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Chapter 12: Problem 26

A sealed flask has \(0.541\) atm of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\). The following equilibrium is established. $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the partial pressure of oxygen is measured to be \(0.216\) atm. Calculate \(K\) for the decomposition of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\).

Short Answer

Expert verified
Question: Calculate the equilibrium constant (K) for the decomposition of SO_3 at 1000 K, given the initial partial pressure of SO_3 as 0.541 atm, and the equilibrium partial pressure of O_2 as 0.216 atm. Answer: The equilibrium constant (K) for the decomposition of SO_3 at 1000 K is approximately 0.418.

Step by step solution

01

Write the equilibrium constant expression

For the given chemical reaction, the equilibrium constant expression (K) can be written in terms of the partial pressures of the reactants and products: $$K_p = \frac{(\mathrm{SO}_2)^2 \cdot (\mathrm{O}_2)}{(\mathrm{SO}_3)^2}$$
02

Set up the ICE table

Use an ICE (Initial, Change, Equilibrium) table to represent the initial partial pressures, the changes during the reaction, and the equilibrium partial pressures of each compound involved in the reaction. Note that we are not given the initial partial pressures of SO_2 and O_2, so we can assume that they are initially 0. | | SO_3 | SO_2 | O_2 | |---------------|------|------|------| | Initial | 0.541| 0 | 0 | | Change | -x | +x | +x | | Equilibrium |0.541-x|x | 0.216|
03

Identify the relationship between x and given partial pressures

Since we know the equilibrium partial pressure of O_2 (0.216 atm), we can now relate it to the change (x): $$ x = 0.216 $$
04

Calculate equilibrium partial pressures of SO_2 and SO_3

Using the value of x, we can find the equilibrium partial pressures of SO_2 and SO_3: For SO_2: $$P_{SO_2} = x = 0.216 \ \mathrm{atm}$$ For SO_3: $$P_{SO_3} = 0.541-x = 0.541-0.216 = 0.325 \ \mathrm{atm}$$
05

Calculate the equilibrium constant (K)

Now that we have the equilibrium partial pressures of each compound, plug them into the equilibrium constant expression to solve for K: \begin{aligned} K_p &= \frac{(\mathrm{SO}_2)^2 \cdot (\mathrm{O}_2)}{(\mathrm{SO}_3)^2} \\ K_p &= \frac{(0.216)^2 \cdot (0.216)}{(0.325)^2} \\ K_p &= 0.418 \end{aligned} The equilibrium constant for the decomposition of SO_3 at 1000 K is approximately 0.418.

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Most popular questions from this chapter

For the system $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ at \(1000 \mathrm{~K}, K=0.45 .\) Sulfur trioxide, originally at \(1.00 \mathrm{~atm}\) pressure, partially dissociates to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) at \(1000 \mathrm{~K}\). What is its partial pressure at equilibrium?

At a certain temperature, \(K\) is \(1.3 \times 10^{5}\) for the reaction $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ What is the equilibrium pressure of hydrogen sulfide if those of hydrogen and sulfur gases are \(0.103\) atm and \(0.417\) atm, respectively?

At \(460^{\circ} \mathrm{C}\), the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g)$$ has \(K=84.7\). All gases are at an initial pressure of \(1.25\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

At \(627^{\circ} \mathrm{C}, K=0.76\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ Calculate \(K\) at \(627^{\circ} \mathrm{C}\) for (a) the synthesis of one mole of sulfur trioxide gas. (b) the decomposition of two moles of \(\mathrm{SO}_{3}\).

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{ccccccc}\hline \text { Time (s) } & 0 & 30 & 45 & 60 & 75 & 90 \\\ P_{\mathrm{A}} \text { (atm) } & 0.500 & 0.390 & 0.360 & 0.340 & 0.325 & 0.325 \\\ P_{\text {B }} \text { (atm) } & 0.000 & 0.220 & 0.280 & 0.320 & 0.350 & 0.350 \\\\\hline\end{array}$$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after 45 s? After 90 s?
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