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Chapter 12: Problem 56

The rate law for the reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ at some temperature is $$ \text {Rate} =-\frac{\Delta[\text { NOBr }]}{\Delta t}=k[\mathrm{NOBr}]^{2} $$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to 0.100\(M ?\)

Short Answer

Expert verified
a. The rate constant 'k' for the reaction is 0.5556 M\(^{-1}\)s\(^{-1}\). b. The time required for the concentration of NOBr to decrease to 0.100 M is 14.19 seconds.

Step by step solution

01

Identify the reaction order and equations

The given rate law equation relates the rate of the reaction to the concentration of NOBr as follows: Rate = \(−\frac{\Delta[\text{NOBr}]}{\Delta t}\) = \(k[\text{NOBr}]^2\) This is a second order reaction, as the concentration of NOBr is raised to the power of 2. For a second order reaction, the equation relating half-life to the rate constant and initial concentration is: \( t_{1/2} = \frac{1}{k[\text{NOBr}_0]} \)
02

Calculate the value of k

We are given that the half-life of the reaction is 2.00 s and the initial concentration of NOBr is 0.900 M. Let's plug these values into the half-life equation to calculate the rate constant 'k'. \( 2.00 = \frac{1}{k(0.900)} \) Now, solve for k: \( k = \frac{1}{2.00 × 0.900} = \frac{1}{1.80} \) \( k = 0.5556\) M\(^{-1}\)s\(^{-1}\) The rate constant 'k' for this reaction is 0.5556 M\(^{-1}\)s\(^{-1}\).
03

Determine the time for NOBr concentration to decrease to 0.100 M

Now that we have the rate constant, we can use the second-order integrated rate law to find the time required for the concentration of NOBr to decrease to 0.100 M. The second-order integrated rate law equation is: \( \frac{1}{[\text{NOBr}]} - \frac{1}{[\text{NOBr}_0]} = kt \) Let's plug in the known values: \( \frac{1}{0.100} - \frac{1}{0.900} = 0.5556 \times t \) Solve for 't': \( 9.00 - 1.11 = 0.5556t \) \( t = \frac{7.89}{0.5556} = 14.19 \, \text{s} \) The time required for the concentration of NOBr to decrease to 0.100 M is 14.19 seconds.

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Most popular questions from this chapter

The rate law of a reaction can only be determined from experiment. Two experimental procedures for determining rate laws were outlined in Chapter 12. What are the two procedures and how are they used to determine the rate laws?

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $2.50 \times 10^{-3} M ?$

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. $\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)$ c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of \(\Delta \mathrm{E}\) . c. The half-life for a reaction, aA \(\longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0} .\) d. The half-life for a second-order reaction increases as the reaction proceeds.

Consider the following initial rate data for the decomposition of compound AB to give A and B: Determine the half-life for the decomposition reaction initially having 1.00$M \mathrm{AB}$ present.
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