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Chapter 14: Problem 40

For the reaction \(A \longrightarrow 2 B+C\), the following data are obtained for \([\mathrm{A}]\) as a function of time: \(t=0 \mathrm{min}\) \([\mathrm{A}]=0.80 \mathrm{M} ; 8 \mathrm{min}, 0.60 \mathrm{M} ; 24 \mathrm{min}, 0.35 \mathrm{M} ; 40 \mathrm{min}\) \(0.20 \mathrm{M}\) (a) By suitable means, establish the order of the reaction. (b) What is the value of the rate constant, \(k ?\) (c) Calculate the rate of formation of \(\mathrm{B}\) at \(t=30 \mathrm{min}\).

Short Answer

Expert verified
By plotting the \(ln[A]\) against time, a straight line indicates first order reaction. The absolute value of the slope of the line provides the rate constant \(k\). Then use the rate equation for the formation of \(B\), \(Rate = 2k[A]\) to find the rate at \(t=30 \mathrm{min}\).

Step by step solution

01

Establish the order of the reaction

In the given data, the concentration of \(A\) is reducing over time, so the reaction order can be determined by plotting the log of concentration of \(A\) against time (a linear relationship suggests a first order reaction). A semi-log plot can be drawn, plotting the natural logarithm (ln) of concentration of \(A\) against time \(t\). If the plot yields a straight line, the reaction is first-order.
02

Determine the rate constant, \(k\)

The slope of the semi-log plot \(ln[A]\) vs \(t\) is equal to \(-k\) (negative because it is a reactant). Calculate the slope of the line and the rate constant \(k\) can be determined by taking absolute value of the slope.
03

Calculate rate of formation of \(B\)

For a first-order reaction like \(A \longrightarrow 2 B+C\), the rate of formation of \(B\) is twice the rate constant times the concentration of \(A\). So at \(t=30 \mathrm{min}\), interpolate the concentration of \(A\) from the given data, multiply it by twice the rate constant to get the rate of formation of \(B\) at \(t=30 \mathrm{min}\).

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Key Concepts

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

Reaction Order
The reaction order is an essential concept in reaction kinetics that helps us understand how the concentration of reactants influences the rate of a chemical reaction. To determine the order of a reaction, we need to observe how the concentration of the reactant changes over time. If the concentration of the reactant and time data produce a straight line when plotted using a logarithmic scale, we can confirm that the reaction is of first-order.

In this exercise, we have a reaction involving a reactant \( A \) gradually converting to products over time. By taking the natural logarithm of the concentration of \( A \) and plotting these values against time, a linear relationship indicates a first-order reaction.

  • A first-order reaction means that the rate of reaction is directly proportional to the concentration of the reactant.
  • For first-order reactions, the half-life is independent of the initial concentration of the reactant.
  • The slope of the plot on a semi-log scale provides further insight into the reaction kinetics.
Rate Constant
The rate constant, \( k \), is a crucial parameter that provides insights into the speed of a chemical reaction. It is a unique value for every reaction at a given temperature and can be influenced by factors like temperature and catalyst presence. For a first-order reaction, the rate constant is determined from the slope of the natural log of the concentration of the reactant plotted against time.

In the example given, \( k\) is calculated from the slope of the line obtained in the semi-logarithmic plot of \( \ln[A] \) versus time. Since the line is straight, indicating a first-order reaction, the slope of the line is equal to \( -k \). Taking the absolute value of the slope gives us the rate constant.

  • The rate constant provides information on how fast or slow a reaction proceeds.
  • For first-order reactions, the unit of \( k \) is \( \text{time}^{-1} \).
  • The value of \( k \) remains constant as long as the conditions (such as temperature) remain unchanged.
Rate of Formation
The rate of formation is an important concept that tells us how quickly products are generated in a chemical reaction. Specifically, for the formation of a product like \( B \) in a reaction, it is crucial to understand how it relates to the reactant consumption.

In a reaction such as \( A \rightarrow 2B + C \), the appearance of \( B \) happens twice as fast as the disappearance of \( A \), since two moles of \( B \) are formed for every mole of \( A \) that reacts. To calculate the formation rate of \( B \), we need to know the concentration of \( A \) at the time in question and the rate constant. The rate of formation at any given time, such as 30 minutes in this case, can be calculated with the formula: Rate = \( 2k[A] \).

  • The rate of formation tells us the speed at which a specific product is produced.
  • For first-order reactions, the rate of formation can be easily estimated once \( k \) and concentration of the reactant are known.
  • The factor of "2" in this context depends on the stoichiometry of the reaction, which shows how many units of \( B \) are formed per unit of \( A \) reacted.

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Most popular questions from this chapter

In the reaction \(A \longrightarrow\) products, 4.40 min after the reac- tion is started, \([\mathrm{A}]=0.588 \mathrm{M}\). The rate of reaction at this point is rate \(=-\Delta[\mathrm{A}] / \Delta t=2.2 \times 10^{-2} \mathrm{M} \mathrm{min}^{-1}.\) Assume that this rate remains constant for a short period of time. (a) What is \([\mathrm{A}] 5.00\) min after the reaction is started? (b) At what time after the reaction is started will \([\mathrm{A}]=0.565 \mathrm{M} ?\)

The following statements about catalysis are not stated as carefully as they might be. What slight modifications would you make in them? (a) A catalyst is a substance that speeds up a chemical reaction but does not take part in the reaction. (b) The function of a catalyst is to lower the activation energy for a chemical reaction.

A kinetic study of the reaction \(A \longrightarrow\) products yields the data: \(t=0 \mathrm{s},[\mathrm{A}]=2.00 \mathrm{M} ; 500 \mathrm{s}, 1.00 \mathrm{M}; 1500 \mathrm{s}, 0.50 \mathrm{M} ; 3500 \mathrm{s}, 0.25 \mathrm{M} .\) Without performing detailed calculations, determine the order of this reaction and indicate your method of reasoning.

The reaction \(A \longrightarrow\) products is second order. The initial rate of decomposition of \(A\) when \([\mathrm{A}]_{0}=0.50 \mathrm{M}\) is \((\mathrm{a})\) the same as the initial rate for any other value of \([\mathrm{A}]_{0} ;\) (b) half as great as when \([\mathrm{A}]_{0}=1.00 \mathrm{M} ;(\mathrm{c})\) five times as great as when \([\mathrm{A}]_{0}=[\mathrm{A}]_{0}=0.25 \mathrm{M}.\)

Suppose that the reaction in Example 14-8 is first order with a rate constant of \(0.12 \mathrm{min}^{-1}\). Starting with \([\mathrm{A}]_{0}=1.00 \mathrm{M},\) will the curve for \([\mathrm{A}]\) versus \(t\) for the first-order reaction cross the curve for the second-order reaction at some time after \(t=0 ?\) Will the two curves cross if \([\mathrm{A}]_{0}=2.00 \mathrm{M} ?\) In each case, if the curves are found to cross, at what time will this happen?
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