/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 A sample of air with a mole rati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of air with a mole ratio of \(\mathrm{N}_{2}\) to \(\mathrm{O}_{2}\) of 79: 21 is heated to 2500 K. When equilibrium is established in a closed container with air initially at 1.00 atm, the mole percent of \(\mathrm{NO}\) is found to be \(1.8 \% .\) Calculate \(K_{\mathrm{p}}\) for the reaction. $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$$

Short Answer

Expert verified
The calculation gives \(K_{\mathrm{p}} = 0.00171\).

Step by step solution

01

Calculate the initial mole fractions

The mole fractions of nitrogen and oxygen given by the ratio 79: 21 sum up to 100. Thus the mole fractions are \(0.79\) for \(\mathrm{N}_{2}\) and \(0.21\) for \(\mathrm{O}_{2}\). Remember that the mole fraction is defined as the amount of a specific component divided by the total amount.
02

Calculate the partial pressures at equilibrium

The total pressure at equilibrium is the sum of the partial pressures of each gas. Since the initial pressure is 1.00 atm and no gas was added or removed (only the proportions changed), the total pressure at equilibrium is still 1.00 atm. Use the mole percent of nitric oxide at equilibrium to find its partial pressure. A \(1.8 \%\) mole percent corresponds to a partial pressure of \(0.018 \times 1.00 \, \mathrm{atm} = 0.018 \, \mathrm{atm}\). The reaction consumes equal moles of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to produce \(\mathrm{NO}\), so the partial pressures of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) decrease by the same amount, \(0.018 \, \mathrm{atm}\). The final partial pressures are therefore \(0.79 \times 1.00 \, \mathrm{atm} - 0.018 \, \mathrm{atm} = 0.772 \, \mathrm{atm}\) for \(\mathrm{N}_{2}\) and \(0.21 \times 1.00 \, \mathrm{atm} - 0.018 \, \mathrm{atm} = 0.192 \, \mathrm{atm}\) for \(\mathrm{O}_{2}\).
03

Use the equilibrium equation to find \(K_{\mathrm{p}}\)

The equilibrium constant, \(K_{\mathrm{p}}\), for the reaction is defined as \[K_{\mathrm{p}}= \frac{\left(\mathrm{P}_{\mathrm{NO}}\right)^2}{\mathrm{P}_{\mathrm{N}_{2}} \cdot \mathrm{P}_{\mathrm{O}_{2}}}\] where \(\mathrm{P}_{\mathrm{NO}}\), \(\mathrm{P}_{\mathrm{N}_{2}}\), and \(\mathrm{P}_{\mathrm{O}_{2}}\) are the partial pressures of \(\mathrm{NO}\), \(\mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) at equilibrium. Substituting the calculated pressures, we get \[K_{\mathrm{p}}= \frac{(0.018 \, \mathrm{atm})^2}{0.772 \, \mathrm{atm} \cdot 0.192 \, \mathrm{atm}}\]
04

Calculate \(K_{\mathrm{p}}\)

