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Chapter 11: Problem 97

A reaction of the form $$aA \longrightarrow Products$$gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach \(22.9 \%\) completion.

Short Answer

Expert verified
The time required for the reaction to reach \(22.9\%\) completion is \(36.1 \,\mathrm{s}\).

Step by step solution

01

Calculate the reaction rate constant

The slope of the given plot equals the negative value of the reaction rate constant, so: \[k = -(-7.35 \times 10^{-3})\] \[k = 7.35 \times 10^{-3} \,\mathrm{s^{-1}}\]
02

Determine the concentration of A at 22.9% completion

To find the concentration of A at \(22.9\%\) completion, we'll multiply the initial concentration with the remaining percentage: \[[\mathrm{A}] = [\mathrm{A}]_{0} \times \left(1 - \frac{22.9}{100}\right)\] \[[\mathrm{A}] = 0.0100 \,\mathrm{M} \times \left(1- 0.229\right)\] \[[\mathrm{A}] = 0.0100 \,\mathrm{M} \times 0.771\] \[[\mathrm{A}] = 0.00771 \,\mathrm{M}\]
03

Calculate the time required for 22.9% completion

Now, we can insert the values of \(k, [\mathrm{A}]_{0}\), and \([\mathrm{A}]\) into the formula for the first-order reaction, and solve for the time, \(t\): \[\ln\frac{[\mathrm{A}]_{0}}{[\mathrm{A}]} = kt\] \[\ln\frac{0.0100}{0.00771} = (7.35 \times 10^{-3}) t\] \[0.265 = 7.35 \times 10^{-3}t\] Now we'll solve for \(t\): \[t = \frac{0.265}{7.35 \times 10^{-3}}\] \[t = 36.1 \,\mathrm{s}\] So, it takes 36.1 seconds for the reaction to reach \(22.9\%\) completion.

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Most popular questions from this chapter

Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \operatorname{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of \(\operatorname{In}^{+}(a q)\) after \(1.25 \mathrm{h}\) if the initial solution of \(\operatorname{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{g}\) InCl \((s)\) in dilute HCl to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{h} ?\)

What are the units for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

Consider the following statements: "In general, the rate of a chemical reaction increases a bit at first because it takes a while for the reaction to get "warmed up.' After that, however, the rate of the reaction decreases because its rate is dependent on the concentrations of the reactants, and these are decreasing." Indicate everything that is correct in these statements, and indicate everything that is incorrect. Correct the incorrect statements and explain.

Draw a rough sketch of the energy profile for each of the following cases: a. \(\Delta E=+10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{kJ} / \mathrm{mol}\) b. \(\Delta E=-10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\) c. \(\Delta E=-50 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\)

A certain first-order reaction is \(45.0 \%\) complete in 65 s. What are the values of the rate constant and the half-life for this process?
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