The following data were measured for the reaction $\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{F}_{3} \mathrm{BNH}_{3}(g):$ $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{BF}_{3}\right](M)} & {\left[\mathrm{NH}_{3}\right](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.250 & 0.250 & 0.2130 \\ 2 & 0.250 & 0.125 & 0.1065 \\ 3 & 0.200 & 0.100 & 0.0682 \\ 4 & 0.350 & 0.100 & 0.1193 \\ 5 & 0.175 & 0.100 & 0.0596 \\ \hline \end{array} $$ (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when \(\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?\)

Step by step solution

01

Write the general form of the rate law

The general form of the rate law for the given reaction can be written as: Rate = k[BF3]^m[NH3]^n Where k is the rate constant, m and n are the orders of the reaction with respect to BF3 and NH3, respectively.

02

Analyze the effect of varying [BF3] on the initial rate

Compare experiments 3 and 5, where the concentration of NH3 is constant at 0.100 M. We can find the order of reaction with respect to BF3 by setting up the following ratio: \(\frac{\text{Rate}_{5}}{\text{Rate}_{3}} = \frac{k[BF3_{5}]^m[NH3_{5}]^n}{k[BF3_{3}]^m[NH3_{3}]^n}\) Substituting the given data into the equation: \(\frac{0.0596}{0.0682} = \frac{k(0.175)^m(0.100)^n}{k(0.200)^m(0.100)^n}\) Since the concentration of NH3 is constant, it can be canceled out: \(\frac{0.0596}{0.0682} = \frac{(0.175)^m}{(0.200)^m}\) Now, solve for the value of m: \(m = \log_{\frac{0.175}{0.200}}\left(\frac{0.0596}{0.0682}\right) = -1\) So, the order of reaction with respect to BF3 is -1.

03

Analyze the effect of varying [NH3] on the initial rate

Compare experiments 2 and 3, where the concentration of BF3 is constant at 0.250 M. We can find the order of reaction with respect to NH3 by setting up the following ratio: \(\frac{\text{Rate}_{3}}{\text{Rate}_{2}} = \frac{k[BF3_{3}]^m[NH3_{3}]^n}{k[BF3_{2}]^m[NH3_{2}]^n}\) Substituting the given data into the equation: \(\frac{0.0682}{0.1065} = \frac{k(0.200)^m(0.100)^n}{k(0.250)^m(0.125)^n}\) Since we found that m = -1, you can simplify the equation: \(\frac{0.0682}{0.1065} = \frac{(0.100)^n}{(0.125)^n}\) Now, solve for the value of n: \(n = \log_{\frac{0.100}{0.125}}\left(\frac{0.0682}{0.1065}\right) = 1\) So, the order of reaction with respect to NH3 is 1.

04

Determine the rate law, overall order and calculate the rate constant

We have found the order of reaction with respect to BF3 to be -1 and with respect to NH3 to be 1. Therefore, the rate law is: Rate = k[BF3]^{-1}[NH3] The overall order of reaction = (-1) + 1 = 0. Now, to find the rate constant (k), substitute any experiment's data into the rate law and solve for k: \(0.2130 = k(0.250)^{-1}(0.250)\) \(k \approx 213 \, \text{M}^{-1}\, \text{s}^{-1}\)

05

Calculate the rate for the given concentrations of reactants

Given, [BF3] = 0.100 M and [NH3] = 0.500 M. Substitute the values into the rate law equation and calculate the rate: Rate = \(213 \, \text{M}^{-1}\, \text{s}^{-1}(0.100)^{-1}(0.500)\) Rate = 1.065 M/s Summary of the answers: (a) Rate law: Rate = k[BF3]^{-1}[NH3] (b) Overall order of the reaction: 0 (c) Rate constant, k = 213 M⁻¹s⁻¹ (d) Rate = 1.065 M/s when [BF3] = 0.100 M and [NH3] = 0.500 M

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