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Chapter 12: Problem 104

Consider the reaction $$\mathbf{P}_{4}(g) \longrightarrow 2 \mathbf{P}_{2}(g)$$, where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{K},\) the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium.

Short Answer

Expert verified
The equilibrium pressures of \(P_{4}(g)\) and \(P_{2}(g)\) are approximately 0.844 atm and 0.312 atm, respectively. The fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium is 0.156.

Step by step solution

01

Write the balanced chemical equation for the reaction

The given reaction is: \(P_4(g) \longrightarrow 2P_2(g)\)
02

Write the expression for the equilibrium constant

For the given reaction, the expression for the equilibrium constant (\(K_p\)) is: \[K_p = \frac{[P_2]^2}{[P_4]}\]
03

Set up the ICE table

Set up an ICE table to represent the initial concentrations, the change in concentrations, and the equilibrium concentrations of the reactants and products. ``` [P_4] [P_2] Initial P_4_0 0 Change -x +2x Equilibrium P_4_0 - x 2x ``` Initial pressures are unknown except for total pressure, so let the initial pressure of \(P_4(g)\) be \(P_{4_0}\) and the initial pressure of \(P_2(g)\) is 0. As the reaction progresses, the pressure of \(P_4(g)\) decreases by x, and the pressure of \(P_2(g)\) increases by 2x. At equilibrium, the pressure of \(P_4(g)\) is \(P_{4_0} - x\) and the pressure of \(P_2(g)\) is \(2x\).
04

Substitute the equilibrium pressures into the equilibrium constant expression

Replace the concentrations in the \(K_p\) expression by equilibrium pressures from the ICE table: \[K_p = \frac{(2x)^2}{(P_{4_0} - x)}\] It is given that the total pressure at equilibrium is 1 atm, so \((P_{4_0} - x) + 2x = 1\). Also, given that \(K_p=1.00 \times 10^{-1}\).
05

Solve the equations for x and the equilibrium pressures

Solve the equations obtained in step 4 for x and substitute the value of x back into the equilibrium pressure expressions to find the equilibrium pressures of the reactants and products. From the total pressure equation, we get \(P_{4_0} + x = 1\). Substituting this into the \(K_p\) equation, we get: \[\frac{(2x)^2}{(1-x)}= 1.00 \times 10^{-1}\] Solving for x, we get \(x \approx 0.156\) Now, substitute the value of x back into the equilibrium pressure expressions: \[P_{4} = P_{4_0} - x \approx 1 - 0.156 \approx 0.844\ \text{atm}\] \[P_{2} = 2x \approx 2 \times 0.156 \approx 0.312\ \text{atm}\]
06

Calculate the fraction of \(P_{4}(g)\) that has dissociated

To find the fraction of \(P_4(g)\) dissociated, divide the change in pressure due to dissociation by the initial pressure of \(P_4(g)\): Fraction dissociated = \(\frac{x}{P_{4_0}} = \frac{0.156}{1} = 0.156\) The equilibrium pressures of \(P_{4}(g)\) and \(P_{2}(g)\) are approximately 0.844 atm and 0.312 atm, respectively. The fraction (by moles) of \(P_{4}(g)\) that has dissociated to reach equilibrium is 0.156.

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Most popular questions from this chapter

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(aq)+\mathrm{H}_{2}(a q)$$. =Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10},\) what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

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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. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Consider an equilibrium mixture of four chemicals \((\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D},\) all gases) reacting in a closed flask according to the equation:$$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$. a. You add more \(A\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more \(\mathrm{D}\) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.
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