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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

Diethylhydrazine reacts with iodine according to the following equation: $$\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}(l)+\mathrm{I}_{2}(a q) \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}_{2}(l)+2 \mathrm{HI}(a q)$$ The rate of the reaction is followed by monitoring the disappearance of the purple color due to iodine. The following data are obtained at a certain temperature. $$ \begin{array}{cccc} \hline \text { Expt. } & {\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]} & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\ \hline 1 & 0.150 & 0.250 & 1.08 \times 10^{-4} \\ 2 & 0.150 & 0.3620 & 1.56 \times 10^{-4} \\ 3 & 0.200 & 0.400 & 2.30 \times 10^{-4} \\ 4 & 0.300 & 0.400 & 3.44 \times 10^{-4} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to diethylhydrazine, iodine, and overall? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. (d) What must \(\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]\) be so that the rate of the reaction is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{h}\) when \(\left[\mathrm{I}_{2}\right]=0.500 ?\)

Express the rate of the reaction $$2 \mathrm{~N}_{2} \mathrm{H}_{4}(l)+\mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)$$ in terms of (a) \(\Delta\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]\) (b) \(\Delta\left[\mathrm{N}_{2}\right]\)

The decomposition of sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), to sulfur dioxide and chlorine gases is a first-order reaction. $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ At a certain temperature, the half-life of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(7.5 \times 10^{2} \mathrm{~min}\). Consider a sealed flask with \(122.0 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (a) How long will it take to reduce the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in the sealed flask to \(45.0 \mathrm{~g}\) ? (b) If the decomposition is stopped after \(29.0 \mathrm{~h}\), what volume of \(\mathrm{Cl}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(1.00\) atm is produced?

In the first-order decomposition of acetone at \(500^{\circ} \mathrm{C}\), $$\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{3}(g) \longrightarrow \text { products }$$ it is found that the concentration is \(0.0300 \mathrm{M}\) after \(200 \mathrm{~min}\) and \(0.0200 \mathrm{M}\) after 400 min. Calculate the following. (a) the rate constant (b) the half-life (c) the initial concentration

Express the rate of the reaction $$2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ in terms of (a) \(\Delta\left[\mathrm{C}_{2} \mathrm{H}_{6}\right]\) (b) \(\Delta\left[\mathrm{CO}_{2}\right]\)
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