Finally, perform the multiplication and division to obtain the value of \(K_{\mathrm{p}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure
In the context of gases, the concept of partial pressure is crucial for understanding how different gases contribute to the total pressure of a mixture. Partial pressure is defined as the pressure that a gas in a mixture would exert if it alone occupied the entire volume of the mixture at the same temperature. To find the partial pressure of a particular gas, you multiply its mole fraction by the total pressure of the gas mixture. For example, in our problem, nitrogen (\( \mathrm{N}_2 \)) and oxygen (\( \mathrm{O}_2 \)) have mole fractions of \(0.79\) and \(0.21\), respectively. If the total pressure is \(1.00 \, \text{atm}\), the partial pressures of \(\mathrm{N}_2\) and \(\mathrm{O}_2\) initially are \(0.79 \, \text{atm}\) and \(0.21 \, \text{atm}\).
This concept helps us in many calculations, including those involving chemical equilibria, as the change in partial pressures during reactions gives us the information needed to find equilibrium constants.
Equilibrium Constant
The equilibrium constant, \(K_p\), for a gaseous reaction is calculated using the partial pressures of the reactants and products at equilibrium. It reflects the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients, at equilibrium. For the reaction \(\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\), this means \(K_p\) is calculated by the formula:
  • \(K_p = \frac{(P_{\mathrm{NO}})^2}{P_{\mathrm{N}_2} \cdot P_{\mathrm{O}_2}}\)
Knowing the equilibrium constant for a reaction is valuable because it indicates the extent to which reactants convert to products under given conditions.
In the example provided, substituting the equilibrium partial pressures of \(P_{\mathrm{NO}} = 0.018 \, \text{atm}\), \(P_{\mathrm{N}_2} = 0.772 \, \text{atm}\), and \(P_{\mathrm{O}_2} = 0.192 \, \text{atm}\) into the formula gives us the equilibrium constant for the formation of NO at 2500 K.
Gaseous Reactions
Gaseous reactions often involve multiple reactants and products that exist in a gas phase. They have notable features and behaviors, primarily dictated by the principles of gas laws and equilibrium. In a closed system, they reach a state where the rate of the forward reaction equals the rate of the reverse reaction, creating what is known as a dynamic equilibrium. When dealing with gaseous reactions, partial pressures are used as a measure of the concentration of gases, and they play a vital role in calculating the equilibrium state, given by \(K_p\).
Gaseous reactions are highly influenced by changes in temperature and pressure. At higher temperatures, such as 2500 K in our example, molecules move more rapidly, potentially affecting the extent of the reaction. The equilibrium achieved is specific to the conditions present, and shifts can occur if temperature or pressure are altered. Understanding these reactions through the concept of equilibrium and the associated mathematics enables predictions of how systems react to changes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Would you expect that the amount of \(\mathrm{N}_{2}\) to increase, decrease, or remain the same in a scuba diver's body as he or she descends below the water surface?

A mixture of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) and \(\mathrm{CH}_{4}(\mathrm{g})\) in the mole ratio 2: 1 was brought to equilibrium at \(700^{\circ} \mathrm{C}\) and a total pressure of 1 atm. On analysis, the equilibrium mixture was found to contain \(9.54 \times 10^{-3} \mathrm{mol} \mathrm{H}_{2} \mathrm{S} .\) The \(\mathrm{CS}_{2}\) pre- sent at equilibrium was converted successively to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and then to \(\mathrm{BaSO}_{4} ; 1.42 \times 10^{-3} \mathrm{mol} \mathrm{BaSO}_{4}\) was obtained. Use these data to determine \(K_{\mathrm{p}}\) at \(700^{\circ} \mathrm{C}\) for the reaction $$\begin{aligned} 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+& 4 \mathrm{H}_{2}(\mathrm{g}) \\\ & K_{\mathrm{p}} \text { at } 700^{\circ} \mathrm{C}=? \end{aligned}$$

The Deacon process for producing chlorine gas from hydrogen chloride is used in situations where \(\mathrm{HCl}\) is available as a by-product from other chemical processes. $$\begin{aligned} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-114 \mathrm{kJ} \end{aligned}$$ A mixture of \(\mathrm{HCl}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{Cl}_{2}\) is brought to equilibrium at \(400^{\circ} \mathrm{C}\). What is the effect on the equilibrium amount of \(\mathrm{Cl}_{2}(\mathrm{g})\) if (a) additional \(\mathrm{O}_{2}(\mathrm{g})\) is added to the mixture at constant volume? (b) \(\mathrm{HCl}(\mathrm{g})\) is removed from the mixture at constant volume? (c) the mixture is transferred to a vessel of twice the volume? (d) a catalyst is added to the reaction mixture? (e) the temperature is raised to \(500^{\circ} \mathrm{C} ?\)

What effect does increasing the volume of the system have on the equilibrium condition in each of the following reactions? (a) \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CaCO}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Determine values of \(K_{c}\) from the \(K_{p}\) values given. (a) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=2.9 \times 10^{-2} \mathrm{at} 303 \mathrm{K}\) (b) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=1.48 \times 10^{4} \mathrm{at} 184^{\circ} \mathrm{C}\) (c) \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(K_{\mathrm{p}}=0.429\) at \(713 \mathrm{K}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.