/*! 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 21 The decomposition of ammonia is:... [FREE SOLUTION] | 91Ó°ÊÓ

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The decomposition of ammonia is: \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) .\) If the pressure of ammonia is \(1.0 \times 10^{-3} \mathrm{~atm}\), and the pressures of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are each \(0.20 \mathrm{~atm}\), what is the value for \(\mathrm{Kp}\) at \(400^{\circ} \mathrm{C}\) for the reverse reaction? a. \(6.2 \times 10^{-4}\) b. \(2.6 \times 10^{-4}\) c. \(-6.2 \times 10^{-4}\) d. \(4.2 \times 10^{-4}\)

Short Answer

Expert verified
The value for \( \mathrm{Kp} \) at \( 400^{\circ} \mathrm{C} \) for the reverse reaction is approximately \( 6.2 \times 10^{-4} \).

Step by step solution

01

Write the Expression for Kp

For the given reverse reaction, the expression for the equilibrium constant in terms of pressure, \( K_p \), is \[ K_p = \frac{{P_{ ext{N}_2} \times P_{ ext{H}_2}^3}}{{P_{ ext{NH}_3}^2}} \]where \( P_{ ext{N}_2} \), \( P_{ ext{H}_2} \), and \( P_{ ext{NH}_3} \) are the partial pressures of nitrogen, hydrogen, and ammonia, respectively.
02

Substitute Known Values into the Equation

Using the known pressures, substitute into the expression: \[ K_p = \frac{{(0.20) \times (0.20)^3}}{{(1.0 \times 10^{-3})^2}} \] where \( P_{ ext{N}_2} = 0.20 \text{ atm} \), \( P_{ ext{H}_2} = 0.20 \text{ atm} \), and \( P_{ ext{NH}_3} = 1.0 \times 10^{-3} \text{ atm} \).
03

Solve the Expression

Calculate the expression:\[ K_p = \frac{{0.20 \times 0.008}}{{0.000001}} \] which simplifies to \[ K_p = \frac{{0.0016}}{{0.000001}} \] calculating further gives \[ K_p = 1600 \].
04

Calculate Kp for the Reverse Reaction

The given Kp calculated above is for the forward reaction, but the question asks for the Kp of the reverse reaction. Therefore, find the reciprocal of the forward reaction Kp to get the reverse Kp:\[ K_p(\text{reverse}) = \frac{1}{K_p(\text{forward})} = \frac{1}{1600} \].This simplifies to \[ K_p(\text{reverse}) = 6.25 \times 10^{-4} \].

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Key Concepts

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

Decomposition Reaction
In chemical reactions, a decomposition reaction is an essential type where a single compound breaks down into two or more simpler substances. For the decomposition of ammonia given in the exercise, the reaction can be represented as: \[ 2 \text{NH}_3(\text{g}) \rightleftharpoons \text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \]Here, ammonia \((\text{NH}_3)\) decomposes into nitrogen \((\text{N}_2)\) and hydrogen \((\text{H}_2)\). Breaking into simpler molecules, such as atoms or smaller compounds, often requires energy in the form of heat or light.
  • Decomposition helps understand the conversion processes in chemical engineering and environmental science.
  • This process makes it possible to extract valuable components like nitrogen and hydrogen gases from ammonia.
Comprehending decomposition reactions through the lens of equilibrium helps predict product formation under varying conditions, which is vital for industrial applications.
Equilibrium Expression
An equilibrium expression mathematically represents the state of balance in a reversible chemical reaction. In this balanced state, the rate at which reactants form products equals the rate at which products revert back into reactants. For our specific reaction, the equilibrium expression for the decomposition of ammonia in terms of partial pressures is given by: \[ K_p = \frac{{P_{\text{N}_2} \times P_{\text{H}_2}^3}}{{P_{\text{NH}_3}^2}} \]
  • The constant \(K_p\) indicates the ratio of the partial pressures of products over reactants at equilibrium.
  • It remains constant for a particular reaction at a given temperature.
Understanding the equilibrium expression is crucial for calculating reaction tendencies and predicting the products or reactants concentrations under specified conditions.
Partial Pressure
In a gas mixture, partial pressure pertains to the pressure exerted by a single type of gas within a mixture. Total pressure is the sum of the partial pressures of all gases present. In the ammonia decomposition example, the partial pressures are:
  • \(P_{\text{NH}_3} = 1.0 \times 10^{-3} \text{ atm}\)
  • \(P_{\text{N}_2} = 0.20 \text{ atm}\)
  • \(P_{\text{H}_2} = 0.20 \text{ atm}\)
Gases tend to mix and distribute evenly, and this concept helps predict how gases behave when contained. It's critical for:- Calculating equilibrium constants in gas reactions.- Understanding behavior in fields such as meteorology and respiratory physiology.FAQs often arise about the role of pressure in determining the direction and extent of chemical reactions.
Reverse Reaction
A reverse reaction occurs when the products of a chemical reaction revert to the original reactants. In terms of equilibrium, it denotes the balance between the forward and backward reactions.For the decomposition of ammonia: \[ 2 \text{NH}_3(\text{g}) \rightleftharpoons \text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \]If you have calculated \(K_p\) for the forward reaction, you can find \(K_p\) for the reverse reaction using the reciprocal, as shown below:\[ K_p(\text{reverse}) = \frac{1}{K_p(\text{forward})} \]
  • A critical concept in kinetics and equilibrium studies.
  • Helps predict conditions that favor product or reactant formation.
Understanding reverse reactions allows chemists to develop strategies to favor desired product formation in industrial and laboratory settings.

