrium in the expression for \(K_p\): $$0.25 = \frac{(2(4.5-x))^2}{x}$$
04

Solve for x

Solve the equation for the value of x: $$0.25x = (2(4.5-x))^2$$ $$x = 2.25$$
05

Calculate the equilibrium partial pressures

Now that we have the value of x, we can determine the equilibrium partial pressures of both N2O4 and NO2: $$[\mathrm{N}_{2}\mathrm{O}_{4}] = 2.25\,\text{atm}$$ $$[\mathrm{NO}_{2}] = 2 \times (4.5 - 2.25) = 4.5\,\text{atm}$$ b. Calculate the equilibrium partial pressures of N2O4 and NO2 with initial pressure of NO2.
06

Define the variable for equilibrium pressures

Let the partial pressure of NO2 at equilibrium be y atm. Since 1 mole of N2O4 is produced for every 2 moles of NO2 consumed, the partial pressure of N2O4 at equilibrium will be (9 - y)/2 atm.
07

Write the equation for \(K_p\) at equilibrium

Substitute the partial pressures at equilibrium in the expression for \(K_p\): $$0.25 = \frac{y^2}{\frac{9-y}{2}}$$
08

Solve for y

Solve the equation for the value of y: $$0.25(9-y) = 2y^2$$ $$y = 4.5$$
09

Calculate the equilibrium partial pressures

Now that we have the value of y, we can determine the equilibrium partial pressures of both N2O4 and NO2: $$[\mathrm{N}_{2}\mathrm{O}_{4}] = \frac{9 - 4.5}{2} = 2.25\,\text{atm}$$ $$[\mathrm{NO}_{2}] = 4.5\,\text{atm}$$ c. Conclusion. The equilibrium pressures in both cases a and b are the same: $$[\mathrm{N}_{2}\mathrm{O}_{4}] = 2.25\,\text{atm}$$ $$[\mathrm{NO}_{2}] = 4.5\,\text{atm}$$ Thus, it does not matter from which direction equilibrium position is reached. The equilibrium partial pressures of the gases will be the same irrespective of the initial conditions.

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Most popular questions from this chapter

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{K}\) at an initial pressure of 1.00 atm, where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$. At equilibrium, the partial pressure of \(S_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{K}\)

The gas arsine, AsH_, decomposes as follows:$$2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g)$$.In an experiment at a certain temperature, pure \(\mathrm{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr.After 48 hours the pressure in the flask was observed to be constant at 488.0 torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(g)\) b. Calculate \(K_{\mathrm{p}}\) for this reaction.

Consider the following reactions:\(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g) \quad\) and \(\quad \mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\).List two property differences between these two reactions that relate to equilibrium.

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 \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\).
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