/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 It is true that:(a) A second ord... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

It is true that:(a) A second order reaction is always a multistep reaction (b) A zero order reaction is a multistep reaction (c) A first order reaction is always a single step reaction (d) A zero order reaction is a single step reaction

Short Answer

Expert verified
All statements are false based on the explanation of reaction order and mechanisms.

Step by step solution

01

Understanding Reaction Orders

Reaction order refers to the dependence of the rate of reaction on the concentration of reactants. For zero-order reactions, the rate is constant and does not depend on the concentration of reactants. First-order reactions depend linearly on one reactant's concentration, while second-order reactions depend on either two reactants or the square of one reactant.
02

Analyzing Zero-Order Reactions

Zero-order reactions have a rate that is independent of the concentration of reactants. As there is no concentration dependency, these reactions often involve a surface or catalyst. They can be either single-step or multi-step based on the mechanism involved.
03

Examining Second-Order Reactions

Second-order reactions involve the concentration of two reactants or the square of a single reactant. This does not imply they are necessarily multistep reactions. Some second-order reactions can proceed in a single step, albeit this is less common.
04

Understanding First-Order Reactions

First-order reactions have a direct linear relationship with the concentration of a single reactant. They are often, but not always, single-step processes because the reaction rate directly corresponds to one reactant's concentration.
05

Evaluating Statements

(a) A second-order reaction is always a multistep reaction - False, as some may be single-step. (b) A zero order reaction is a multistep reaction - False, it can be single or multi-step. (c) A first order reaction is always a single step reaction - False, as mechanisms can vary. (d) A zero order reaction is a single step reaction - Not necessarily true; can be either.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Zero-Order Reactions
Zero-order reactions are quite unique because the rate of reaction is constant and does not depend on the concentration of reactants. This means that even if you increase the amount of reactants, the reaction rate stays the same. Often, zero-order reactions are observed in processes involving a limiting factor like a surface or a catalyst, which controls the rate.

These reactions are represented by the rate law equation: \[ \text{Rate} = k \] where \( k \) is the rate constant. One classic example of a zero-order reaction is the decomposition of ammonia on a platinum surface. It's important to note that zero-order reactions can be either single-step or multi-step, depending on the mechanism and conditions involved.

Overall, the characteristic feature is the constant rate over time, which distinctly sets zero-order reactions apart from other order reactions.
First-Order Reactions
First-order reactions have a distinct feature in that the reaction rate is directly proportional to the concentration of a single reactant. Essentially, if the concentration of the reactant doubles, the reaction rate also doubles. This linear relationship is a defining characteristic of first-order reactions.

The rate law for first-order reactions can be expressed as: \[ \text{Rate} = k[A] \] where \( [A] \) is the concentration of the reactant, and \( k \) is the rate constant. Radioactive decay is a well-known example of a first-order reaction, where the decay rate of a radioactive substance is proportional to the amount of the substance present.

While it is common to think of first-order reactions as single-step processes, this is not always the case. The mechanism can sometimes involve more steps, yet the overall rate still hinges on the concentration of one reactant. First-order reactions are frequently encountered in both chemical and biological systems.
Second-Order Reactions
Second-order reactions have a more complex dependency when it comes to reaction rates. They can depend on the concentration of two different reactants or the square of the concentration of a single reactant. This dependency makes the analysis of second-order reactions a bit more intricate.

The general rate law for a second-order reaction can be given as: \[ \text{Rate} = k[A]^{2} \text{ or } \text{Rate} = k[A][B] \] where \( k \) is the rate constant, and \( [A] \) and \( [B] \) are reactant concentrations. A typical example is the reaction between hydrogen atoms and bromine molecules to form hydrogen bromide.

One might assume that the complexity of second-order reactions means they are always multi-step in nature. However, while many second-order reactions do involve multiple steps, this is not a strict rule. Some can occur in a single step, although this scenario is not the usual narrative.

Second-order reactions are integral to understanding reaction kinetics, especially when dealing with reactions in solutions where two reactants come into play.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A nuclear explosion has taken place leading to increase in concentration of \(C^{14}\) in nearby areas. \(C^{14}\) concentration is \(C_{1}\) in nearby areas and \(C_{2}\) in areas far away. If the age of the fossil is determined to be \(T_{1}\) and \(T_{2}\) at the respective places then (a) The age of the fossil will increase at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (b) The age of the fossil will decrease at the place where explosion has taken place and \(T_{1}-T_{2}=\frac{1}{\lambda} \ln \frac{C_{1}}{C_{2}}\) (c) The age of fossil will be determined to be same (d) \(\frac{T_{1}}{T_{2}}=\frac{C_{1}}{C_{2}}\)

The reaction \(X \rightarrow Y\) is an exothermic reaction. Activation energy of the reaction for \(\mathrm{X}\) into \(\mathrm{Y}\) is \(150 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Enthalpy of reaction is 135 \(\mathrm{kJ} \mathrm{mol}^{-1}\). The activation energy for the reverse reaction, \(\mathrm{Y} \rightarrow \mathrm{X}\) will be : (a) \(280 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(285 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(270 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(15 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The rate of a first-order reaction is \(0.04\) mol litre \(^{-1} \mathrm{~s}^{-1}\) at 10 minutes and \(0.03\) mol litre \(^{-1} \mathrm{~s}^{-1}\) at 20 minutes after initiation. Find the half-life of the reaction. [2001 - 5 Marks]

A positron is emitted from \({ }_{11}^{23} \mathrm{Na}\). The ratio of the atomic mass and atomic number of the resulting nuclide is (a) \(22 / 10\) (b) \(22 / 11\) (c) \(23 / 10\) (d) \(23 / 12\)

Assertion : Nuclide \(\frac{30}{13} \mathrm{Al}\) is less stable than \({ }_{20}^{40} \mathrm{Ca}\) Reason : Nuclides having odd number of protons and neutrons are generally unstable. (a) If both assertion and reason are correct, and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct, but reason is not the correct explanation of the assertion. (c) If assertion is correct but reason is incorrect. (d) If assertion is incorrect but reason is correct.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.