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Most popular questions from this chapter

What is the correct sequence of active masses in increasing order in gaseous mixture containing one gram per litre of each of the following? I. \(\mathrm{NH}_{3}\) II. \(\mathrm{N}_{2}\) III. \(\mathrm{H}_{2}\) IV. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: a. III, I, IV, II b. III, IV, II, I c. II, I, IV, III d. IV, II, I, III

At constant temperature, the equilibrium constant (Kp) for the decomposition reaction \(\mathrm{N}_{2} \mathrm{O}_{4} \leftrightarrow 2 \mathrm{NO}_{2}\) is expressed by \(\mathrm{Kp}=\left(4 \mathrm{x}^{2} \mathrm{P}\right) /\left(1-\mathrm{x}^{2}\right)\), where \(\mathrm{P}=\) pressure \(x=\) extent of decomposition. Which one of the following statements is true? a. Kp increases with increase of \(\mathrm{P}\) b. Kp increases with increase of \(\mathrm{x}\) c. Kp increases with decrease of \(x\) d. Kp remains constant with change in P and \(x\).

The equilibrium constant \((\mathrm{Kp})\) equals \(3.40\) for the isomerization reaction: cis-2-butene \(\rightleftharpoons\) trans-2-butene If a flask initially contains \(0.250 \mathrm{~atm}\) of cis-2-butene and \(0.125\) atm of trans-2-butene, what is the equilibrium pressure of each gas? a. \(\mathrm{P}(\mathrm{cis}-2\)-butene \()=0.085 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.290 \mathrm{~atm}\) b. \(\mathrm{P}(\) cis-2-butene \()=0.058 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.230 \mathrm{~atm}\) c. \(\mathrm{P}(\mathrm{cis}-2\)-butene \()=0.028 \mathrm{~atm}\), \(\mathrm{P}\) (trans-2-butene) \(=0.156 \mathrm{~atm}\) d. \(\mathrm{P}\) (cis-2-butene) \(=0.034 \mathrm{~atm}\), \(\mathrm{P}(\) trans-2-butene \()=0.128 \mathrm{~atm}\)

(A): When \(\mathrm{Q}=\mathrm{K}_{\mathrm{c}}\), reaction is at equilibrium (R) At equilibrium \(\Delta \mathrm{G}\) is 0 .

A mixture of carbon monoxide, hydrogen and methanol is at equilibrium. The balanced chemical equation is \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) At \(250^{\circ} \mathrm{C}\), the mixture contains \(0.0960 \mathrm{M} \mathrm{CO}, 0.191 \mathrm{M}\) \(\mathrm{H}_{2}\), and \(0.150 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}\). What is the value for \(\mathrm{K}_{\mathrm{C}}\) ? a. \(4.52\) b. \(42.8\) c. \(52.9\) d. \(0.581\)

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