Question

For the reaction: \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{~s}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}), K_{\mathrm{p}}=3.2 \times 10^{-5} \mathrm{~atm}^{3}\) The total pressure of the gaseous products when sufficient amount of reactant is allowed to achieve equilibrium, is (a) \(0.02 \mathrm{~atm}\) (b) \(0.04\) atm (c) \(0.06 \mathrm{~atm}\) (d) \(0.095 \mathrm{~atm}\)

Short answer

The total pressure of the gaseous products at equilibrium is 0.06 atm.

Step-by-step solution

Write the Equlibrium Expression

For the given reaction, the equilibrium constant expression based on partial pressures (Kp) can be written as: \[K_{p} = \frac{(P_{NH_{3}})^2 \cdot P_{CO_{2}}}{1}\]Since the reactant is a solid and does not appear in the expression.

Assume Initial Pressures and Changes at Equilibrium

Assume that the initial pressures of NH3 and CO2 are zero and let the change at equilibrium be x. This means at equilibrium, the partial pressures will be:\[P_{NH_{3}} = 2x\] and \[P_{CO_{2}} = x\].

Substitute the Equilibrium Pressures into the Kp Expression

Substituting the equilibrium partial pressures into the Kp expression gives:\[K_{p} = \frac{(2x)^2 \cdot x}{1} = 4x^3\].Now, plug in the value of Kp:\[3.2 \times 10^{-5} = 4x^3\].

Solve for x

Solve the above equation for x:\[x^3 = \frac{3.2 \times 10^{-5}}{4}\]\[x^3 = 8 \times 10^{-6}\]\[x = \sqrt[3]{8 \times 10^{-6}}\]\[x = 2 \times 10^{-2}\].

Calculate the Total Pressure at Equilibrium

The total pressure at equilibrium is the sum of the partial pressures of NH3 and CO2:\[P_{total} = P_{NH_{3}} + P_{CO_{2}} = 2x + x\]\[P_{total} = 3x = 3 \times (2 \times 10^{-2})\]\[P_{total} = 0.06 \text{ atm}\].

Key concepts

Equilibrium Constant Expression

In physical chemistry, understanding equilibrium is fundamental, and a key component of this understanding is the equilibrium constant expression. This mathematical expression quantifies the state of chemical equilibrium between reactants and products. It is a ratio of product concentrations to reactant concentrations, each raised to the power of their respective coefficients as derived from the balanced chemical equation. In the context of gases, concentrations are often replaced by partial pressures, leading to the expression known as the equilibrium constant for partial pressures, or Kp.

In the solved problem, we see how Kp is calculated using the partial pressures of ammonia (NH3) and carbon dioxide (CO2). Note that solid reactants, like NH2COONH4 in our example, are omitted from the Kp expression since their quantity doesn't change the equilibrium state; this is a critical detail not to overlook. When dealing with equilibrium constant expressions, it's essential to remember that these are calculated at a particular temperature. Changing the temperature would change the Kp value, even if the equilibrium position remains the same.

Partial Pressure Calculations

Partial pressure calculations are vital in the realm of chemical equilibrium involving gases. The partial pressure of a gas is the pressure it would exert if it alone occupied the volume of the mixture at the same temperature. Recognizing this concept is crucial when solving equilibrium problems like the one provided.

To solve for total pressure in equilibrium systems, one must sum the individual partial pressures of each gas present. During the calculation, as seen in our textbook solution, the introduction of a variable 'x' representing the change in pressures at equilibrium simplifies the process. Then, we apply algebraic methods to solve for 'x' and finally sum up the partial pressures. It's a useful skill, especially when tackling questions in competitive exams, to not only perform the calculations correctly but to understand their physical meaning: each 'x' represents the amount by which each gas contributes to the total pressure of the system.

Le Chatelier's Principle

Le Chatelier's Principle is a cornerstone concept in chemical equilibrium. It helps us predict how a system at equilibrium will respond to changes in concentration, temperature, or pressure. Simply put, the principle states: if an external stress is applied to a system at equilibrium, the system adjusts to partially counteract the imposed change and a new equilibrium is established.

This principle can be illustrated in our textbook problem: if, for example, the pressure over the equilibrium mixture were increased, the system would respond by favoring the reaction that produces fewer mole(s) of gas—shifting the equilibrium position. Conversely, if the pressure were decreased, the equilibrium would shift toward producing more gaseous product. Additionally, temperature changes can also affect the system; increasing the temperature favors the endothermic direction of the reaction, while decreasing it favors the exothermic direction. Understanding Le Chatelier's Principle empowers students to handle more complex equilibrium problems where multiple dynamic factors are in play.

Physical Chemistry Competitive Exams

Competitive exams in the physical sciences, including physical chemistry, often feature problems related to chemical equilibrium to test a student's understanding of this challenging concept. The problems are designed to assess a range of skills: conceptual understanding, mathematical proficiency, and problem-solving strategies.

In preparing for such exams, it is imperative to master calculations involving equilibrium constants, partial pressures, and to utilize Le Chatelier's Principle effectively. Our solved problem represents a typical question that candidates might encounter. Students should practice various scenarios involving shifts in equilibrium, interpretation of constant expressions, and calculations under different conditions. Furthermore, attention should be given to the meticulous presentation of data and stepwise problem-solving tactics as demonstrated in the provided solution. This meticulous approach is appreciated in competitive exams and is often a deciding factor in a student's performance.