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For the reaction \(A \longrightarrow\) products, the following data give \([\mathrm{A}]\) as a function of time \(t=0 \mathrm{s},[\mathrm{A}]=0.600 \mathrm{M};100 \mathrm{s}, 0.497 \mathrm{M} ; 200 \mathrm{s}, 0.413 \mathrm{M} ; 300 \mathrm{s}, 0.344 \mathrm{M} ; 400 \mathrm{s}\) \(0.285 \mathrm{M} ; 600 \mathrm{s}, 0.198 \mathrm{M} ; 1000 \mathrm{s}, 0.094 \mathrm{M}.\) (a) Show that the reaction is first order. (b) What is the value of the rate constant, \(k ?\) (c) What is \([\mathrm{A}]\) at \(t=750 \mathrm{s} ?\)

Short Answer

Expert verified
Detailed answer will be given after doing calculations. The reaction order is dependent upon the plot of the natural log of concentration versus time, the rate constant (\(k\)) is dependent upon the slope of that plot, and the concentration of A at a particular time is determined using the first order reaction equation with the derived rate constant, and initial concentration.

Step by step solution

01

Determining Reaction Order

Plot the natural logarithm of the concentration, ln([\(\mathrm{A}\])), against time. If a straight line is obtained, it's evidence that the reaction is first order. So for each given time and concentration, calculate ln([\(\mathrm{A}\)]). Then, plot these values and check if the plot is approximately a straight line.
02

Calculate Rate Constant \(k\)

Assuming that the reaction is first order, the slope of the straight line plot will equal to \(-k\). Since, based on the formula for the equation of a line \(y=mx+b\), \(m\) represents the slope of the line. Therefore, \(-k = m\). Calculate the slope of line and take its absolute value to find the value of \(k\).
03

Determining the concentration at a certain time

Use the calculated rate constant \(k\) and the first order reaction equation: \(ln[A]_t = -kt + ln[A]_0 \) to determine concentration of A at \(t=750s\). First substitute the values for \(k\) and the initial concentration \([A]_0\) into the equation. Then solve the equation to find \([A]_t\) at \(t=750s\). The solution should be converted back from a logarithm to a concentration by taking the exponent of the solution: \([A]_t = e^{(ln[A]_t)}\).

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Key Concepts

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

First Order Reaction
In the realm of reaction kinetics, understanding the order of a reaction is crucial in analyzing how the concentration of reactants influences the rate of the reaction. A first order reaction is characterized by its reliance on the concentration of a single reactant. The rate of such a reaction is directly proportional to the concentration of one reactant. This implies that if the concentration of the reactant doubles, the reaction rate will also double.
The identification process involves plotting the natural logarithm of the reactant concentration against time. If this plot results in a straight line, you have evidence that the reaction follows first order kinetics. For instance, in the given exercise, by calculating and plotting the values of ln([A]) at various time intervals, a straight line should emerge, confirming the first order nature of the reaction.
  • First order reaction rate law: \( \mathrm{Rate} = k[A] \)
  • Straight line plot: Indicates ln([A]) vs. time is linear.
Rate Constant
The rate constant, denoted by \( k \), is a crucial figure in the rate law equation. It reflects the proportionality factor linking the reaction rate to the concentration of reactants. For first order reactions, once you verify that the reaction is indeed first order, \( k \) can be directly extracted from the slope of the line obtained in your plot of ln([A]) vs. time.
The slope of this line, represented as \(-k\), will give you the rate constant when its absolute value is taken. Utilizing this slope is essential for calculating how fast the reaction proceeds under given conditions.
  • Rate constant calculation involves measuring the slope from ln([A]) vs. time plot.
  • The slope as \(-k\) indicates the rate at which concentration decreases.
Concentration vs Time
The relationship between concentration and time in first order reactions is beautifully described by the first order integrated rate equation: \( \ln[A]_t = -kt + \ln[A]_0 \). This can be used to calculate the concentration of a reactant at any point in time, given you know the initial concentration and the rate constant.
In the provided exercise, you can determine \([A]_t\) for any future time by substituting the known values of \([A]_0\) and \( k \) into this equation. To find \([A]\) at \( t = 750 s \), the calculation involves determining the ln([A]) at this time and then using the exponential function to convert back from logarithmic to real concentration terms:
  • Equation: \( \ln[A]_t = -kt + \ln[A]_0 \)
  • To find \([A]_t\), convert back using \( [A]_t = e^{(\ln[A]_t)} \).
These calculations are foundational in predicting reactant concentration at any given time during the reaction.

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Most popular questions from this chapter

In the reaction \(A \longrightarrow\) products, at \(t=0\), the \([\mathrm{A}]=0.1565 \mathrm{M} .\) After \(1.00 \mathrm{min},[\mathrm{A}]=0.1498 \mathrm{M},\) and after \(2.00 \mathrm{min},[\mathrm{A}]=0.1433 \mathrm{M}\) (a) Calculate the average rate of the reaction during the first minute and during the second minute. (b) Why are these two rates not equal?

For the reaction \(A+2 B \longrightarrow 2 C\), the rate of reaction is \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) at the time when \([\mathrm{A}]=0.3580 \mathrm{M}.\) (a) What is the rate of formation of \(\mathrm{C}\) ? (b) What will \([\mathrm{A}]\) be 1.00 min later? (c) Assume the rate remains at \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) How long would it take for \([\mathrm{A}]\) to change from 0.3580 to \(0.3500 \mathrm{M} ?\)

The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

The half-lives of both zero-order and second-order reactions depend on the initial concentration, as well as on the rate constant. In one case, the half- life gets longer as the initial concentration increases, and in the other it gets shorter. Which is which, and why isn't the situation the same for both?

In the first-order decomposition of substance A the following concentrations are found at the indicated times: \(t=0 \mathrm{s},[\mathrm{A}]=0.88 \mathrm{M} ; t=50 \mathrm{s},[\mathrm{A}]=0.62 \mathrm{M} ; t=100 \mathrm{s},[\mathrm{A}]=0.44 \mathrm{M} ; t=150 \mathrm{s},[\mathrm{A}]=0.31 \mathrm{M}.\) Calculate the instantaneous rate of decomposition at \(t=100 \mathrm{s}.\)

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