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Because partial pressures are proportional to concentrations, rate laws for gas-phase reactions can also be expressed in terms of partial pressures, for instance, as Rate \(=\) \(k P_{\mathrm{X}}\) for a first-order reaction of a gas \(\mathrm{X}\). What are the units for the rate constants when partial pressures are expressed in torr and time is expressed in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second-order reactions?

Short Answer

Expert verified
The units for the rate constants k are: (a) torr/second for zero-order reactions; (b) 1/second \text{(s^-1)} for first-order reactions; (c) 1/(torr·second) for second-order reactions.

Step by step solution

01

Analyze the Units for a Zero-Order Reaction

For a zero-order reaction, the rate of reaction is given by Rate = k. Since Rate is given in terms of change in pressure over time, for a zero-order reaction, the rate units would be the same as the units of partial pressure, which is torr. Thus, the units for k in a zero-order reaction would be torr/second.
02

Analyze the Units for a First-Order Reaction

For a first-order reaction, the rate of reaction is given by Rate = k \(P_X\). Here, Rate has units of torr/second, and \(P_X\) has units of torr. To get units of torr/second for Rate, the units for k must be 1/second (s^-1), because the torr unit from \(P_X\) cancels out with the torr in the rate expression.
03

Analyze the Units for a Second-Order Reaction

For a second-order reaction, the rate of reaction is given by Rate = k \(P_X^2\). In this case, Rate is in torr/second, and \(P_X^2\) has units of torr^2. To balance the units, k must have units of 1/(torr·second) to yield a Rate in torr/second after multiplying by \(P_X^2\) in torr^2.

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Key Concepts

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

Partial Pressures
Understanding partial pressures is essential when studying gas-phase reactions, especially when we need to convert concentration-based rate laws into a form that uses pressure. This concept hinges on Dalton's Law, which says that each gas in a mixture exerts pressure independently as if it were alone in the volume. The partial pressure of a particular gas is proportional to its molar fraction in the mixture.

When we look at reaction rates in terms of partial pressures, we're often talking about closed systems where volume remains constant. Under these conditions, the changes in partial pressure directly reflect the changes in concentration.

In the context of the exercise, we can see that the unit measurements for the rate of an reaction (torr/second) involve partial pressures, providing a straightforward approach to understanding how the concentration of gas X changes over time, and emphasizing the practical application of partial pressures in measuring the progress of a reaction.
Rate Constants
The rate constant is a proportionality factor in the rate law of a chemical reaction and is essential for quantitatively describing the speed of a chemical reaction. Its value is determined experimentally and provides insights into the reaction dynamics.

For rate constants, units are not one-size-fits-all; they depend on the reaction order. In the exercise provided, we explore how to ascertain the correct units for rate constants when reactions are zero-order, first-order, or second-order, each requiring a different approach due to the way they each relate to the concentration (or partial pressure) of reactants.

To pinpoint the units of the rate constant, we must ensure that when the rate constant and the partial pressures (measured in torr) are multiplied together, the result matches the units of the reaction rate (torr/second). This results in different units for the rate constant depending on the order of the reaction, ensuring the final units align correctly with the rate of pressure change in the reaction.
Gas-Phase Reactions
Gas-phase reactions describe the processes involving reactants in the gaseous state. These reactions are influenced by various factors such as temperature, pressure, and volume, which can considerably affect reaction rates and mechanisms.

In gas-phase kinetics, we often use partial pressures to express concentrations of reactants, influenced by the ideal gas law, which correlates the pressure, volume, and temperature of a gas. When we measure rates in terms of partial pressure, we directly relate the pressure exerted by the reacting gas to its concentration, making it easier to connect the experimental observations to the underlying kinetic model.

As demonstrated in the exercise, understanding the units associated with gas-phase reaction rate constants relative to partial pressures is crucial, not just for solving textbook problems, but for real-world chemical kinetics where accurate descriptions of reaction speeds are necessary for both academic and industrial applications.

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Most popular questions from this chapter

The half-life for the second-order reaction of a substance A is \(50.5 \mathrm{~s}\) when \([\mathrm{A}]_{0}=0.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) one- sixteenth; (b) onefourth; (c) one-fifth of its original value.

Determine which of the following statements about catalysts are true. If the statement is false, explain why. (a) A heterogeneous catalyst works by binding one or more of the molecules undergoing reaction to the surface of the catalyst. (b) Enzymes are naturally occurring proteins that serve as catalysts in biological systems. (c) The equilibrium constant for a reaction is greater in the presence of a catalyst. (d) A catalyst changes the pathway of a reaction in such a way that the reaction becomes more exothermic.

The pre-equilibrium and the steady-state approximations are two different approaches to deriving a rate law from a proposed mechanism. For the following mechanism, determine the rate law (a) by the pre-equilibrium approximation and (b) by the steady-state approximation. (c) Under what conditions do the two methods give the same answer? (d) What will the rate laws become at high concentrations of \(\mathrm{Br}^{-}\)? $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {(fast equilibrium) } \\ &\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} \text { (slow) } \end{aligned} $$

(a) From the following mechanism, derive Eq. 19a, which Michaelis and Menten proposed to represent the rate of formation of products in an enzyme-catalyzed reaction. (b) Show that the rate is independent of substrate concentration at high concentrations of substrate. $$ \begin{aligned} &\mathrm{E}+\mathrm{S} \rightleftarrows \mathrm{ES} \quad k_{1}, k_{1}^{\prime} \\ &\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad k_{2} \end{aligned} $$ where \(E\) is the free enzyme, \(S\) is the substrate, ES is the enzyme-substrate complex, and \(P\) is the product. Note that the steady-state concentration of free enzyme will be equal to the initial concentration of the enzyme less the amount of enzyme that is present in the enzyme-substrate complex: \([\mathrm{E}]=[\mathrm{E}]_{0}-[\mathrm{ES}]\)

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) The equilibrium constant for a reaction equals the rate constant for the forward reaction divided by the rate constant for the reverse reaction. (b) In a reaction that is a series of equilibrium steps, the overall equilibrium constant is equal to the product of all the forward rate constants divided by the product of all the reverse rate constants. (c) Increasing the concentration of a product increases the rate of the reverse reaction, and so the rate of the forward reaction must then increase, too.

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