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Chapter 7: Problem 48

If the units for rate are \(\mathrm{M} \mathrm{s}^{-1}\), what are the units for the rate constant \((\mathrm{k})\), if the overall order of the reaction is three? a. \(\mathrm{M}^{-1} \mathrm{~s}^{-1}\) b. \(\mathrm{M}^{-2} \mathrm{~s}^{-1}\) c. \(\mathrm{s}^{-1}\) d. \(\mathrm{M}^{2} \mathrm{~s}^{-1}\)

Short Answer

Expert verified
b. \( \mathrm{M}^{-2} \mathrm{~s}^{-1} \)

Step by step solution

01

Understand the rate law expression

For a reaction of B \text{Order} = n \u007D, the rate law is typically written as \( \text{rate} = k [A]^x [B]^y ... \) where the sum of x, y, etc., equals the overall order \( n \). Here, you are given \( n = 3 \).
02

Identify the units for rate

The units for the rate are given as \( \mathrm{M} \mathrm{s}^{-1} \). This tells us how concentration (in molarity, M) changes over time (in seconds, s).
03

Set up the dimensional equation for the rate constant

The dimensional formula for the rate law is \( [\text{rate}] = [k] [\text{concentration}]^n \). Here, \([\text{rate}] = \mathrm{M} \mathrm{s}^{-1}\), and \(n = 3\), so \( [k] = \frac{[\text{rate}]}{[\text{concentration}]^3} \).
04

Substitute and solve

Substitute the known units into the equation: \( \text{Units of } k = \frac{\mathrm{M} \mathrm{s}^{-1}}{\mathrm{M}^3} \). Simplify to find the units of \( k \).
05

Simplify the expression

The expression simplifies as follows: \( \frac{\mathrm{M} \mathrm{s}^{-1}}{\mathrm{M}^3} = \mathrm{M}^{-2} \mathrm{s}^{-1} \). This shows that the correct units for \( k \) are \( \mathrm{M}^{-2} \mathrm{s}^{-1} \).

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Key Concepts

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

Rate Law
The rate law in chemistry describes the relationship between the rate of a chemical reaction and the concentration of its reactants. It's an important equation that helps in predicting how fast a reaction occurs. For a generic reaction, the rate law can be expressed as:
  • \[\text{rate} = k [A]^x [B]^y \ldots\]
Where:
  • \( k \) is the rate constant, unique to each reaction.
  • \([A] \) and \([B] \) are the concentrations of the reactants \( A \) and \( B \).
  • \( x \) and \( y \) are the reaction orders with respect to \( A \) and \( B \), respectively.
The sum \( x + y + \ldots \) gives the overall order of reaction. The rate law is determined experimentally and varies for different reactions. Understanding it allows us to comprehend how changes in concentration affect the reaction rate.
Overall Order of Reaction
The overall order of a reaction is the sum of the powers of the concentration terms in the rate law expression. This number gives insight into how the concentration of reactants impacts the reaction rate. For example, if a reaction has an overall order of three, expressed as:
  • \[\text{rate} = k [A]^1 [B]^2\]
The overall order is calculated as \( 1 + 2 = 3 \). Each term in the rate equation influences the rate in exponential proportions, signifying that small changes in concentration can have large effects on the reaction rate.
The overall order helps predict how a system's rate will change when reactant concentrations are altered, crucial for controlling industrial and laboratory reaction conditions.
Dimensional Analysis
Dimensional analysis is a mathematical method used to convert units and solve equations by assessing their dimensions. In chemistry, it's particularly useful for identifying the units of the rate constant \( k \) in a rate law equation. Given the rate, expressed in units of \( \mathrm{M} \mathrm{s}^{-1} \), and an overall order of three, you can determine \( k \)'s units as follows:
  • First, set up the dimensional equation: \[[\text{rate}] = [k] [\text{concentration}]^n\]
  • Plug in the values: \[[k] = \frac{\mathrm{M} \mathrm{s}^{-1}}{(\mathrm{M})^3}\]
  • Simplify: \[[k] = \mathrm{M}^{-2} \mathrm{s}^{-1}\]
Dimensional analysis is vital in calculating the correct units for various quantities, ensuring consistency and correctness in scientific calculations. It helps in checking the physical feasibility of equations and offers a way to cross-verify results by focusing strictly on the units used.

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Most popular questions from this chapter

The reaction for the decomposition of dinitrogen monoxide gas to form an oxygen radical is: \(\mathrm{N}_{2} \mathrm{O}\) (g) \(\rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g})\). If the activation energy is 250 \(\mathrm{kJ} / \mathrm{mol}\) and the frequency factor is \(8.0 \times 10^{11} \mathrm{~s}^{-1}\), what is the rate constant for the first order reaction at \(1000 \mathrm{~K} ?\) a. \(7.0 \times 10^{-2} \mathrm{~s}^{-1}\) b. \(3.7 \times 10^{-2} \mathrm{~s}^{-1}\) c. \(0.71 \times 10^{-3} \mathrm{~s}^{-1}\) d. \(9.7 \times 10^{-6} \mathrm{~s}^{-1}\)

The first order isomerization reaction: Cyclopropane \(\rightarrow\) Propene, has a rate constant of \(1.10 \times 10^{-4} \mathrm{~s}^{-1}\) at \(470^{\circ} \mathrm{C}\) and \(5.70 \times 10^{-4} \mathrm{~s}^{-1}\) at \(500^{\circ} \mathrm{C}\). What is the activation energy (Ea) for the reaction? a. \(340 \mathrm{~kJ} / \mathrm{mol}\) b. \(260 \mathrm{~kJ} / \mathrm{mol}\) c. \(160 \mathrm{~kJ} / \mathrm{mol}\) d. \(620 \mathrm{~kJ} / \mathrm{mol}\)

For the first order reaction, \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\), what is the concentra- tion of \(\mathrm{N}_{2} \mathrm{O}\) after 3 half-lives if \(0.15 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}\) is initially placed into in a \(1.00\) litre reaction vessel? a. \(9.1 \times 10^{-2} \mathrm{M}\) b. \(0.9 \times 10^{-2} \mathrm{M}\) c. \(2.68 \times 10^{-2} \mathrm{M}\) d. \(1.9 \times 10^{-2} \mathrm{M}\)

The rate constant for an isomerization reaction \(\mathrm{A} \rightarrow\) \(\mathrm{B}\), is \(4.5 \times 10^{-3} \mathrm{~min}^{-1}\). If the initial concentration of \(\mathrm{A}\) is \(1 \mathrm{M}\). Calculate the rate of reaction after 1 hour. a. \(0.34354 \mathrm{M} \mathrm{min}^{-1}\) b. \(0.034354 \mathrm{M} \mathrm{min}^{-1}\) c. \(0.0034354 \mathrm{M} \mathrm{min}^{-1}\) d. \(0.0003454 \mathrm{M} \mathrm{min}^{-1}\)

Hydrogen iodide decomposes at \(800 \mathrm{~K}\) via a second order process to produce hydrogen and iodine according to the following chemical equation. \(2 \mathrm{HI}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) At \(800 \mathrm{~K}\) it takes 142 seconds for the initial concentration of HI to decrease from \(6.75 \times 10^{-2} \mathrm{M}\) to \(3.50 \times 10^{-2} \mathrm{M}\). What is the rate constant for the reaction at this temperature?a. \(6.69 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) b. \(7.96 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) c. \(19.6 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) d. \(9.69 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\)
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