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The rate of the reaction \(4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\) was studied by charging \(\mathrm{PH}_{3}(g)\) into a constant-volume reaction vessel and measuring the total pressure.

Short Answer

Expert verified
The given reaction takes place in a constant-volume container: \[4\mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g) + 6\mathrm{H}_{2}(g)\] Using the ideal gas law, we can relate pressure to moles in the container: \(P = \frac{nRT}{V}\). Initially, let the moles of PH3 be \(n_0\) and the moles of P4 and H2 be zero. After some time, when x moles of PH3 have reacted, the moles of PH3 decrease to \(n_0 - 4x\), and the moles of P4 and H2 become \(x\) and \(6x\), respectively. Since the pressure is proportional to the total number of moles, we have: \[P_{total} = k(n_0 + 3x)\] By measuring \(P_{total}\) as the reaction progresses, we can determine the change in moles of PH3 and analyze the rate of the reaction.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the given reaction is already provided: \[4\mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g) + 6\mathrm{H}_{2}(g)\]
02

Understand the relationship between moles and pressure

Since the reaction takes place in a constant-volume container, we can use the ideal gas law to express the relationship between pressure, moles, and temperature: \[P = \frac{nRT}{V}\] Here, \(P\) represents the pressure, \(n\) represents the number of moles, \(R\) is the ideal gas constant, \(T\) is the temperature, and \(V\) is the volume. In a constant-volume container, the total pressure is proportional to the total number of moles of gas.
03

Analyze the rate of the reaction

Let's analyze the rate of the reaction with respect to the concentration of reactants. At the beginning of the reaction, let the moles of PH3 be \(n_0\), the moles of P4 and H2 be zero. After some time, let's say that x moles of PH3 have reacted to form P4 and H2. Based on the balanced chemical equation, we can calculate the moles of P4 and H2 formed: PH3: Moles = \(n_0 - 4x\) P4: Moles = \(x\) H2: Moles = \(6x\) The total pressure, \(P_{total}\), in the container at this point is proportional to the total number of moles: \[P_{total} = k(n_0 - 4x + x + 6x)\] Here, \(k\) is a constant of proportionality, equal to \(\frac{RT}{V}\). \[P_{total} = k(n_0 + 3x)\] As the reaction progresses, we can measure the total pressure to determine the change in moles of PH3. This will help us in understanding the rate of the reaction.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation that describes the behavior of gases in terms of pressure, volume, temperature, and the number of moles. It is represented mathematically as:\[ P = \frac{nRT}{V} \]Where:
  • P is the pressure of the gas.
  • V is the volume of the gas.
  • n is the number of moles of gas.
  • R is the ideal gas constant.
  • T is the temperature in Kelvin.
In situations where the volume is held constant, like in a constant-volume reaction vessel, the pressure becomes directly proportional to the number of moles and the temperature. This means, as the reaction proceeds and the number of moles changes, the pressure will also vary. If the temperature is kept constant, changes in pressure will directly reflect changes in the number of moles, allowing us to study the reaction kinetics based on pressure measurements.
Balanced Chemical Equation
A balanced chemical equation is essential for correctly representing a chemical reaction. It ensures that the same number of each type of atom is present on both sides of the equation, adhering to the conservation of mass principle. In the equation provided:\[4\mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g) + 6\mathrm{H}_{2}(g)\]We see that:
  • Four molecules of \(\mathrm{PH}_{3}\) decompose to produce one molecule of \(\mathrm{P}_{4}\) and six molecules of \(\mathrm{H}_{2}\).
  • The atoms of phosphorus (P) and hydrogen (H) are balanced. There are 4 phosphorus atoms and 12 hydrogen atoms on both sides.
Balanced equations are vital in calculating the stoichiometry of the reaction, helping determine how much product is formed from a given amount of reactant. In a kinetic study, such as the one described in the exercise, the balanced equation allows us to relate the changes in pressure measured in the vessel to changes in the amount of each gas participating in the reaction.
Constant-volume Reaction Vessel
A constant-volume reaction vessel is specifically designed to maintain the same volume throughout a chemical reaction. This type of setup is crucial in experiments where one needs to accurately measure changes in pressure without the added variable of changing volume. By keeping the volume constant, any observed changes in pressure are directly attributable to changes in the number of gas moles or temperature.
In reaction kinetics, monitoring the pressure in a constant-volume vessel offers a simpler means to analyze the progression of the reaction. As the reactants are consumed to form products, the number of gas molecules changes, which in turn affects the pressure. By using the ideal gas law, the relationship between pressure and moles is understood, allowing us to track the rate of the reaction.
Constant-volume setups are particularly useful in gas-phase reactions where reactants or products involve gaseous species. This control allows for precise studies in chemical kinetics, providing insights into how quickly reactions proceed and the factors that influence rate changes.

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Most popular questions from this chapter

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right) .(\mathbf{a})\) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

For a first order reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C},\) if the half-life of \(\mathrm{A}\) at \(25^{\circ} \mathrm{C}\) is \(3.05 \times 10^{4} \mathrm{~s},\) what is the rate constant \(k\) at this temperature? What percentage of A will not have reacted after one day?

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) As the temperature increases, does the reaction rate increase or decrease? (c) As a reaction proceeds, does the instantaneous reaction rate increase or decrease?

The human body is characterized by an extremely complex system of interrelated chemical reactions. A large number of enzymes are necessary for many of these reactions to occur at suitable rates. Enzymes are very selective in the reactions they catalyze, and some are absolutely specific. Use the lock-and- key model to account for the specificity of an enzyme.

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\), with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. \((\mathbf{b})\) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

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