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Chapter 14: Problem 37

Consider the gas-phase reaction between nitric oxide and bromine at $273^{\circ} \mathrm{C}: 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{NOBr}(g)$ The following data for the initial rate of appearance of NOBr were obtained: $$ \begin{array}{lccc} \hline \text { Experiment } & {[\mathrm{N} 0](M)} & {\left[\mathrm{Br}_{2}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.10 & 0.20 & 24 \\ 2 & 0.25 & 0.20 & 150 \\ 3 & 0.10 & 0.50 & 60 \\ 4 & 0.35 & 0.50 & 735 \\ \hline \end{array} $$ (a) Determine the rate law. (b) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets. (c) How is the rate of appearance of NOBr related to the rate of disappearance of \(\mathrm{Br}_{2} ?(\mathbf{d})\) What is the rate of disappearance of \(\mathrm{Br}_{2}\) when \([\mathrm{NO}]=0.075 M\) and \(\left[\mathrm{Br}_{2}\right]=0.25 M ?\)

Short Answer

Expert verified
In summary, the rate law for the given reaction is Rate = \(k[\mathrm{NO}]^2[\mathrm{Br}_2]\), with an average rate constant k of 1200 M^{-2}s^{-1}. The rate of disappearance of Br2 is -0.5 times the rate of appearance of NOBr. Finally, when [NO] = 0.075 M and [Br2] = 0.25 M, the rate of disappearance of Br2 is approximately -8.4375 M/s.

Step by step solution

01

Determine the order with respect to NO and Br2

We will compare Experiments 1 and 2 to find the order with respect to NO. In Experiment 1, [NO] = 0.10 M and the initial rate is 24 M/s. In Experiment 2, [NO] = 0.25 M and the initial rate is 150 M/s. Hence, we have: \( \frac{150}{24} = \frac{0.25^n}{0.10^n} \) Solving for n with trial and error, we find its value to be 2 (approximately). Thus, the reaction is second order with respect to NO. Next, we will compare Experiments 1 and 3 to find the order with respect to Br2. In Experiment 1, [Br2] = 0.20 M and the initial rate is 24 M/s. In Experiment 3, [Br2] = 0.50 M and the initial rate is 60 M/s. Now we get: \( \frac{60}{24} = \frac{0.50^m}{0.20^m} \) Solving for m with trial and error, we find its value to be 1 (approximately). Thus, the reaction is first order with respect to Br2.
02

Determine the rate law

Now that we know the orders for NO and Br2, we can determine the rate law. The rate law is given by: Rate = \(k[\mathrm{NO}]^2[\mathrm{Br}_2]\)
03

Calculate the average rate constant

Using the rate law from Step 2, we can solve for the rate constant, k, for each data set. Experiment 1: \( k = \frac{24}{(0.10)^2(0.20)} = 1200 \, M^{-2}s^{-1} \) Experiment 2: \( k = \frac{150}{(0.25)^2(0.20)} = 1200 \, M^{-2}s^{-1} \) Experiment 3: \( k = \frac{60}{(0.10)^2(0.50)} = 1200 \, M^{-2}s^{-1} \) Experiment 4: \( k = \frac{735}{(0.35)^2(0.50)} \approx 1200 \, M^{-2}s^{-1} \) The average rate constant is 1200 M^{-2}s^{-1}.
04

Relate the rate of appearance of NOBr to the rate of disappearance of Br2

Looking at the balanced equation, we can see that for every 2 moles of NOBr formed, 1 mole of Br2 is consumed. Hence, we can write, Rate of disappearance of Br2 = -0.5 × Rate of appearance of NOBr
05

Calculate the rate of disappearance of Br2 for given concentrations

Now, we can use the rate law, average rate constant, and the given concentrations of NO and Br2 to find the rate of disappearance of Br2. Rate of appearance of NOBr = 1200 × \( (0.075)^2 \) × (0.25) ≈ 16.875 M/s Now using the relationship derived in Step 4, Rate of disappearance of Br2 = -0.5 × 16.875 ≈ -8.4375 M/s.

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Most popular questions from this chapter

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-}\). This rapid reaction gives the following rate data: $$ \begin{array}{lll} \hline\left[\mathrm{OCl}^{-}\right](M) & {\left[I^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\)

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2} .\) The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5} .\) At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?\) (c) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M} ?\) (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to \(0.0120 \mathrm{M} ?\)

Sketch a graph for the generic first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) that has concentration of \(\mathrm{A}\) on the vertical axis and time on the horizontal axis. (a) Is this graph linear? Explain. (b) Indicate on your graph the half-life for the reaction.

(a) Explain the importance of enzymes in biological systems. (b) What chemical transformations are catalyzed (i) by the enzyme catalase, \((i i)\) by nitrogenase? (c) Many enzymes follow this generic reaction mechanism, where \(\mathrm{E}\) is enzyme, \(\mathrm{S}\) is substrate, ES is the enzyme-substrate complex (where the substrate is bound to the enzyme's active site), and \(\mathrm{P}\) is the product: 1\. \(\mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES}\) 2\. \(\mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P}\) What assumptions are made in this model with regard to the rate of the bound substrate being chemically transformed into bound product in the active site?

The rate of the reaction $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q)+\mathrm{OH}^{-}(a q) & \longrightarrow \\ \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \end{aligned} $$ was measured at several temperatures, and the following data were collected: $$ \begin{array}{ll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \boldsymbol{k}\left(\boldsymbol{M}^{-1} \mathrm{~s}^{-1}\right) \\ \hline 15 & 0.0521 \\ 25 & 0.101 \\ 35 & 0.184 \\ 45 & 0.332 \\ \hline \end{array} $$ Calculate the value of \(E_{a}\) by constructing an appropriate graph.
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