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Chapter 15: Problem 75

Consider the hypothetical reaction \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) .\) A flask is charged with 0.75 atm of pure \(A\), after which it is allowed to reach equilibrium at \(0^{\circ} \mathrm{C}\). At equilibrium the partial pressure of \(\mathrm{A}\) is 0.36 atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?\) (c) What could we do to maximize the yield of \(\mathrm{B}\) ?

Short Answer

Expert verified
The total pressure in the flask at equilibrium is 1.14 atm. The value of \(K_p\) for the reaction is 1.68. To maximize the yield of B, we should increase the pressure of the system, as it will shift the equilibrium towards the production of B according to Le Chatelier's Principle.

Step by step solution

01

Write the reaction and the equilibrium constant expression

The given reaction is: \( A(g) \rightleftharpoons 2B(g) \) We can now write the equilibrium constant expression for this reaction in terms of partial pressures, Kp, as follows: \( K_p = \frac{P_B^2}{P_A} \) where \(P_A\) and \(P_B\) represent the partial pressures of A and B gases, respectively, at equilibrium.
02

Calculate the change in partial pressures at equilibrium

Initially, the flask has a pressure of 0.75 atm for A and 0 atm for B. Let x be the drop in pressure of A at equilibrium. Therefore, the increase in pressure of B will be twice this change (since there are 2 moles of B produced for each mole of A that reacts). So, \( P_A = 0.75 - x \) \( P_B = 0 \Rightarrow P_B = 2x \) At equilibrium, we are given that the pressure of A is 0.36 atm. Therefore, we can now calculate the change in pressure (x) and the pressure of B at equilibrium.
03

Calculate equilibrium pressures of A and B

Since we have the equilibrium pressure of A, \( P_A = 0.36 \, atm \) We can calculate the change in pressure (x): \( x = 0.75 - P_A = 0.75 - 0.36 = 0.39 \, atm \) Now, we can calculate the equilibrium pressure of B: \( P_B = 2x = 2 * 0.39 = 0.78 \, atm \)
04

Calculate the total pressure at equilibrium

To find the total pressure at equilibrium, we simply add the partial pressures of A and B: \( P_{total} = P_A + P_B = 0.36 + 0.78 = 1.14 \, atm \) So, the total pressure in the flask at equilibrium is 1.14 atm.
05

Calculate the value of Kp

Now, we can substitute the equilibrium pressures of A and B into the equilibrium constant expression: \( K_p = \frac{P_B^2}{P_A} = \frac{(0.78)^2}{0.36} = 1.68 \) Hence, the value of Kp for this reaction is 1.68.
06

Suggest a method to maximize the yield of B

To maximize the yield of B, we can use Le Chatelier's Principle, which states that a change in external conditions (pressure, temperature, and concentration) will cause a shift in the equilibrium to counteract the change. In this case, since the formation of B is associated with a decrease in the number of moles of gas (1 mole of A produces 2 moles of B), increasing the pressure of the system will cause the equilibrium to shift towards the side with fewer moles of gas, i.e., towards the production of B. Therefore, to maximize the yield of B, we should increase the pressure of the system.

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Most popular questions from this chapter

The following graph represents the yield of the compound \(\mathrm{AB}\) at equilibrium in the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \longrightarrow \mathrm{AB}(g)\) at two different pressures, \(x\) and \(y\), as a function of temperature. (a) Is this reaction exothermic or endothermic? (b) Is \(P=x\) greater or smaller than $P=y ?

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? Explain. (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is estab- lished at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\), respectively. (a) Calculate \(K_{p}\) for this reaction at \(500.0 \mathrm{~K}\). (b) If the vessel has a volume of \(5.00 \mathrm{~L}\) calculate \(K_{c}\) at this temperature.

As shown in Table \(15.2, K_{p}\) for the equilibrium $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.51 \times 10^{-5}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C}\). If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) \(98 \mathrm{~atm} \mathrm{NH}_{3}, 45 \mathrm{~atm} \mathrm{~N}_{2}, 55 \mathrm{~atm} \mathrm{H}_{2}\) (b) \(57 \mathrm{~atm} \mathrm{NH}_{3}, 143 \mathrm{~atm} \mathrm{~N}_{2},\) no \(\mathrm{H}_{2}\) (c) \(13 \mathrm{~atm} \mathrm{NH}_{2} 27 \mathrm{~atm} \mathrm{~N}_{2} 82 \mathrm{~atm} \mathrm{H}_{2}\)

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24{ }^{\circ} \mathrm{C}\). The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C}\) ?
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