Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

When heated, ammonium carbamate decomposes as $$ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ At a certain temperature the equilibrium pressure of the system is 0.318 atm. Calculate \(K_{P}\) for the reaction.

Short Answer

Expert verified
The equilibrium constant, \(K_p\), for the reaction is 0.032 atm.

Step by step solution

01

Understanding Kp

The equilibrium constant for pressure, Kp, is calculated by taking the partial pressures of the products to their stoichiometric coefficients, divided by the partial pressures of the reactants raised to their stoichiometric coefficients. However, solid and liquids are not included in this calculation as their concentrations do not change under equilibrium conditions. The equation for the equilibrium constant, \(K_P\), is represented by \(K_p = \frac{{(P_{NH_3})^2 \cdot (P_{CO_2})}}{(P_{NH_{4}CO_{2}NH_{2}})\), where the \(P\) stands for partial pressure. However, as NH4CO2NH2 is a solid, the equation simplifies to \(K_p = (P_{NH_3})^2 \cdot (P_{CO_2})\).
02

Identify given parameters

The equilibrium pressure of the system is given as 0.318 atm. This is the total pressure at equilibrium for all gases combined. Since there are 3 moles of gas on the product side (2 NH3 + 1 CO2), the pressure of each of these is 0.318 atm.
03

Calculate Kp

Substitute the equilibrium pressures of NH3 and CO2 into the simplified equation. So, \(K_p = (0.318)^2 \cdot (0.318) = 0.032 atm\).

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 Vaia!

Key Concepts

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

Equilibrium Constant (Kp)
The equilibrium constant expressed in terms of partial pressures is denoted as Kp. It's a useful measure for gauging the extent of a chemical reaction when dealing with gases at equilibrium. The mathematical formula for Kp takes on the general form:
\[ K_p = \frac{(P_{\text{products}})^{\text{stoichiometry of products}}}{(P_{\text{reactants}})^{\text{stoichiometry of reactants}}} \]
For reactions involving solids or liquids, it's important to note that their concentrations remain constant under equilibrium conditions and therefore do not figure into the calculation of Kp. This specificity certainly aids in simplifying calculations for many equilibrium systems. Remembering that the partial pressure of each gas is related to the mole fraction and the total pressure can also assist in understanding the dynamics of the system.
Partial Pressure
Partial pressure is a term that describes the pressure contributed by a single gas in a mixture of gases. Total pressure is the sum of these individual pressures. At equilibrium, the total pressure provided is the collective pressure of all gaseous reactants and products.
The concept of partial pressure is essential not only in calculating Kp but also in understanding gas behavior and applying the ideal gas law. Each gas in a mixture behaves as if it were alone in the container, influencing the reaction dynamics in accordance with its concentration or partial pressure. This notion is critical to comprehend, as changes in partial pressures can signify shifts in equilibrium.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the context of equilibrium reactions and Kp, stoichiometry determines the exponents in the equilibrium constant expression. This reflects the ratio of moles of each substance as indicated in the balanced chemical equation.
Understanding stoichiometry allows one to convert between masses, moles, and particles, and to calculate yields and purities — vital skills in both practical and theoretical chemistry. A clear grasp of stoichiometry is essential in determining partial pressures, which are pivotal when working with equilibrium systems.
Equilibrium Reactions
Equilibrium reactions are chemical processes that occur in both the forward and reverse directions. At equilibrium, the rate of the forward reaction equals that of the reverse, leading to a constant concentration of reactants and products. However, this doesn't imply that the reaction has stopped, but rather that it's dynamically balanced.
Grasping the nature of equilibrium is fundamental in predicting how a system will react to external changes. Le Chatelier's principle, for instance, suggests that a system will adapt so as to counteract these changes, thereby maintaining its equilibrium. Such insights are crucial in predicting the behavior of reactants and products under varying conditions and are particularly relevant in the study of equilibrium constant expressions like Kp.

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

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$ 2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g) $$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A})\). Suppose the catalyst is suddenly exposed to the equilibrium system as shown below. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

Use Le Châtelier's principle to explain why the equilibrium vapor pressure of a liquid increases with increasing temperature.

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.

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation $$ \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{CO}(g) $$ At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

A quantity of 0.20 mole of carbon dioxide was heated at a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Under this condition, the average molar mass of the gases was found to be \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is the \(K_{P}\) for the equilibrium if the total pressure was 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

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.

Sign-up for free