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The dimerization of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) to \(\mathrm{C}_{4} \mathrm{~F}_{\mathrm{s}}\) has a rate constant \(k=0.045 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(450 \mathrm{~K}\). (a) Based on the unit of \(k\), what is the reaction order in \(\mathrm{C}_{2} \mathrm{~F}_{4} ?(\mathbf{b})\) If the initial concentration of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) is \(0.100 \mathrm{M}\), how long would it take for the concentration to decrease to \(0.020 \mathrm{M}\) at \(450 \mathrm{~K} ?\)

Short Answer

Expert verified
The reaction order is second, and it takes about 888.89 seconds for the concentration to decrease from 0.100 M to 0.020 M.

Step by step solution

01

Identify Reaction Order from Rate Constant

The units of the rate constant for a reaction help determine its order. Here, the unit of the rate constant \(k\) is \(\text{M}^{-1}\text{ s}^{-1}\). For a rate constant with units of \(\text{M}^{1-n}\text{ s}^{-1}\), the reaction order \(n\) can be found by setting the expression \(1-n = -1\). Solving this gives \(n = 2\). Thus, the reaction is of second order with respect to \(\mathrm{C}_{2} \mathrm{F}_{4}\).
02

Apply Second Order Kinetics Equation

For a second-order reaction, the integrated rate law is \[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \] where \([A]\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant. Here, \([A]_0 = 0.100\, \text{M}\), \([A] = 0.020\, \text{M}\), and \(k = 0.045\, \text{M}^{-1}\text{ s}^{-1}\).
03

Solve for Time t

Substituting known values into the integrated rate law gives: \[ \frac{1}{0.020} - \frac{1}{0.100} = 0.045 \times t \]. Calculate \(\frac{1}{0.020} = 50\) and \(\frac{1}{0.100} = 10\). The equation becomes: \[ 50 - 10 = 0.045t \]. Simplifying gives \[ 40 = 0.045t \]. Divide both sides by \(0.045\) to solve for \(t\): \[ t = \frac{40}{0.045} \approx 888.89 \] seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Order Reactions
Second order reactions are fascinating because their rates depend on the concentration of one reactant squared or on the products of the concentrations of two different reactants. In this exercise, we examined a second-order reaction involving the dimerization of \( \mathrm{C}_{2} \mathrm{~F}_{4} \).

The unique trait of second-order reactions is their rate equation. For a reaction \( A + B \rightarrow \, \text{products} \), the rate is proportional to \( [A][B] \) or \( [A]^2 \). This means that doubling \([A]\) or \([B]\) will quadruple the reaction rate. In single-reactant reactions, this equation simplifies to a rate proportional to \( [A]^2 \).

This dependency on concentration explains why second-order reactions can initially proceed rapidly but slow as \([A]\) decreases during the reaction. The units of the rate constant \(k\) are crucial for identifying reaction order. In this problem, the rate constant \(k = 0.045 \text{ M}^{-1}\text{ s}^{-1}\) indicates a second-order reaction, aligning with the formula for determining the order from rate constant units.
Rate Constant
The rate constant, typically denoted as \(k\), is a critical part of reaction kinetics. It provides invaluable information about the speed of a reaction and its temperature dependence. In our example, the given rate constant \(k = 0.045 \text{ M}^{-1}\text{ s}^{-1}\) at \(450 \text{ K}\) tells us not only about the reaction speed but also confirms the order of the reaction.

The units of \(k\) matter significantly: they differ based on whether a reaction is zeroth, first, or second order. For second-order reactions, \(k\) typically has units of \( \text{M}^{-1}\text{ s}^{-1} \). This unit guides us in factoring in concentration and how it impacts reaction speed.

A key point to remember is that while the rate constant helps identify reaction order, it also adjusts based on temperature changes. Understanding this allows one to calculate how long a reaction takes under certain temperature conditions, as we see in the ambiguity of reactions occurring at various conditions.
Integrated Rate Law
Integrated rate laws are pivotal for predicting concentrations over time in reactions. These laws allow us to compute the time required for a reactant concentration to change or for a reaction to reach a certain completion level.

For a second-order reaction, the integrated rate law is expressed as:
  • \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \)
Here, \([A]\) is the concentration at time \(t\), \([A]_0\) the initial concentration, and \(k\) the rate constant.

In our exercise, we applied this law successfully to determine how long it takes for \([C_2F_4]\) to decrease from \(0.100 \text{ M}\) to \(0.020 \text{ M}\). By knowing \([A]_0\), final \([A]\), and \(k\), you can solve for \(t\), revealing the reaction time as approximately 888.89 seconds.

Integrated rate laws thus are excellent predictive tools, enabling scientists and students alike to linearize the intricate dynamics of chemical reactions and calculate essential parameters like time and concentration at future points.

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Most popular questions from this chapter

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are \(154 \mathrm{~kJ} / \mathrm{mol}\) and \(136 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

Consider the following reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{Cl}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\right]=4.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=2.5 \times 10^{-2} \mathrm{M},\) the reaction rate at 310 \(\mathrm{K}\) is \(5.20 \times 10^{-2} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant?

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) \(25^{\circ} \mathrm{C},(\mathbf{b}) 125^{\circ} \mathrm{C} ?\)

In a hydrocarbon solution, the gold compound \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) decomposes into ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and a different gold compound, \(\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3} .\) The following mechanism has been proposed for the decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}:\) Step 1: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \stackrel{\mathrm{k}_{\mathrm{L}}}{2}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3}\) Step 2: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{h}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad\) (slow) Step 3: \(\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{\mathrm{k}_{\mathrm{j}}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}\) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} ?\)

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

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