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

Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{~A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{~B}(g) K_{c}\) and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

Short Answer

Expert verified
The short answer is: Statement (a) is false because Kp and Kc are not numerically the same for this reaction. Statement (b) is true because the units for Kc (moles per liter) and Kp (atmospheres) are different, allowing us to distinguish between them. Statement (c) is true because, according to Le Chatelier's principle, increasing pressure causes the system to shift towards the side with fewer gas moles, leading to a higher value of Kc.

Step by step solution

01

Statement (a): Comparing Kc and Kp for the given reaction

For the reaction 2A(g) + B(g) ⇋ A2B(g), we can write the expressions for Kc and Kp as follows: Kc = \(\frac{[A2B]}{[A]^2[B]}\) Kp = \(\frac{(P_{A2B})}{(P_A)^2(P_B)}\) Here, [A], [B], and [A2B] represent the molar concentrations of A, B, and A2B at equilibrium, while \(P_A\), \(P_B\), and \(P_{A2B}\) represent their partial pressures at equilibrium. The relationship between Kc and Kp is given by the expression: Kp = Kc(RT)^Δn where R is the gas constant, T is the temperature, and Δn is the change in the number of moles of gas between the reactants and products. For this reaction, Δn = 1 - (2+1) = -2. Therefore, Kp and Kc are not numerically the same for this reaction. Statement (a) is false.
02

Statement (b): Distinguishing Kc and Kp by comparing units

The units of Kc are related to the concentrations of the substances in the reaction. In this case, the units for Kc are moles per liter (M). The units of Kp are related to the partial pressures of the substances in the reaction, typically expressed in atmospheres (atm). Since the units for Kc and Kp are different, it is possible to distinguish Kc from Kp by comparing the units used to express the equilibrium constants. Statement (b) is true.
03

Statement (c): Effect of increasing pressure on Kc

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions (such as pressure, temperature, or concentration), the system will shift its position to counteract the change. Applying this principle to the given reaction, we can analyze the effect of increasing pressure on Kc. When the pressure increases, the system tries to compensate for the change by shifting towards the side with fewer gas moles. In this reaction, the reactant side has more gas moles (2 moles of A and 1 mole of B) than the product side (1 mole of A2B). Therefore, under increased pressure, the system will shift towards the formation of more A2B, which leads to a higher value of Kc. Statement (c) is true. In summary: - Statement (a) is false. - Statement (b) is true. - Statement (c) is true.

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 Constants
In chemistry, equilibrium constants are vital for understanding the balance between reactants and products in a reversible reaction. They provide a quantitative measure of a reaction's position at equilibrium, indicating the extent to which reactants are converted into products.

Equilibrium constants come in different forms, notably as the concentration quotient (\( K_c \)) using molar concentrations or the pressure quotient (\( K_p \)) using partial pressures. The value of these constants is fixed at a given temperature; thus, they do not change unless the temperature changes. When the system is at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, meaning no net change in the concentrations of reactants and products occurs.

It is also important to understand that equilibrium constants have no units; they are just numbers. Any change in conditions such as concentration, temperature, or pressure does not alter the equilibrium constants but rather shifts the position of equilibrium, which can be predicted using Le Chatelier's principle.
Le Chatelier's Principle
Le Chatelier's principle is a concept that helps predict the response of an equilibrium system when it is disturbed by external changes. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts itself to minimize the impact of this change.

External changes can include variations in temperature, pressure, or concentration of either reactants or products. For example, increasing the pressure on a gaseous equilibrium will favor the side with fewer moles of gas, as seen in the exercise where an increase in pressure shifts the equilibrium towards the formation of product A2B in the reaction. In essence, Le Chatelier's principle is an invaluable tool that aids in predicting how a chemical system behaves when subjected to various changes.
Partial Pressure
Partial pressure is a fundamental concept when dealing with gases. It is the pressure that one gas in a mixture of gases would exert if it alone occupied the entire volume of the mixture at the same temperature. The partial pressure of a gas is directly proportional to its mole fraction in the gas mixture.

This concept is crucial when calculating the equilibrium constant in terms of pressure (\( K_p \)). In the given exercise, the partial pressures of gases A, B, and A2B at equilibrium (\( P_A \), \( P_B \), and \( P_{A2B} \)) play an integral role in determining the value of \( K_p \) for the reaction. In summary, partial pressures allow us to understand how individual gases within a mixture contribute to the overall pressure and how their concentrations change as a reaction reaches equilibrium.
Reaction Quotient
The reaction quotient (\( Q \)) is a term that can cause some confusion when first encountered, but it is a convenient way to determine the direction in which a reaction will proceed to reach equilibrium. It has the same form as the equilibrium constant expression but is calculated using the initial concentrations or pressures of the reactants and products, not the concentrations or pressures at equilibrium.

If the reaction quotient (\( Q \)) equals the equilibrium constant (\( K \)), the system is at equilibrium. A higher reaction quotient than the equilibrium constant (\( Q > K \)) indicates that the reaction will proceed in the reverse direction to reach equilibrium. Conversely, if the reaction quotient is less, (\( Q < K \)), the reaction will go forward. By comparing \( Q \) to \( K \), you can predict the changes that will occur within a system to achieve equilibrium.

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

Suppose that the gas-phase reactions \(A \longrightarrow B\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) ? (b) Which is greater at equilibrium, the partial pressure of \(A\) or the partial pressure of \(B\) ?

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\) (f) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (g) \(2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+2 \mathrm{OH}(a q)\)

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00- \(\mathrm{L}\) flask at \(303 \mathrm{~K}\), \(56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at \(303 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture were transferred to a \(15.00\)-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00\) - \(\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

The equilibrium constant \(K_{c}\) for \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is \(1.9\) at \(1000 \mathrm{~K}\) and \(0.133\) at \(298 \mathrm{~K}\). (a) If excess \(\mathrm{C}\) is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a \(3.00\) - \(\mathrm{L}\) vessel at 1000 \(\mathrm{K}\), how many grams of \(\mathrm{CO}\) are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(\mathrm{~g})\). A \(7.5-\mathrm{L}\) gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g)\), which is allowed to equilibrate at 450 \(\mathrm{K}\). At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.157 \mathrm{~atm}\), and \(P_{\mathrm{PCl}_{5}}=1.30 \mathrm{~atm}\). (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at \(450 \mathrm{~K}\).

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.

Sign-up for free