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Match the following: List I List II 1\. zero order reaction (p) mole- \({ }^{1} \mathrm{Lt} \sec -1\) 2\. first order reaction (q) \(\mathrm{mole}-{ }^{2}\) Lt2 \(\mathrm{sec}-1\) 3\. second order reaction (r) mole Lt- \({ }^{-1} \sec -1\) 4\. third order reaction (s) \(\sec -1\)

Short Answer

Expert verified
1-p, 2-s, 3-r, 4-q.

Step by step solution

01

Interpretation of Matching Rules

To solve this match-the-following exercise, we need to first understand the rate laws of each order of reaction and match them to the correct units in List II.
02

Rate Law for Zero Order Reaction

A zero order reaction has a rate law: \( \text{Rate} = k \). The rate constant \( k \) here has units that result in the rate of reaction units \( \text{mole} \, \text{L}^{-1} \, \text{sec}^{-1} \). Hence the unit that matches is \( \text{mole}-^{1} \text{Lt} \sec^{-1} \) which corresponds to list item (p).
03

Rate Law for First Order Reaction

A first order reaction has a rate law: \( \text{Rate} = k[A] \). The rate constant \( k \) has units \( \sec^{-1} \) to ensure the resultant rate has units \( \text{mole} \, \text{L}^{-1} \, \text{sec}^{-1} \). This corresponds to list item (s).
04

Rate Law for Second Order Reaction

A second order reaction can be written as \( \text{Rate} = k[A][B] \) or \( \text{Rate} = k[A]^2 \). The rate constant \( k \) has units \( \text{mole}^{-1} \text{L} \sec^{-1} \), thereby producing rate units of \( \text{mole} \, \text{L}^{-1} \, \text{sec}^{-1} \). Therefore, it should match with item (r).
05

Rate Law for Third Order Reaction

A third order reaction typically written as \( \text{Rate}=k[A]^2[B] \) or similar combinations. Its rate constant \( k \) will have units \( \text{mole}^{-2} \text{L}^2 \sec^{-1} \), ensuring the expression remains dimensionally consistent. This matches with item (q).
06

Final Matched Results

Based on the calculated rate law units for each order, the matches are: 1. Zero order reaction matches with (p) 2. First order reaction matches with (s) 3. Second order reaction matches with (r) 4. Third order reaction matches with (q).

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Key Concepts

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

Rate Law
Understanding the rate law of a reaction is essential for grasping the concept of reaction order and rates. The rate law expresses the rate of a chemical reaction as a function of the concentration of its reactants. In simpler terms, it shows how the speed of a reaction depends on the substances involved.

  • **Zero-order reactions**: The rate is independent of the concentration of the reactant(s). The rate law is expressed as: \( \text{Rate} = k \). Here, \( k \) is the rate constant with units of \( \text{mole}^-1 \text{Lt} \sec^{-1} \).

  • **First-order reactions**: The rate is directly proportional to the concentration of one reactant. It follows the law: \( \text{Rate} = k[A] \). The unit for \( k \) is \( \sec^{-1} \), ensuring the rate has units of \( \text{mole} \, \text{L}^{-1} \, \text{sec}^{-1} \).

  • **Second-order reactions**: These reactions depend on the concentration of one reactant squared, or two reactants. The rate law is: \( \text{Rate} = k[A]^2 \) or \( \text{Rate} = k[A][B] \). Here, the unit for \( k \) is \( \text{mole}^{-1} \text{Lt} \sec^{-1} \).

  • **Third-order reactions**: Dependent on either the concentration tripled or combinations like \( A^2[B] \). The rate law defines \( k \) with units of \( \text{mole}^{-2} \text{Lt}^2 \sec^{-1} \).
Reaction Kinetics
Reaction kinetics explores the rate at which a reaction progresses and the factors impacting this rate. It delves into the pathways and steps within a chemical reaction. Knowing the kinetics is crucial for predicting how a reaction behaves over time.

