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Chapter 11: Problem 55

The rate constant for the second-order reaction $$\mathrm{NOBr}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)$$ is \(48 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{min}\) at a certain temperature. How long will it take to decompose \(90.0 \%\) of a \(0.0200 \mathrm{M}\) solution of nitrosyl bromide?

Short Answer

Expert verified
Based on the given rate constant (k) and initial concentration of the reactant, it will take approximately 9.38 minutes for 90% of the reactant to decompose in a second-order reaction.

Step by step solution

01

Write down the second-order rate law equation

The rate law for a second-order reaction is given by: $$\frac{1}{[A]_{t}}-\frac{1}{[A]_{0}}=kt$$ Where: - \([A]_{t}\) = concentration of the reactant A at time t - \([A]_{0}\) = initial concentration of A - k = rate constant - t = time
02

Find the concentration of the reactant after 90% decomposition

Since 90% of the reactant has decomposed, only 10% is left. Thus, the final concentration of the reactant at time "t" is: $$[A]_{t} = 0.10 \times [A]_{0}$$
03

Substitute the given values in the rate law equation

We are given the initial concentration, \([A]_{0} = 0.0200 \mathrm{M}\), and the rate constant, k = \(48 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{min}\). Substituting these values, along with the expression for \([A]_{t}\), we get: $$\frac{1}{0.10 \times 0.0200}-\frac{1}{0.0200}= 48\times t$$
04

Solve for t (time)

Now, we need to solve this equation for time. First, calculate the fractions: $$\frac{1}{0.002}-\frac{1}{0.0200}= 48\times t$$ Next, subtract and solve for t: $$500-50= 48\times t$$ $$450 = 48 \times t$$ $$t= \frac{450}{48} \mathrm{min}$$ $$t \approx 9.38 \mathrm{min}$$ Therefore, it will take approximately 9.38 minutes to decompose 90% of the 0.0200 M solution of nitrosyl bromide.

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Most popular questions from this chapter

The equation for the iodination of acetone in acidic solution is $$\mathrm{CH}_{3} \mathrm{COCH}_{3}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q)$$ The rate of the reaction is found to be dependent not only on the concentration of the reactants but also on the hydrogen ion concentration. Hence the rate expression of this reaction is $$\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]^{m}\left[\mathrm{I}_{2}\right]^{n}\left[\mathrm{H}^{+}\right]^{p}$$ The rate is obtained by following the disappearance of iodine using starch as an indicator. The following data are obtained: $$ \begin{array}{cccc} \hline\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right] & \left.\mathrm{[H}^{+}\right] & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 0.80 & 0.20 & 0.001 & 4.2 \times 10^{-6} \\ 1.6 & 0.20 & 0.001 & 8.2 \times 10^{-6} \\ 0.80 & 0.40 & 0.001 & 8.7 \times 10^{-6} \\ 0.80 & 0.20 & 0.0005 & 4.3 \times 10^{-6} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to each reactant? (b) Write the rate expression for the reaction. (c) Calculate \(k\). (d) What is the rate of the reaction when \(\left[\mathrm{H}^{+}\right]=0.933 M\) and \(\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]=3\left[\mathrm{H}^{+}\right]=10\left[\mathrm{I}^{-}\right] ?\)

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) hydrolyzes into glucose and fructose. The hydrolysis is a first-order reaction. The half-life for the hydrolysis of sucrose is \(64.2\) min at \(25^{\circ} \mathrm{C}\). How many grams of sucrose in \(1.25 \mathrm{~L}\) of a \(0.389 \mathrm{M}\) solution are hydrolyzed in \(1.73\) hours?

Argon- 41 is used to measure the rate of gas flow. It has a decay constant of \(6.3 \times 10^{-3} \mathrm{~min}^{-1}\). (a) What is its half-life? (b) How long will it take before only \(1.00 \%\) of the original amount of Ar- 41 is left?

The decomposition of phosphine, \(\mathrm{PH}_{3}\), to \(\mathrm{P}_{4}(g)\) and \(\mathrm{H}_{2}(g)\) is firstorder. Its rate constant at a certain temperature is \(1.1 \mathrm{~min}^{-1}\). (a) What is its half-life in seconds? (b) What percentage of phosphine is decomposed after \(1.25 \mathrm{~min}\) ? (c) How long will it take to decompose one fifth of the phosphine?

WEB When boron trifluoride reacts with ammonia, the following \(T\) reaction occurs: for $$\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{BF}_{3} \mathrm{NH}_{3}(g)$$ The following data are obtained at a particular temperature: $$\begin{array}{cccc}\hline \text { Expt. } & {\left[\mathrm{BF}_{3}\right]} & {\left[\mathrm{NH}_{3}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\\\\hline 1 & 0.100 & 0.100 & 0.0341 \\ 2 & 0.200 & 0.233 & 0.159 \\ 3 & 0.200 & 0.0750 & 0.0512 \\ 4 & 0.300 & 0.100 & 0.102 \\\\\hline\end{array}$$
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