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Chapter 11: Problem 26

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: $$\begin{array}{cccc}\hline \text { Expt. } & {\left[\mathrm{BF}_{3}\right]} & {\left[\mathrm{NH}_{3}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\\\\hline 1 & 0.100 & 0.100 & 0.0341 \\ 2 & 0.200 & 0.233 & 0.159 \\ 3 & 0.200 & 0.0750 & 0.0512 \\ 4 & 0.300 & 0.100 & 0.102 \\\\\hline\end{array}$$

Short Answer

Expert verified
Based on the given experimental data, the rate law for the reaction between boron trifluoride (BF3) and ammonia (NH3) is: $$Rate = (3.41 \ \mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1})([\mathrm{BF}_{3}])([\mathrm{NH}_{3}])$$ with both reactants having a reaction order of 1 and the rate constant (k) being 3.41 L·mol⁻¹·s⁻¹.

Step by step solution

01

Find the reaction order with respect to the reactants

Compare Experiments 1 and 3 to find the reaction order for BF3: Initial rate exp. 3 = \(\frac {1}{2}\) (Initial rate exp. 1) \(\frac{0.0512}{0.0341} = \frac{2^x}{1^x} \implies 1.5 = 2^x \implies x = 0.58496\) We can approximate this value as the reaction order for BF3 to 1. Now, let's find the reaction order for NH3 by comparing Experiments 1 and 4: Initial rate exp. 4 = \(3\) (Initial rate exp. 1) \(\frac{0.102}{0.0341} = \frac{1^x}{3^y} \implies 2.99 \approx 3^y\) We can approximate the reaction order for NH3 to 1 as well.
02

Calculate the Rate Constant

Now that we have both reaction orders, let's use any of the given experiment data to determine the rate constant (k): Using Experiment 1 data to calculate the rate constant: $$Rate = k[\mathrm{BF}_{3}]^1[\mathrm{NH}_{3}]^1$$ $$0.0341 = k(0.100)(0.100) \implies k = 3.41 \ \mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}$$
03

Write the Rate Law based on Reaction Orders and Rate Constant

We will now write the rate law using the reaction orders for both reactants and the rate constant k: $$Rate = (3.41 \ \mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1})([\mathrm{BF}_{3}])([\mathrm{NH}_{3}])$$ This expression represents the rate law for the given reaction.

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Most popular questions from this chapter

Nitrosyl bromide (NOBr) decomposes to nitrogen oxide and bromine. Use the following data to determine the order of the decomposition reaction of nitrosyl bromide. $$\begin{array}{cccccc}\hline \text { Time (s) } & 0 & 6 & 12 & 18 & 24 \\ \text { [NOBr] } & 0.0286 & 0.0253 & 0.0229 & 0.0208 & 0.0190 \\\\\hline\end{array}$$

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?

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

A reaction has two reactants \(\mathrm{A}\) and \(\mathrm{B}\). What is the order with respect to each reactant and the overall order of the reaction described by each of the following rate expressions? (a) rate \(=k_{1}[\mathrm{~A}]^{3}\) (b) rate \(=k_{2}[\mathrm{~A}] \times[\mathrm{B}]\) (c) rate \(=k_{3}[\mathrm{~A}] \times[\mathrm{B}]^{2}\) (d) rate \(=k_{4}[\mathrm{~B}]\)

For the reaction between hydrogen and iodine, $$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})$$ the experimental rate expression is rate \(=k\left[\mathrm{H}_{2}\right] \times\left[\mathrm{I}_{2}\right] .\) Show that this expression is consistent with the mechanism $$\begin{array}{cc}\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) & \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) & \text { (slow) }\end{array}$$
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