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Chapter 13: Problem 114

Consider the reaction $$\mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g)$$ 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 \(\mathrm{P}_{4}(g)\) 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 \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

Short Answer

Expert verified
The equilibrium pressures of $\mathrm{P}_{4}(g)$ and $\mathrm{P}_{2}(g)$ are approximately 0.9084 atm and 0.1832 atm, respectively. The fraction of $\mathrm{P}_{4}(g)$ that has dissociated to reach equilibrium is approximately 0.0916 or 9.16%.

Step by step solution

01

Write down the Kp expression and the equilibrium pressure expressions

For the given reaction: $$\mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g)$$ The expression for the equilibrium constant Kp is given by: $$K_{\mathrm{p}}=\frac{[\mathrm{P}_{2}]^{2}}{[\mathrm{P}_{4}]}$$ Let the initial pressure of P4(g) be P, and since there is no P2(g) initially, the initial pressure of P2(g) is 0. At equilibrium, the pressure of P4(g) decreases by x atm, and the pressure of P2(g) increases by 2x atm. Therefore, the equilibrium pressures of P4(g) and P2(g) are (P-x) atm and 2x atm, respectively.
02

Use the given total pressure and the equilibrium pressure expressions to create a system of equations

We are given that the total pressure at equilibrium is 1 atm. Therefore, we can write an equation relating the equilibrium pressures of P4(g) and P2(g): $$(P - x) + 2x = 1$$ Now, we can use the Kp expression and the equilibrium pressures to write another equation: $$K_{\mathrm{p}}=\frac{(2x)^{2}}{(P-x)}$$ We are given that Kp = 1.00 × 10⁻¹, so we can substitute that in the equation: $$1.00 \times 10^{-1}=\frac{(2x)^{2}}{(P-x)}$$ Now, we have a system of two equations with two unknowns (P and x).
03

Solve the system of equations for P and x

First, we solve the equation (P-x) + 2x = 1 for P in terms of x: $$P = 1 - x$$ Now substitute this expression for P into the Kp equation: $$1.00 \times 10^{-1}=\frac{(2x)^{2}}{(1-x)}$$ Rearrange the equation to solve for x: $$(2x)^{2} = 1.00 \times 10^{-1} (1-x)$$ Solve this quadratic equation to obtain x: $$x\approx 0.0916$$ Now, we find the equilibrium pressures of P4(g) and P2(g) using the values of P and x: $$\mathrm{P}_{4} = P - x = 1 - 0.0916 \approx 0.9084 \:\text{atm}$$ $$\mathrm{P}_{2} = 2x \approx 0.1832 \:\text{atm}$$
04

Calculate the fraction of P4(g) that has dissociated

Finally, we will find the fraction of P4(g) that has dissociated to reach equilibrium. The initial pressure of P4(g) is P, and the decrease in pressure is x. So, the fraction of dissociated P4(g) is the ratio of the decrease in pressure to the initial pressure: $$\frac{\text{Dissociated P}_4}{\text{Initial P}_4} = \frac{x}{P}$$ Substitute the values of x and P: $$\frac{0.0916}{1} = 0.0916$$ Hence, the fraction of P4(g) that has dissociated to reach equilibrium is approximately 0.0916, or 9.16%.

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Most popular questions from this chapter

Le Chatelier's principle is stated (Section 13.7\()\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$ is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. $2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$ b. $\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)$ c. $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)$ d. $4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)$

A sample of \(\mathrm{S}_{8}(g)\) 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 \(\mathrm{S}_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at 1325 \(\mathrm{K}\) .

Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the following reaction: $$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \quad K=2.2$$ For the following mixtures (a-d), will the concentration of $\mathrm{H}_{2} \mathrm{O}$ increase, decrease, or remain the same as equilibrium is established? a. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.10 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.010 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.010 M$ b. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.22 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.0020 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.0020 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=0.10 M$ c. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=0.88 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.12 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.044 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=6.0 M$ d. $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=4.4 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=4.4 M$ $\quad\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.88 M,\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=10.0 M$ e. What must the concentration of water be for a mixture with $\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\right]=2.0 M,\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=0.10 M,$ and \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right]=5.0 M\) to be at equilibrium? f. Why is water included in the equilibrium expression for this reaction?

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: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q)$$ 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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