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Chapter 13: Problem 104

Consider the reaction $$ \mathrm{P}_{4}(g) \longrightarrow 2 \mathrm{P}_{2}(g) $$ 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 \(\mathrm{P}_{4}(\mathrm{~g})\) and \(\mathrm{P}_{2}(g)\) has a total pressure of \(1.00 \mathrm{~atm} .\) Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

Short Answer

Expert verified
The equilibrium pressures of P_4(g) and P_2(g) are approximately 0.8942 atm and 0.2116 atm, respectively, and the mole fraction of P_4(g) that has dissociated at equilibrium is 0.1058.

Step by step solution

01

Write the balanced chemical equation

The given balanced chemical equation for the reaction is: \[ \mathrm{P}_{4}(g) \longrightarrow 2 \mathrm{P}_{2}(g) \]
02

Set up the Equilibrium Constant expression

Since we have K_p given, we can write the equilibrium constant expression as: \[ K_p = \frac{(\mathrm{P}_{2})^2}{\mathrm{P}_{4}} \]
03

Set the initial condition for P_4(g) and dissociation

Let x be the moles of P_4(g) that dissociate to reach equilibrium. Then the initial pressures of P_4(g) and P_2(g) are: P_4 = P - x P_2 = 2x P_total is given as 1 atm.
04

Calculate the equilibrium pressures using K_p

Substitute the equilibrium pressures to the K_p expression: \[ 1.00 \times 10^{-1} = \frac{(2x)^2}{(1-x)} \] We can now solve for x: \[ 1.00 \times 10^{-1} = \frac{4x^2}{(1-x)} \] \[ 0.1 (1-x) = 4x^2 \]
05

Solve the equation for x

Rearrange and solve the quadratic equation for x: \[ 4x^2 + 0.1x - 0.1 = 0 \] Using the quadratic formula to solve for x, we get: \[ x = \frac{-0.1 \pm \sqrt{(-0.1)^2 - 4(4)(-0.1)}}{2(4)} \] Solving for the positive root (as x cannot be negative), we get: \[ x \approx 0.1058 \]
06

Calculate equilibrium pressures and mole fraction

Now, we can calculate equilibrium pressures using the values of x: P_4 = 1 - x = 1 - 0.1058 ≈ 0.8942 atm P_2 = 2x ≈ 0.2116 atm Finally, we can calculate the fraction (by moles) of P_4(g) that has dissociated to reach equilibrium: \[ \text{Mole fraction} = \frac{x}{1} = 0.1058 \] #Conclusion# The equilibrium pressures of P_4(g) and P_2(g) are approximately 0.8942 atm and 0.2116 atm, respectively, and the mole fraction of P_4(g) that has dissociated at equilibrium is 0.1058.

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Most popular questions from this chapter

For the reaction: $$ 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g) $$ \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 M\), calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

For the reaction $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3}\) \(M,\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{~N}_{2}(g)\right]=5.3 \times 10^{-2} M\), and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

Consider the following exothermic reaction at equilibrium: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8}\). What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{~K}\) at an initial pressure of \(1.00 \mathrm{~atm}\), 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 \(\mathrm{S}_{\mathrm{g}}\) is \(0.25 \mathrm{~atm} .\) Calculate \(K\). for this reaction at \(1325 \mathrm{~K}\).
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