/*! 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 47 Chlorine reacts with chloroform ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 47

Chlorine reacts with chloroform according to the reaction given below: \(\mathrm{Cl}_{2}+\mathrm{CHCl}_{3} \rightarrow \mathrm{CCl}_{4}+\mathrm{HCl}\) When the initial concentration of \(\mathrm{Cl}_{2}\) is doubled the reaction rate increases by a factor of \(1.41\). What is the order of the reaction with respect to \(\mathrm{Cl}_{2} ?\) a. \(1 / 2\) \mathbf{b} . ~ \(-1 / 2\) c. \(-1\) d. 2

Short Answer

Expert verified
The order of the reaction with respect to \(\mathrm{Cl}_{2}\) is \(1/2\).

Step by step solution

01

Understand the Rate Law Expression

In a chemical reaction, the rate law expresses the rate of reaction as a function of the concentration of the reactants, each raised to a power called the order of reaction. For the reaction \(\mathrm{Cl}_{2}+\mathrm{CHCl}_{3} \rightarrow \mathrm{CCl}_{4}+\mathrm{HCl}\), the general rate law can be written as: \[ \text{Rate} = k \cdot [\mathrm{Cl}_{2}]^m \cdot [\mathrm{CHCl}_{3}]^n \]where \(k\) is the rate constant, and \(m\) and \(n\) are the orders of reaction with respect to \(\mathrm{Cl}_{2}\) and \(\mathrm{CHCl}_{3}\), respectively. We need to determine \(m\).
02

Analyze the Impact of Doubling Concentration

According to the problem, when the initial concentration of \(\mathrm{Cl}_{2}\) is doubled, the reaction rate increases by a factor of 1.41. Mathematically, this can be expressed as:\[ \frac{\text{New Rate}}{\text{Old Rate}} = \frac{k \cdot (2[\mathrm{Cl}_{2}])^m \cdot [\mathrm{CHCl}_{3}]^n}{k \cdot ([\mathrm{Cl}_{2}])^m \cdot [\mathrm{CHCl}_{3}]^n} = 1.41 \]This simplifies to:\[ 2^m = 1.41 \]
03

Solve the Equation for m

To find the value of \(m\), take the logarithm of both sides:\[ \log(2^m) = \log(1.41) \]By logarithmic property, this becomes:\[ m \cdot \log(2) = \log(1.41) \]Now, solve for \(m\):\[ m = \frac{\log(1.41)}{\log(2)} \approx \frac{0.1492}{0.3010} \approx 0.5 \]
04

Interpret the Result

The result \(m \approx 0.5\) indicates that the reaction is half-order with respect to \(\mathrm{Cl}_{2}\). From the given choices, option a (1/2) is correct, which matches our calculated value. This means that the rate of reaction changes by the square root of the ratio of change in concentration when \(\mathrm{Cl}_{2}\) is altered.

