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Chapter 13: 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 \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

Short Answer

Expert verified
For part a, the equilibrium partial pressures for N2O4 and NO2 are 3.97 atm and 1.06 atm, respectively. For part b, they are 1.097 atm and 6.806 atm, respectively. Comparing the results, the ratios of equilibrium partial pressures for both cases are approximately the same (0.267), indicating that it doesn't matter from which direction the equilibrium position is reached, as long as the reaction is at equilibrium and has the same Kp value.

Step by step solution

01

Write the expression for Kp

For the given reaction, the expression for Kp can be written as: \[ K_{\mathrm{p}} = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2}\mathrm{O}_{4}]} \] #a#
02

Set up the ICE table for part a

Let x be the change in concentration of N2O4 and NO2 during the reaction. The ICE (Initial, Change, Equilibrium) table for part a will look like this: | | N2O4 | NO2 | |------|--------------|--------------| |Initial | 4.5 | 0 | |Change | -x | + 2x | |Equilibrium | 4.5 - x | 2x |
03

Substitute the given Kp value and set up the equation

Now, we plug in the values from the ICE table into the Kp expression: \[ 0.25 = \frac{(2x)^2}{(4.5 - x)} \]
04

Solve the equation for x

Solve the quadratic equation to find the value of x to get the equilibrium partial pressures: \[ 0.25(4.5 - x) = 4x^2 \Rightarrow x^2 = 0.28125\] \[ x = \pm 0.53\] Since the concentration cannot be negative, x = 0.53.
05

Calculate the equilibrium partial pressures for part a

Now, substitute the value of x in the expressions for equilibrium partial pressures of N2O4 and NO2: \[ [\mathrm{N}_{2}\mathrm{O}_{4}] = 4.5 - x = 3.97 \,atm \] \[ [\mathrm{NO}_{2}] = 2x = 1.06\, atm \] The equilibrium partial pressures of N2O4 and NO2 for part a are 3.97 atm and 1.06 atm, respectively. #b#
06

Set up the ICE table for part b

For part b, the initial pressure of NO2 is given. The ICE table for part b will look like this: | | N2O4 | NO2 | |------|--------------|--------------| |Initial | 0 | 9 | |Change | +x/2 | -x | |Equilibrium | x/2 | 9 - x |
07

Substitute the values and set up the equation

Plug in the values from the ICE table into the Kp expression: \[ 0.25 = \frac{[(9 - x)]^2}{(x/2)} \]
08

Solve the equation for x

Solve the quadratic equation to find the value of x and get the equilibrium partial pressures: \[ 0.25\frac{x}{2} = (9 - x)^2 \Rightarrow x^2 = 4.815625\] \[ x = \pm 2.194\] Since the concentration cannot be negative, x = 2.194.
09

Calculate the equilibrium partial pressures for part b

Now, substitute the value of x in the expressions for equilibrium partial pressures of N2O4 and NO2: \[ [\mathrm{N}_{2}\mathrm{O}_{4}] = x/2 = 1.097\,atm \] \[ [\mathrm{NO}_{2}] = 9 - x = 6.806\,atm \] The equilibrium partial pressures of N2O4 and NO2 for part b are 1.097 atm and 6.806 atm, respectively. #c#
10

Compare the results

Comparing the results obtained in part a and part b, we can see that the ratios of equilibrium partial pressures for both cases is the same. Part a: \[ \frac{[\mathrm{NO}_{2}]}{[\mathrm{N}_{2}\mathrm{O}_{4}]} = \frac{1.06}{3.97} \approx 0.267 \] Part b: \[ \frac{[\mathrm{NO}_{2}]}{[\mathrm{N}_{2}\mathrm{O}_{4}]} = \frac{6.806}{1.097} \approx 0.267 \] This indicates that it doesn't matter from which direction the equilibrium position is reached, as long as the reaction is at equilibrium and has the same Kp value.

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Most popular questions from this chapter

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=0.041 M\), \(\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calculate the value of \(K\) for the reaction.

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 \(0.256\) -mol sample of naphthalene has a mass of \(32.8 \mathrm{~g}\). 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 enclosed space with a volume of \(5.00 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\), what percentage of the naphthalene will have sublimed once equilibrium has been established?

At \(25^{\circ} \mathrm{C}\), gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that \(12.5 \%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is \(0.900\) atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ A 1.00-L flask containing \(10.0 \mathrm{~g} \mathrm{NaHCO}_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C}\). a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) . \mathrm{A}\) 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 B?" What do you tell your friend?
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