Several factors affect reaction kinetics, including:
  • **Concentration of reactants**: Higher concentrations generally increase the rate of reaction due to more frequent collisions between molecules.

  • **Temperature**: Raising the temperature speeds up reactions by providing more energy, which increases the frequency and intensity of collisions.

  • **Catalysts**: These substances lower the activation energy needed for a reaction, increasing the rate without being consumed in the process.

  • **Surface area**: More surface area allows more collisions and, as a result, a faster reaction rate.
Understanding these factors helps chemists to control and optimize reactions for desired outcomes.
Reaction Order Units
Reaction order is a classification of reactions based on how the concentration of reactants affects the rate. For each order, the rate law is different and has specific units for the rate constant to maintain consistency.

The units of the rate constant (\( k \)) vary depending on the reaction order:
  • **Zero-order**: Units of \( k \) are \( \text{mole}^1 \text{Lt}^{-1} \sec^{-1} \). This indicates that the rate is independent of the concentration of reactants.

  • **First-order**: The units are \( \sec^{-1} \), reflecting a direct proportionality to a single reactant's concentration.

  • **Second-order**: For these reactions, \( k \)'s units are \( \text{mole}^{-1} \text{Lt} \sec^{-1} \). This accounts for either a single reactant squared or two reactants.

  • **Third-order**: Here, the units are \( \text{mole}^{-2} \text{Lt}^2 \sec^{-1} \), accounting for reactant combinations such as \( [A]^2[B] \).
By understanding the units of the rate constant, chemists can deduce the order of the reaction and predict the reaction's behavior under various conditions.

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Most popular questions from this chapter

The equation tris(1,10-phenanthroline) iron(II) in acid solution takes place according to the equation: \(\mathrm{Fe}(\text { phen })_{3}^{2+}+3 \mathrm{H}_{3} \mathrm{O}^{+}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\) $$ \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}+3 \text { (Phen) } \mathrm{H}^{+} $$ If the activation energy (Ea) is \(126 \mathrm{~kJ} / \mathrm{mol}\) and the rate constant at \(30^{\circ} \mathrm{C}\) is \(9.8 \times 10^{-3} \mathrm{~min}^{-1}\), what is the frequency factor (A)? a. \(9.5 \times 10^{18} \mathrm{~min}^{-1}\) b. \(2.5 \times 10^{19} \mathrm{~min}^{-1}\) c. \(55 \times 10^{19} \mathrm{~min}^{-1}\) d. \(5.0 \times 10^{19} \mathrm{~min}^{-1}\)

In the following question two statements Assertion (A) and Reason (R) are given Mark. a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. If \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of \(\mathrm{A}\); c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. \(\mathrm{A}\) is false but \(\mathrm{R}\) is true, e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false. (A): The rate constant increases exponentially with the increase in temperature. ( \(\mathbf{R}\) ): With the rise in temperature, the average kinetic energy of the molecules increases.

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-4} \mathrm{~s}^{-1}\). If start made with \(1.0 \mathrm{~mol} \mathrm{~L}^{-1}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\), calculate the rate of formation of \(\mathrm{NO}_{2}\) at the moment of the reaction when concentration of \(\mathrm{O}_{2}\) is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\). a. \(1.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) b. \(3.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) c. \(9.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) d. \(4.8 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)

Consider a reaction \(\mathrm{aG}+\mathrm{bH} \rightarrow\) Products. When concentration of both the reactants \(\mathrm{G}\) and \(\mathrm{H}\) is doubled, the rate increases by eight times. However when concentration of \(\mathrm{G}\) is doubled keeping the concentration of \(\mathrm{H}\) fixed, the rate is doubled. The overall order of the reaction is a. 0 b. 1 c. 2 d. 3

Consider the following statements: (1) Rate of a process is directly proportional to its free energy change. (2) The order of an elementary reaction step can be determined by examining the stoichiometry. (3) The first order reaction describe exponential time coarse. Of the statements a. 1 and 2 are correct b. 1 and 3 are correct c. 2 and 3 are correct d. 1,2 and 3 are correct

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