/*! 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 99 For the reaction $$ \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ show that $$ K_{c}=K_{p}(R T)^{2} $$ Do not use the formula \(K_{p}=K_{c}(R T)^{\Delta n}\) given in the text. Start from the fact that \(P_{i}=[i] R T\), where \(P_{i}\) is the partial pressure of substance \(i\) and \([i]\) is its molar concentration. Substitute into \(K_{c}\).

Short Answer

Expert verified
The relationship is confirmed: \( K_c = K_p (RT)^2 \).

Step by step solution

01

Identify Relevant Expressions

The reaction given is \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \). The equilibrium constants involved are \( K_c \) for concentrations and \( K_p \) for partial pressures. Start with the concentration equilibrium constant expression: \[ K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \].
02

Express Partial Pressures in Terms of Concentrations

We use the relation \( P_i = [i]RT \) for the partial pressure \( P_i \) of a gas. Thus, for each component: \[ P_{\mathrm{NH}_3} = [\mathrm{NH}_3] RT, \; P_{\mathrm{N}_2} = [\mathrm{N}_2] RT, \; P_{\mathrm{H}_2} = [\mathrm{H}_2] RT. \]
03

Plug Partial Pressure Expressions into Kp Formula

The expression for the partial pressure equilibrium constant, \( K_p \), is: \[ K_p = \frac{(P_{\mathrm{NH}_3})^2}{P_{\mathrm{N}_2}(P_{\mathrm{H}_2})^3} \]. Substitute the expressions from Step 2: \[ K_p = \frac{([\mathrm{NH}_3] RT)^2}{([\mathrm{N}_2] RT)([\mathrm{H}_2] RT)^3} \].
04

Simplify the Expression for Kp

Expand the expression from Step 3: \[ K_p = \frac{[\mathrm{NH}_3]^2 (RT)^2}{[\mathrm{N}_2] (RT) ([\mathrm{H}_2]^3 (RT)^3)} \]. Veirify that after canceling common factors and powers of \( RT \): \[ K_p = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \times \frac{1}{(RT)^{(1+3-2)}} = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \times (RT)^{-2}. \]
05

Conclusion

Comparing the above derived expressions with the original \( K_c \), we find that \( K_c = K_p \times (RT)^2 \), confirming the given relation.

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Key Concepts

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

Partial Pressure
Partial pressure is a vital concept when dealing with gases in chemical reactions. It refers to the pressure that a gas would exert if it alone occupied the entire volume that the gas mixture currently occupies. In the context of equilibrium reactions involving gases, each gas contributes to the total pressure in proportion to its amount. This is the essence of Dalton's Law of Partial Pressures.

When an equilibrium expression includes gases, chemists often convert concentrations to partial pressures. For instance, in the equation provided, the reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) consists of gases. The equilibrium constant in terms of partial pressure, Kₚ, is derived from the partial pressures of each gas:
  • The partial pressure of NH₃, denoted as Pₙₕ₃, is linked to its molar concentration and the gas constant.
  • Similarly, Pₙ₂ and Pâ‚•â‚‚ are the partial pressures of nitrogen and hydrogen, respectively.
This transition from concentrations to partial pressures is accomplished using the equation Páµ¢ = [i]RT, highlighting the role of temperature and the ideal gas constant (R) in determining these pressures.
Molar Concentration
Molar concentration, often expressed as [i], represents the concentration of a compound in a solution, quantified in moles per liter (mol/L). It is a foundational concept tying the amounts of substances directly to the reactions they participate in. In chemical equilibrium problems, especially those involving gases, it's crucial to understand how molar concentrations translate into other useful forms for calculations.

In equilibrium expressions, molar concentration plays a pivotal role in defining the equilibrium constant Kₚ using the relation between concentration and partial pressure. For example, in the reaction highlighted, we initially start with the concentration-based equilibrium constant (Kₙ). The equation:
  • Kâ‚™ = ([NH₃]²) / ([Nâ‚‚][Hâ‚‚]³)
This shows how the concentration of each gas component dictates the equilibrium conditions, arrangements, and calculations. Then, transformations using Páµ¢ = [i]RT help pivot from concentrations to partial pressures, bridging the relationship between different equilibrium constants.
Gas Constant
The gas constant, R, is an essential factor in chemistry, particularly when studying the behavior of gases. R connects variables in the ideal gas law equation, PV = nRT, where it appears alongside pressure (P), volume (V), moles of gas (n), and temperature (T). The gas constant serves as a bridge between molar concentration and partial pressure in reactions involving gases.

In the context of the given equilibrium problem, R emerges as critical when using the formula Páµ¢ = [i]RT. For each component of the gaseous reaction:
  • The gas constant assists in scaling concentration from units of moles per liter to the partial pressures in equilibrium constant expressions.
  • Additionally, its role further extends in correlations, such as converting Kâ‚™ to Kₚ using the transformation Kₚ = Kâ‚™ (RT)Δn.
Understanding R's role thus equips students to navigate problems where gases are subjected to equilibrium conditions, showcasing its ubiquitous nature in calculations linking different states of matter in chemistry.

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Most popular questions from this chapter

During the commercial preparation of sulfuric acid, sulfur dioxide reacts with oxygen in an exothermic reaction to produce sulfur trioxide. In this step, sulfur dioxide mixed with oxygen-enriched air passes into a reaction tower at about \(420^{\circ} \mathrm{C}\), where reaction occurs on a vanadium(V) oxide catalyst. Discuss the conditions used in this reaction in terms of its effect on the yield of sulfur trioxide. Are there any other conditions that you might explore in order to increase the yield of sulfur trioxide?

An equilibrium mixture of dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\), and nitrogen dioxide, \(\mathrm{NO}_{2}\), is \(65.8 \% \mathrm{NO}_{2}\) by mass at \(1.00 \mathrm{~atm}\) pressure and \(25^{\circ} \mathrm{C}\). Calculate \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ at \(450^{\circ} \mathrm{C}\) is \(0.159 .\) Calculate the equilibrium composition when \(1.00 \mathrm{~mol} \mathrm{~N}_{2}\) is mixed with \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) in a \(5.00-\mathrm{L}\) vessel.

The amount of nitrogen dioxide formed by dissociation of dinitrogen tetroxide, $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ increases as the temperature rises. Is the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) endothermic or exothermic?

An equilibrium mixture of \(\mathrm{SO}_{3}, \mathrm{SO}_{2}\), and \(\mathrm{O}_{2}\) at \(727^{\circ} \mathrm{C}\) is \(0.0160 \mathrm{MSO}_{3}, 0.0056 \mathrm{MSO}_{2}\), and \(0.0021 \mathrm{M} \mathrm{O}_{2}\). What is the value of \(K_{c}\) for the following reaction? $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$

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