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Chapter 13: Problem 75

The gas arsine, \(\mathrm{AsH}_{3}\), decomposes as follows: $$ 2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g) $$ In an experiment at a certain temperature, pure \(\operatorname{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of \(392.0\) torr. After 48 hours the pressure in the flask was observed to be constant at \(488.0\) torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(\mathrm{~g})\). b. Calculate \(K_{\mathrm{p}}\) for this reaction.

Short Answer

Expert verified
The equilibrium pressure of H2(g) is found to be 144 torr, and the equilibrium constant Kp for this reaction is 0.163.

Step by step solution

01

a. Calculate the equilibrium pressure of H2(g)

The initial pressure of AsH3 is 392.0 torr. After reaching equilibrium, the total pressure (including H2) is 488.0 torr. Let's denote the decrease in pressure of AsH3 as x. Since 2 moles of AsH3 form 3 moles of H2, the pressure of H2 formed will be 3x/2. From the total pressure, we can write: Initial Pressure_(AsH3) - x + (3x/2) = Final Pressure 392.0 - x + (3x/2) = 488.0 Now, solve for x.
02

b. Solve for x

Solving the equation 392.0 - x + (3x/2) = 488.0, we get: 3x/2 - x = 488 - 392 3x - 2x = 2(96) x = 96 torr So, the change in pressure of AsH3 is 96 torr. Now we will find the equilibrium pressure of H2.
03

c. Find the equilibrium pressure of H2

We know that 2 moles of AsH3 form 3 moles of H2. So, if the pressure of AsH3 decreased by 96 torr, the pressure of H2 will be: Pressure_(H2) = (3/2) * 96 = 144 torr The equilibrium pressure of H2 is 144 torr.
04

d. Calculate Kp for the reaction

Now that we have the equilibrium pressures of the reactants and products, we will calculate the equilibrium constant, Kp. Kp = [Pressure_(H2)^3] / [Pressure_(AsH3)^2] The equilibrium pressure of AsH3 is (Initial Pressure - x) = (392 - 96) = 296 torr. Now we can calculate Kp: Kp = (144^3) / (296^2)
05

e. Find the value of Kp

Calculating Kp for the reaction, we get: Kp = (144^3) / (296^2) = 0.163 The equilibrium constant, Kp, for this reaction is 0.163.

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Most popular questions from this chapter

An important reaction in the commercial production of hydrogen is $$ \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) $$ How will this system at equilibrium shift in each of the five following cases? a. Gaseous carbon dioxide is removed. b. Water vapor is added. c. In a rigid reaction container, the pressure is increased by adding helium gas. d. The temperature is increased (the reaction is exothermic). e. The pressure is increased by decreasing the volume of the reaction container.

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is \(1.5\) atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C} .\) b. If the volume of the cylinder is increased until the total pressure is \(1.0 \mathrm{~atm}\) (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\). c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure = I.00 atm)?

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of \(4.5\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of \(9.0\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ What is the partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of \(0.80\) and \(0.20\) atm, respectively?

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ The initial density of the system was recorded as \(4.495 \mathrm{~g} / \mathrm{L}\). After equilibrium was reached, the density was noted to be \(4.086 \mathrm{~g} / \mathrm{L}\) a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K\) ? Explain each answer.
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