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The rate constant of a first-order reaction is \(66 \mathrm{~s}^{-1}\) What is the rate constant in units of minutes?

Short Answer

Expert verified
The rate constant in units of minutes is \(1.1 \mathrm{~min}^{-1}\).

Step by step solution

01

Identify the Conversion Factor

There are 60 seconds in a minute. Therefore, to convert from seconds to minutes, you divide the number of seconds by 60. The conversion factor between seconds and minutes is therefore \( \frac{1}{60} \) or 0.01667.
02

Apply the Conversion Factor

To change the units of the rate constant from seconds to minutes, multiply the rate constant in seconds by the conversion factor. In mathematical terms, this would be \( 66 \mathrm{~s}^{-1} \times \frac{1}{60} = 1.1 \mathrm{~min}^{-1} \).
03

Write the Final Answer

The rate constant of the first-order reaction in units of minutes is therefore \( 1.1 \mathrm{~min}^{-1} \).

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Key Concepts

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

Chemical Kinetics
Understanding chemical kinetics is akin to figuring out how a clock ticks, but instead of seconds and minutes, chemists are looking at how fast reactions proceed. At its core, chemical kinetics deals with studying the speed or rate at which chemical reactions occur. This rate is influenced by various factors such as temperature, concentration of reactants, and the presence of a catalyst.

For instance, if we take a first-order reaction where the rate depends linearly on the concentration of one reactant, knowing the rate constant—a number that quantifies the speed of the reaction—becomes essential. This constant is unique for every reaction and conditions under which the reaction occurs. In an analogy for ease of understanding, if the reaction were a race, the rate constant would represent the average speed of runners, influencing how quickly they reach the finish line.

A higher rate constant would mean a faster reaction, analogous to a faster runner. And just like planning race strategies changes with the distance of the race, chemists might employ different techniques or consider different conditions based on the reaction's rate constant to control or predict the progression of the reaction.
Unit Conversion
When we talk about unit conversion in chemistry, we're looking at the translation of quantities from one measurement system to another, ensuring that we're comparing apples to apples, instead of apples to oranges. Applied to our first-order reaction rate constant, it's vital to ensure that the units we use are consistent with the rest of the calculations in the problem or experimental setup.

Units such as seconds \(\mathrm{s^{-1}}\) and minutes \(\mathrm{min^{-1}}\) are essential to understand the time scale of a reaction. For those new to unit conversion, it may seem challenging, but it follows a logical method. Using the correct conversion factors, as in our example of converting seconds to minutes, plays a crucial role in ensuring accurate calculations.

It's just like converting currency when traveling to a different country; you want to know precisely how much value you're getting after the exchange. Similarly, scientists need to be fluent in converting units to make sure their experiments 'currency' aligns with their expectations and understanding of the reaction in question.
Rate Laws
Rate laws are equations that quantify the relationship between the concentration of reactants and the rate of a chemical reaction. They are the formulas that let chemists put numbers to their predictions and solutions. Understanding a rate law is crucial because it helps predict how concentrations will change over time and under various conditions.

For a first-order reaction, the rate law can be expressed simply: the rate is directly proportional to the concentration of one reactant. This means that if you double the amount of reactant, the rate of reaction will also double. The rate law for a first-order reaction is typically written as \(\text{rate} = k[\text{A}]\), where \(k\) is the first-order rate constant and \[\text{A}\] represents the concentration of the reactant.

In our example, understanding the rate law allows us to seamlessly convert the rate constant from one set of time units to another, preserving the integrity of the scientific calculations. Just as a recipe might require adjustment when switching from a tablespoon to a teaspoon, chemists must adjust their rate constants to match the units used in their 'recipe' for studying reactions.

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Most popular questions from this chapter

As we know, methane burns readily in oxygen in a highly exothermic reaction. Yet a mixture of methane and oxygen gas can be kept indefinitely without any apparent change. Explain.

Consider the reaction $$ X+Y \longrightarrow Z $$ These data are obtained at \(360 \mathrm{~K}\): $$ \begin{aligned} &\text { Initial Rate of }\\\ &\begin{array}{ccc} \text { Disappearance of } \mathrm{X}(\mathrm{M} / \mathrm{s}) & {[\mathrm{X}]} & {[\mathrm{Y}]} \\ \hline 0.147 & 0.10 & 0.50 \\ 0.127 & 0.20 & 0.30 \\ 4.064 & 0.40 & 0.60 \\ 1.016 & 0.20 & 0.60 \\ 0.508 & 0.40 & 0.30 \end{array} \end{aligned} $$ (a) Determine the order of the reaction. (b) Determine the initial rate of disappearance of \(X\) when the concentration of \(\mathrm{X}\) is \(0.30 \mathrm{M}\) and that of \(\mathrm{Y}\) is \(0.40 \mathrm{M}\)

Consider the following elementary steps for a consecutive reaction $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(\mathrm{A}\).

The following gas-phase reaction was studied at \(290^{\circ} \mathrm{C}\) by observing the change in pressure as a function of time in a constant-volume vessel: $$ \mathrm{ClCO}_{2} \mathrm{CCl}_{3}(g) \longrightarrow 2 \mathrm{COCl}_{2}(g) $$ Determine the order of the reaction and the rate constant based on the following data: $$ \begin{array}{rc} \text { Time }(\mathrm{s}) & \mathrm{P}(\mathrm{mmHg}) \\ \hline 0 & 15.76 \\ 181 & 18.88 \\ 513 & 22.79 \\ 1164 & 27.08 \end{array} $$ where \(P\) is the total pressure.

The integrated rate law for the zero-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) is \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-k t .\) (a) Sketch the follow- ing plots: (i) rate versus \([\mathrm{A}]_{t}\) and (ii) \([\mathrm{A}]_{t}\) versus \(t\). (b) Derive an expression for the half-life of the reaction. (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when \([\mathrm{A}]_{t}=0\)

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