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Chapter 12: Problem 104

A certain substance, initially present at \(0.0800 M,\) decomposes by zero-order kinetics with a rate constant of $2.50 \times 10^{-2} \mathrm{mol} / \mathrm{L}$ . s. Calculate the time (in seconds required for the system to reach a concentration of 0.0210\(M .\)

Short Answer

Expert verified
The time required for the system to reach a concentration of 0.0210 M is 2360 seconds.

Step by step solution

01

Write down the given information

We are given the following information: - Initial concentration \([A]_0 = 0.0800 \mathrm{M}\) - Rate constant \(k = 2.50 \times 10^{-2} \mathrm{mol}/\mathrm{L} \cdot s\) - Final concentration \([A] = 0.0210 \mathrm{M}\)
02

Use the integrated rate law equation for zero-order reactions

The integrated rate law for a zero-order reaction is given by the following formula: \[[A] = [A]_0 - kt\] We need to solve this equation for the time \(t\).
03

Plug in the given values and solve for time \(t\)

Substitute the given concentrations and rate constant into the equation: \begin{align*} 0.0210 &= 0.0800 - (2.50 \times 10^{-2})t \\ \end{align*} Now, solve for \(t\): \begin{align*} (2.50 \times 10^{-2})t &= 0.0800 - 0.0210 \\ t &= \frac{0.0800 - 0.0210}{2.50 \times 10^{-2}} \\ t &= \frac{0.0590}{2.50 \times 10^{-2}} \\ t &= 2360\,\mathrm{s} \end{align*}
04

Present the result

The time required for the system to reach a concentration of 0.0210 M is 2360 seconds.

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Most popular questions from this chapter

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M,\) con- centration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{min}^{-1}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $2.50 \times 10^{-3} M ?$

Rate Laws from Experimental Data: Initial Rates Method. The reaction $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$$ was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained where $$\text { Rate }=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t}$$ $$ \begin{array}{ccc} {[\mathrm{NO}]_{0}} & {\left[\mathrm{Cl}_{2}\right]_{0}} & \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ 0.10 & 0.10 & 0.18 \\ 0.10 & 0.20 & 0.36 \\ 0.20 & 0.20 & 1.45 \end{array} $$ a. What is the rate law? b. What is the value of the rate constant?

A reaction of the form $$ \mathrm{aA} \longrightarrow $$ 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.

Upon dissolving \(\operatorname{In} \mathrm{Cl}(s)\) 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 dis- solving 2.38 \(\mathrm{g} \operatorname{InCl}(s)\) in dilute \(\mathrm{HCl}\) to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution? What mass of In \((s)\) is formed after 1.25 \(\mathrm{h}\) ?

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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