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Chapter 13: Problem 61

At \(1100 \mathrm{K}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ Calculate the equilibrium partial pressures of $\mathrm{SO}_{2}, \mathrm{O}_{2},\( and \)\mathrm{SO}_{3}$ produced from an initial mixture in which \(P_{\mathrm{SO}_{2}}=P_{\mathrm{O}_{2}}=\) 0.50 \(\mathrm{atm}\) and \(P_{\mathrm{so}_{3}}=0 .\) (Hint: If you don't have a graphing calculator, then use the method of successive approximations to solve, as discussed in Appendix \(1.4 . )\)

Short Answer

Expert verified
The equilibrium partial pressures for the given reaction at \(1100K\) with \(K_p = 0.25\) are approximately: - \(P_{\mathrm{SO}_{2}} \approx 0.192 \ \mathrm{atm}\) - \(P_{\mathrm{O}_{2}} \approx 0.346 \ \mathrm{atm}\) - \(P_{\mathrm{SO}_{3}} \approx 0.308 \ \mathrm{atm}\)

Step by step solution

01

Write the Kp expression

First, we'll write the Kp expression for the given reaction. \(K_{p} = \frac{P_{\mathrm{SO}_3}^2}{(P_{\mathrm{O}_2})(P_{\mathrm{SO}_2})^2} \)
02

Define change in partial pressure (x)

Let's define the change in partial pressure as x. The reaction consumes 2 moles of SO₂ and 1 mole of O₂ to produce 2 moles of SO₃. So the change in partial pressures will be: - Decrease in SO₂ pressure: -2x - Decrease in O₂ pressure: -x - Increase in SO₃ pressure: +2x
03

Write the equation for equilibrium partial pressures

Using the initial partial pressures and the changes, we can write the equilibrium partial pressures as follows: - \(P_{\mathrm{SO}_2} = 0.50 -2x\) - \(P_{\mathrm{O}_2} = 0.50 - x\) - \(P_{\mathrm{SO}_3} = 0+ 2x\)
04

Substitute equilibrium partial pressures into Kp expression

Now, we'll substitute the equilibrium partial pressures into the Kp expression: \(K_{p} = 0.25 = \frac{(2x)^2}{(0.50 - x)(0.50 - 2x)^2} \)
05

Solve for x using the successive approximations method

We'll now use the method of successive approximations to get the value of x. 1. Guess a value for x (e.g., x = 0.1) 2. Substitute the value into the equation and solve for x. 3. Continue this process until the value of x converges. Using this method, we get x ≈ 0.154.
06

Calculate the equilibrium partial pressures

Now, we have the value of x (0.154) and can plug it into the expressions for equilibrium partial pressures: \(P_{\mathrm{SO}_2} = 0.50 - 2(0.154) = 0.192 \ \mathrm{atm}\) \(P_{\mathrm{O}_2} = 0.50 - 0.154 = 0.346 \ \mathrm{atm}\) \(P_{\mathrm{SO}_3} = 2(0.154) = 0.308 \ \mathrm{atm}\) The equilibrium partial pressures are: - \(P_{\mathrm{SO}_{2}} \approx 0.192 \ \mathrm{atm}\) - \(P_{\mathrm{O}_{2}} \approx 0.346 \ \mathrm{atm}\) - \(P_{\mathrm{SO}_{3}} \approx 0.308 \ \mathrm{atm}\)

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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.

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.

An equilibrium mixture contains 0.60 g solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of 2.60 atm and 2.89 atm, respectively. Calculate the value of \(K_{\mathrm{p}}\) for the reaction $\mathrm{C}(s)+\mathrm{CO}_{2}(g) \Longrightarrow 2 \mathrm{CO}(g)$

Consider the following reaction at \(725^{\circ} \mathrm{C} :\) $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ At equilibrium, a \(4.50-\mathrm{L}\) container has 2.6 \(\mathrm{g}\) of carbon, \(\mathrm{CO}_{2}\) at a partial pressure of \(0.0020 \mathrm{atm},\) and a total pressure of 0.572 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(725^{\circ} \mathrm{C}\)

For the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ at \(600 . \mathrm{K}\) , the equilibrium constant, \(K_{\mathrm{p}},\) is \(11.5 .\) Suppose that 2.450 \(\mathrm{g} \mathrm{PCl}_{5}\) is placed in an evacuated 500 -mL bulb, which is then heated to \(600 . \mathrm{K}\) . a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of PCl_ at equilibrium?
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