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Key Concepts

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

Reaction Order
Understanding reaction order is crucial in determining how changes in reactant concentrations affect the rate of a reaction. In the given chemical equation, \[ \mathrm{Cl}_{2} + \mathrm{CHCl}_{3} \rightarrow \mathrm{CCl}_{4} + \mathrm{HCl} \]we observe that the order of reaction with respect to chlorine (\( \mathrm{Cl}_{2} \)) is key to predicting how the reaction rate will respond to variations in its concentration.
In this case, the order of reaction with respect to \( \mathrm{Cl}_{2} \) is determined to be 0.5.
This tells us that the rate of reaction is proportional to the square root of the concentration of \( \mathrm{Cl}_{2} \), which is neither proportional to the concentration directly (first-order) nor constant (zero-order).
Understanding this helps in devising strategies to control reaction speeds in industrial and laboratory settings.
Rate Law
The rate law for a chemical reaction is a mathematical expression that relates the rate of reaction to the concentration of its reactants. For the reaction in question, the rate law can be given as:\[ \text{Rate} = k \cdot [\mathrm{Cl}_{2}]^m \cdot [\mathrm{CHCl}_{3}]^n \]Here, \( k \) is the rate constant, while \( m \) and \( n \) represent the reaction orders with respect to \( \mathrm{Cl}_{2} \) and \( \mathrm{CHCl}_{3} \) respectively. The coefficients \( m \) and \( n \) help us understand how sensitive the rate is to concentration changes of the reactants involved.
  • When \( m \) is 0, the rate is independent of \( [\mathrm{Cl}_{2}] \).
  • If \( m \) is 1, the rate changes directly with changes in \( [\mathrm{Cl}_{2}] \).
  • In this problem, \( m \) was found to be 0.5, indicating a half-order with respect to \( \mathrm{Cl}_{2} \).
By exploring these coefficients, scientists can predict reaction behavior under varying conditions.
Chemical Kinetics
Chemical kinetics is the study of factors affecting the speed or rate of chemical reactions. It is key in understanding how different variables such as concentration, temperature, and presence of catalysts influence the rate at which a chemical reaction proceeds.
In this particular exercise, kinetics was used to identify how altering the concentration of \( \mathrm{Cl}_{2} \) impacts the reaction's reaction rate.
This involves analyzing how reactant particles collide and transform into products over time.
  • Experimentally, doubling the concentration of \( \mathrm{Cl}_{2} \) increased the rate by a factor of 1.41.
  • By establishing the relationship between reactant concentrations and reaction rates, we can design experiments and industrial processes with predictable outcomes.
Understanding chemical kinetics allows chemists to control reaction conditions, enhancing efficiency in chemical manufacturing and minimizing undesired side reactions.
Rate Constant
The rate constant, \( k \), is a crucial factor in the rate law equation. It indicates the inherent speed of a reaction when all reactants are at unit concentration.
It is a fixed value specific to a particular reaction at a given temperature.
  • The rate constant does not change with concentration or time but can vary with temperature.
  • In the rate law: \( \text{Rate} = k \cdot [\mathrm{Cl}_{2}]^m \cdot [\mathrm{CHCl}_{3}]^n \), \( k \) provides the proportionality needed to equate the rate to the concentration terms.
  • A larger \( k \) means a faster reaction, while a smaller \( k \) suggests a slower process.
Understanding and calculating the rate constant helps in comparing different reactions and establishing which conditions optimize reaction speed in real-world applications.

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Most popular questions from this chapter

Which of the following is/are experimentally determined? a. Rate law b. Order c. Molecularity d. Rate constant

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. \(620 \mathrm{~kJ} / \mathrm{mol}\)

The following data pertains to the reaction between A and B $$ \begin{array}{llll} \hline \text { S. } & {[\mathrm{A}]} & {[\mathrm{B}]} & \text { Rate } \\ \text { No. } & \mathrm{mol} \mathrm{L}^{-1} & \mathrm{~mol} \mathrm{~L}^{-1} & \mathrm{Mol} \mathrm{L}^{-1} \mathrm{t}^{-1} \\ \hline 1 & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\ 2 & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-4} \\ \hline \end{array} $$ Which of the following inferences are drawn from the above data? (1) Rate constant of the reaction is \(10^{-4}\) (2) Rate law of the reaction is k [A][B] (3) Rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: a. 1 and 3 b. 2 and 3 c. 1 and 2 d. 1,2 and 3

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): For the hydrogen halogen photochemical reaction, the quantum yield for the formation of \(\mathrm{HBr}\), is lower than that of \(\mathrm{HCl}\). (R): \(\mathrm{Br}+\mathrm{H}_{2} \rightarrow \mathrm{HBr}+\mathrm{H}\) has higher activation energy than \(\mathrm{Cl}+\mathrm{H}_{2} \rightarrow \mathrm{HCl}+\mathrm{H}\)

Which of the following statement(s) is/are incorrect? a. A plot of \(\mathrm{P}\) versus \(\mathrm{l} / \mathrm{V}\) is linear at constant temperature. b. A plot of P versus \(1 / \mathrm{T}\) is linear at constant volume. c. A plot of \(\log \mathrm{K}_{\mathrm{p}}\) versus \(1 / \mathrm{T}\) is linear d. A plot of \(\log [x]\) versus time is linear for zero order reaction.
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