/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 The following data were measured... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 31

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{tabular}{lccl} \hline Experiment & {\(\left[\mathrm{BF}_{3}\right](M)\)} & {\(\left[\mathrm{NH}_{3}\right](M)\)} & 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{tabular} (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) What is the value of the rate constant for the reaction? (d) What is the rate when \(\left[\mathrm{BF}_{3}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{NH}_{3}\right]=0.500 \mathrm{M} ?\)

Short Answer

Expert verified
a) The rate law for the reaction is: \(R = k[\mathrm{BF_3}]^{1}[\mathrm{NH_3}]^{2}\) b) The overall order of the reaction is: (1+2) = 3 c) The value of the rate constant for the reaction is: 6.8 M^{-2}s^{-1} d) The rate when \([\mathrm{BF_3}]=0.100 \mathrm{M}\) and \([\mathrm{NH_3}]=0.500 \mathrm{M}\) is approximately 0.850 M/s.

Step by step solution

01

Analyzing initial rates to find reaction orders

To find the individual orders, we will observe how the concentration changes affect the initial rate of the reaction. Let the rate law be: \[R = k[\mathrm{BF_3}]^{m}[\mathrm{NH_3}]^{n}\] We will now compare Experiments 1 and 2 to find the order with respect to \(\mathrm{NH_3}\): \( R_1 = k[\mathrm{BF_3}]^{m}[\mathrm{NH_3}]^{n}\) \( R_2 = k[\mathrm{BF_3}]^{m}[\mathrm{NH_3}]^{n/2}\) \(\frac{R_1}{R_2} = \frac{0.2130}{0.1065} = \frac{[\mathrm{NH_3}]^n}{[\mathrm{NH_3}]^{n/2}}\)
02

Solving for the order with respect to \(\mathrm{NH_3}\)

\(\frac{2 R_1}{R_2} = [\mathrm{NH_3}]^{n/2}\) \( 2 = \mathrm{NH_3}^{n/2}\) \(n = 2\) So, the order of the reaction with respect to \(\mathrm{NH_3}\) is 2. Now let's compare Experiments 2 and 3 to find the order with respect to \(\mathrm{BF_3}\): \(\frac{R_2}{R_3} = \frac{[\mathrm{BF_3}]^{m/5/4}}{[\mathrm{NH_3}]^{n/5}}\)
03

Solving for the order with respect to \(\mathrm{BF_3}\)

\(\frac{R_2}{R_3}\frac{[\mathrm{NH_3}]^{n/5}}{[\mathrm{BF_3}]^{m/5/4}} = 1\) \(m = 1\) So, the order of the reaction with respect to \(\mathrm{BF_3}\) is 1.
04

Writing rate law

Now that we have the orders for both reactants, we can write the rate law: \[R = k[\mathrm{BF_3}]^{1}[\mathrm{NH_3}]^{2}\]
05

Finding the rate constant (k)

We can use any of the experimental data to find the rate constant. Let's use Experiment 1: \[0.2130 = k(0.250)^1(0.250)^2\] Solving for k, we get: \(k \approx 6.8\)
06

Answering part (d) - Calculate the rate with new concentrations

To calculate the rate with given concentrations, we will use the rate law and rate constant: \[R = 6.8[\mathrm{BF_3}]^{1}[\mathrm{NH_3}]^{2}\] \[ R = 6.8 \times (0.100) \times (0.500)^{2}\] \( R \approx 0.850\) #Summary# Here are the answers for each part of the problem: a) The rate law for the reaction is: \(R = k[\mathrm{BF_3}]^{1}[\mathrm{NH_3}]^{2}\) b) The overall order of the reaction is: (1+2) = 3 c) The value of the rate constant for the reaction is: 6.8 M^{-2}s^{-1} d) The rate when \([\mathrm{BF_3}]=0.100 \mathrm{M}\) and \([\mathrm{NH_3}]=0.500 \mathrm{M}\) is approximately 0.850 M/s.

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Most popular questions from this chapter

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C}\), (b) \(125^{\circ} \mathrm{C}\) ?

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) \(\longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(600 \mathrm{~K}\) the half-life for this process is \(2.3 \times 10^{5} \mathrm{~s}\). What is the rate constant at this temperature? (b) At \(320^{\circ} \mathrm{C}\) the rate constant is \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\). What is the half-life at this temperature?

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? Explain your answer. (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\), is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C}\), \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) .\) If the reac- tion is first order with a half-life of \(56.3 \mathrm{~min}\) at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).
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