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Chapter 14: Problem 95

The dimerization of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) to $\mathrm{C}_{4} \mathrm{~F}_{8}\( has a rate constant \)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 in C2F4 is second-order, as the rate constant, k, has units of \(M^{-1} . s^{-1}\). To calculate the time it takes for the concentration of C2F4 to decrease from 0.100 M to 0.020 M at 450 K, the integrated rate law for a second-order reaction is used: \(\frac{1}{[A]_{t}}-\frac{1}{[A]_{0}} = kt\). By plugging in the given values, the time (t) required is approximately 888.89 seconds.

Step by step solution

01

Determine the reaction order

We know that the units of the rate constant vary depending on the reaction order. The reaction orders and their corresponding rate constant units are as follows: - Zero-order: \(M.s^{-1}\) - First-order: \(s^{-1}\) - Second-order: \(M^{-1}.s^{-1}\) The given rate constant, k has the unit \(M^{-1} . s^{-1}\). Comparing this unit to the units listed above, we can determine that the reaction order in C2F4 is second-order.
02

Write the integrated rate law

Since we know that the reaction is second-order, we can write the integrated rate law for a second-order reaction: \[\frac{1}{[A]_{t}}-\frac{1}{[A]_{0}} = kt\] Here, \([A]_{0}\) represents the initial concentration, \([A]_{t}\) represents the final concentration of C2F4, k is the rate constant, and t is the time.
03

Solve for time

We are given the initial concentration \([A]_{0} = 0.100 M\), the final concentration \([A]_{t} = 0.020 M\), and the rate constant \(k = 0.045 M^{-1}.s^{-1}\). We can plug these values into the integrated rate law equation and solve for time (t): \[\frac{1}{0.020} - \frac{1}{0.100} = 0.045 \cdot t\] Now, calculate t: \(t = \frac{\frac{1}{0.020} - \frac{1}{0.100}}{0.045} \) \(t = \frac{50 - 10}{0.045}\) \(t = \frac{40}{0.045}\) \(t \approx 888.89 s\)
04

Result

The time it takes for the concentration of C2F4 to decrease from 0.100 M to 0.020 M at 450 K is approximately 888.89 seconds.

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Most popular questions from this chapter

The human body is characterized by an extremely complex system of interrelated chemical reactions. A large number of enzymes are necessary for many of these reactions to occur at suitable rates. Enzymes are very selective in the reactions they catalyze, and some are absolutely specific. Use the lock-and- key model to account for the specificity of an enzyme.

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[\mathrm{~A}]^{2} .(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.

Indicate whether each statement is true or false. (a) If you measure the rate constant for a reaction al different temperatures, you can calculate the overall enthalpy change for the reaction. (b) Exothermic reactions are faster than endothermic reactions. (c) If you double the temperature for a reaction, you cut the activation energy in half.

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