The rate law for the reaction $$ \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{4}(g) $$ is $$ \text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right] $$ What are the units for \(k\), assuming time in seconds and concentration in \(\mathrm{mol} / \mathrm{L}\) ?

Short Answer

Expert verified
The units of the rate constant k are \(\text{mol}^{1/2}\text{L}^{-1/2}\text{s}^{-1}\).

Step by step solution

01

Rewrite the rate law equation

We are given the rate law as: \[ \text {Rate} = k[\mathrm{Cl}_{2}]^{1/2} [\mathrm{CHCl}_{3}] \]
02

Determine the units of the Rate and concentrations

The time is given in seconds, and the concentrations are in moles per liter (mol/L). Thus, the rate has units of moles per liter per second (mol/L.s).
03

Determine the units of k using dimensional analysis

In order to determine the units of k, we need the rate equation to have the same units on both sides. We have: \[ \frac{\text{mol}}{\text{L}\cdot\text{s}} = k \left(\frac{\text{mol}}{\text{L}}\right)^{1/2}\left(\frac{\text{mol}}{\text{L}}\right) \] Now, we need to find the units of k that make both sides of the equation equal.
04

Solve for the units of k

To find the units of k, we will multiply both sides of the equation by the inverse of the concentrations: \[ k = \frac{\text{mol}}{\text{L}\cdot\text{s}} \cdot \left(\frac{\text{L}}{\text{mol}}\right)^{1/2} \cdot \left(\frac{\text{L}}{\text{mol}}\right) \] Now, by multiplying and simplifying the units, we get: \[ k = \frac{\text{mol}^{1/2}}{\text{L}^{1/2}\cdot\text{s}} \] The units of the rate constant k are \(\text{mol}^{1/2}\text{L}^{-1/2}\text{s}^{-1}\).

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Key Concepts

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

Understanding Chemical Kinetics
Chemical kinetics is the branch of physical chemistry that studies the rates at which chemical reactions occur and the factors that affect these rates. It involves the analysis of how different experimental conditions can influence the speed of a chemical reaction and the determination of the reaction mechanisms. A central idea in chemical kinetics is the reaction rate, which is a measure of how fast a reactant is consumed or a product is formed.

When studying chemical kinetics, students often encounter the rate law, which is a mathematical equation showing the relationship between the concentration of reactants and the reaction rate. The rate law provides valuable insights into the reaction's mechanism, as different reactions will have unique rate laws reflecting steps involved in the chemical process.

For example, a simple rate law might look like this: \[ \text{Rate} = k[\text{A}] \] where \( k \) is the rate constant, \( [\text{A}] \) represents the concentration of reactant A, and the Rate is the speed at which the reaction proceeds. The exponents, like the 1/2 for \( \text{Cl}_2 \) in our original exercise, give the reaction order with respect to each reactant, providing clues about the reaction mechanism.
Deciphering Rate Constant Units
Rate constant units can be perplexing for students initially, but they are simply a consequence of ensuring that the rate law equation is dimensionally consistent. The rate constant, symbolized by \( k \), expresses how the rate of reaction changes with varying reactant concentration. Because the rate constant's units depend on the overall order of the reaction, different reactions will yield different units for their rate constants.

In the context of our original exercise, where the rate law is \( \text{Rate} = k[\mathrm{Cl}_{2}]^{1/2} [\mathrm{CHCl}_{3}] \), determining the units of \( k \) requires solving for \( k \) while accounting for the units of concentration and rate. The final units obtained, in this case, \( \text{mol}^{1/2}\text{L}^{-1/2}\text{s}^{-1} \) reflect the half-order dependence on \( \text{Cl}_2 \), and the first-order dependence on \( \mathrm{CHCl}_{3} \).

This dimensional analysis is crucial for experimentalists who need to calculate the rate constant from their measurements and compare it across different reactions under various conditions, ensuring accurate and consistent calculations in chemical kinetics.
Measuring The Reaction Rate
A fundamental concept within chemical kinetics is the reaction rate, the speed at which a chemical reaction occurs. It's an indicator of how quickly reactants turn into products over time and is directly measurable in the laboratory. For example, the reaction rate may be established by recording the decrease in the concentration of a reactant or the increase in the concentration of a product at regular intervals.

Understanding reaction rates is essential for controlling processes in industrial applications, developing pharmaceuticals, and even elucidating biological pathways. In an academic setting, students frequently conduct experiments to measure reaction rates and use these observed rates to understand better the dynamics of chemical changes.

The reaction rate can be affected by various factors, including temperature, pressure, concentration of reactants, and the presence of a catalyst. Higher temperatures, for instance, usually increase the reaction rate because the reacting molecules have more kinetic energy and collide more forcefully and frequently, enhancing the chances of a reaction occurring.

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Most popular questions from this chapter

A certain substance, initially present at \(0.0800 M\), decomposes by zero-order kinetics with a rate constant of \(2.50 \times 10^{-2}\) \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\). Calculate the time (in seconds) required for the system to reach a concentration of \(0.0210 \mathrm{M}\).

For the reaction $$ \mathrm{O}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ the observed rate law is $$ \text { Rate }=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right] $$ Which of the changes listed below would affect the value of the rate constant \(k ?\) a. increasing the partial pressure of oxygen gas b. changing the temperature c. using an appropriate catalyst

Sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) decomposes to sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) and chlorine \(\left(\mathrm{Cl}_{2}\right)\) by reaction in the gas phase. The following pressure data were obtained when a sample containing \(5.00 \times 10^{-2}\) mol sulfuryl chloride was heated to \(600 . \mathrm{K}\) in a \(5.00 \times 10^{-1}-\mathrm{L}\) container. Defining the rate as \(-\frac{\Delta\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]}{\Delta t}\), a. determine the value of the rate constant for the decomposition of sulfuryl chloride at \(600 . \mathrm{K}\). b. what is the half-life of the reaction? c. what fraction of the sulfuryl chloride remains after \(20.0 \mathrm{~h}\) ?

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t\) : $$ \begin{aligned} -\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ -\frac{\Delta \mathrm{B}}{\Delta t} &=0.0120 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.

Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \operatorname{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$ \operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ This disproportionation follows first-order kinetics with a half-life of \(667 \mathrm{~s}\). What is the concentration of \(\operatorname{In}^{+}(a q)\) after \(1.25 \mathrm{~h}\) if the initial solution of \(\mathrm{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{~g} \operatorname{InCl}(s)\) in dilute \(\mathrm{HCl}\) to make \(5.00 \times 10^{2} \mathrm{~mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{~h}\) ?

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