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Chapter 14: Problem 82

Consider the dissociation of iodine: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ A 1.00-g sample of \(I_{2}\) is heated to \(1200^{\circ} \mathrm{C}\) in a \(500-\mathrm{mL}\) flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in \(14.117(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.]

Short Answer

Expert verified
After following the steps, the value of \(K_P\) calculated should be the final answer.

Step by step solution

01

Calculate Initial Pressure

The first step is to calculate the initial pressure of the iodine gas before dissociation. This can be done using the ideal gas law \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. The number of moles n can be calculated by dividing the mass of iodine by its molar mass. The initial pressure can then be calculated by rearranging the ideal gas law to solve for P: \[ P = \frac{nRT}{V} \]Plug in the values \(n = \frac{1~g}{253.8~g/mol}\), \(R = 0.0821~L.atm/mol.K\), \(T = 1200 + 273.15 = 1473.15~K\), and \(V = 500~mL = 0.5~L\).
02

Compute Degree of Dissociation

Next, calculate the degree of dissociation \(\alpha\) using the hint provided in the exercise. This can be achieved by calculating the ratio of observed pressure over calculated pressure.\[\alpha = \frac{P_{observed}}{P_{initial}}\]Substitute the given observed pressure \(1.51~atm\) and previously calculated initial pressure from step 1.
03

Use Degree of Dissociation to Find Partial Pressures

Now calculate the partial pressures at equilibrium for \(\mathrm{I}_2\) and \(\mathrm{I}\). This can be done by using the equation \(P_{total} = P_{\mathrm{I}_2} + 2P_{\mathrm{I}}\). The pressure of \(I_2\) is \((1-\alpha) \times P_{initial}\) and the pressure of \(\mathrm{I}\) is \(2 \times \alpha \times P_{initial}\). Substitute the values of \(\alpha\) and \(P_{initial}\) obtained from previous steps.
04

Calculate \(K_P\)

Finally, \(K_P\) can be calculated using the expression:\[ K_P = \frac{P_{I}^2}{P_{I_2}} \]Substitute the values of \(P_I\) and \(P_{I_2}\) obtained from step 3 to find the answer.

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Most popular questions from this chapter

Consider this equilibrium reaction in a closed container: $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ What will happen if the following occurs? (a) The volume is increased. (b) Some \(\mathrm{CaO}\) is added to the mixture. (c) Some \(\mathrm{CaCO}_{3}\) is removed. (d) Some \(\mathrm{CO}_{2}\) is added to the mixture. (e) A few drops of a \(\mathrm{NaOH}\) solution are added to the mixture. (f) A few drops of a \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water). (g) Temperature is increased.

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the changes in the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) over time for the following situations: (a) Initially only A is present. (b) Initially only B is present. (c) Initially both A and B are present (with A in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

In the uncatalyzed reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ the pressure of the gases at equilibrium are \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=\) 0.377 and \(P_{\mathrm{NO}_{2}}=1.56 \mathrm{~atm}\) at \(100^{\circ} \mathrm{C} .\) What would happen to these pressures if a catalyst were added to the mixture?

The dissociation of molecular iodine into iodine atoms is represented as $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ At \(1000 \mathrm{~K},\) the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(\mathrm{I}_{2}\) in a 2.30 -L flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?

The equilibrium constant \(K_{P}\) for the following reaction is \(4.31 \times 10^{-4}\) at \(375^{\circ} \mathrm{C}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ In a certain experiment a student starts with 0.862 atm of \(\mathrm{N}_{2}\) and 0.373 atm of \(\mathrm{H}_{2}\) in a constantvolume vessel at \(375^{\circ} \mathrm{C}\). Calculate the partial pressures of all species when equilibrium is reached.
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