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Chapter 15: Problem 72

The equilibrium constant \(K_{P}\) for the following reaction is found to be \(4.31 \times 10^{-4}\) 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 constant-volume vessel at \(375^{\circ} \mathrm{C}\). Calculate the partial pressures of all species when equilibrium is reached.

Short Answer

Expert verified
The exact values of the partial pressures at equilibrium can't be given without solving the equation for \(x\). The methods used to solve the equation can vary depending on the difficulty of the equation and the math skills of the students. However, once \(x\) is determined, the equilibrium pressures are found by simple substitution.

Step by step solution

01

Understand the Problem and Given Information

The reaction given is \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longleftrightarrow 2 \mathrm{NH}_{3}(g)\). The equilibrium constant (\(K_{P}\)) for this reaction at 375°C is given as \(4.31 \times 10^{-4}\). The initial pressures are \[\mathrm{P}_{N_{2}}=0.862 \,atm\,\] and \[\mathrm{P}_{H_{2}}=0.373\,atm\,\]. There is initially no \(\mathrm{NH}_{3}\). The goal is to find the partial pressures of all species at equilibrium.
02

Write the Expression for the Equilibrium Constant (\(K_{P}\))

The expression for the equilibrium constant \(K_{P}\) in terms of the partial pressures of the components is \(K_{P} = \frac{\left(\mathrm{P}_{NH_{3}}\right)^2}{\mathrm{P}_{N_{2}}\cdot \left(\mathrm{P}_{H_{2}}\right)^3}\), where \(\mathrm{P}_{NH_{3}}\), \(\mathrm{P}_{N_{2}}\), and \(\mathrm{P}_{H_{2}}\) are the partial pressures of the \(\mathrm{NH}_{3}\), \(\mathrm{N}_{2}\), and \(\mathrm{H}_{2}\) at equilibrium, respectively.
03

Write the Expression for the Changes in Partial Pressures

For every mole of \(\mathrm{N}_{2}\) that reacts, three moles of \(\mathrm{H}_{2}\) react to produce two moles of \(\mathrm{NH}_{3}\). Therefore, if \(x\) is the decrease in \(\mathrm{P}_{N_{2}}\) and \(\mathrm{P}_{H_{2}}\), then the increase in \(\mathrm{P}_{NH_{3}}\) would be \(2x\). So, the equilibrium pressures can be expressed as:\(\mathrm{P'_{N_{2}}} = \mathrm{P}_{N_{2}}-x\),\(\mathrm{P'_{H_{2}}} = \mathrm{P}_{H_{2}}-3x\),\(\mathrm{P'_{NH_{3}}} = \mathrm{P}_{NH_{3}}+ 2x = 2x\). Note that the initial pressure of \(\mathrm{NH}_{3}\) is zero.
04

Solve for \(x\)

Substitute the equilibrium pressure terms for \(N_{2}\), \(H_{2}\), and \(NH_{3}\) in the \(K_{P}\) expression and solve for \(x\). Thus, we have \(4.31 \times 10^{-4} = \frac{(2x)^{2}}{(0.862-x)(0.373 - 3x)^3}\). Simplifying and solving this equation will give us the value of \(x\).
05

Find the Equilibrium Partial Pressures

Substitute the value of \(x\) in the expressions from step 3 to get the partial pressures at equilibrium. Thus, the equilibrium pressures are:\(\mathrm{P'_{N_{2}}} = \mathrm{P}_{N_{2}}-x\),\(\mathrm{P'_{H_{2}}} = \mathrm{P}_{H_{2}}-3x\),\(\mathrm{P'_{NH_{3}}} = \mathrm{P}_{NH_{3}}+ 2x = 2x\).

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Most popular questions from this chapter

A mixture of 0.47 mole of \(H_{2}\) and 3.59 moles of \(H C l\) is heated to \(2800^{\circ} \mathrm{C}\). Calculate the equilibrium partial pressures of \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\), and \(\mathrm{HCl}\) if the total pressure is 2.00 atm. The \(K_{P}\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)\) is 193 at \(2800^{\circ} \mathrm{C}.\)

Consider this reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ The equilibrium constant \(K_{P}\) for the reaction is \(1.0 \times\) \(10^{-15}\) at \(25^{\circ} \mathrm{C}\) and 0.050 at \(2200^{\circ} \mathrm{C}\). Is the formation of nitric oxide endothermic or exothermic? Explain your answer.

Pure NOCl gas was heated at \(240^{\circ} \mathrm{C}\) in a 1.00 - \(\mathrm{L}\) container. At equilibrium the total pressure was 1.00 atm and the NOCl pressure was 0.64 atm. $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate the partial pressures of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) in the system. (b) Calculate the equilibrium constant \(K_{P}\).

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$ \mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g) $$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

Consider the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ At \(430^{\circ} \mathrm{C}\), an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2} .\) Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.
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