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Chapter 12: Problem 51

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?

Short Answer

Expert verified
The rate constant (k) for this first-order reaction is approximately 0.00899 s⁻¹, and its half-life (t_half) is approximately 77.1 seconds.

Step by step solution

01

Write down the given information

We are given that the reaction is 45.0% complete in 65 seconds. Let's represent the initial concentration of the reactant as [A]_0 and the remaining concentration after 65 seconds as [A]. Since the reaction is 45.0% complete, 55.0% of the reactant is remaining. So, we have the following relationship between [A]_0 and [A]: \[ A = 0.55 * A_0 \] Additionally, we know that the time taken for this change is 65 seconds.
02

Use the first-order reaction formula

For a first-order reaction, the rate law is given by: \[ \ln \frac{[A]}{[A]_0} = -kt \] Where: k is the rate constant, t is the time taken (in seconds), [A] is the remaining concentration of the reactant, [A]_0 is the initial concentration of the reactant. We'll now plug in our known values into the formula to find the rate constant k: \[ \ln \frac{0.55 * [A]_0}{[A]_0} = -k * 65 \]
03

Calculate the rate constant (k)

Now, let's simplify our equation and solve for the rate constant k: \[ \ln (0.55) = -k * 65 \] \[ k = -\frac{\ln (0.55)}{65} \] Using a calculator, we find: \[ k = 0.00899 s^{-1} \] Hence, the rate constant (k) for this reaction is approximately 0.00899 s⁻¹.
04

Calculate the half-life (t_half) using the rate constant

For a first-order reaction, the half-life formula is given by: \[ t_{1/2} = \frac{\ln 2}{k} \] Now, we can plug in the value of the rate constant (k) that we calculated in step 3 to find the half-life (t_half): \[ t_{1/2} = \frac{\ln 2}{0.00899} \] Using a calculator, we find: \[ t_{1/2} ≈ 77.1 s \] Hence, the half-life (t_half) for this reaction is approximately 77.1 seconds. In conclusion, the rate constant (k) for this first-order reaction is approximately 0.00899 s⁻¹, and its half-life (t_half) is approximately 77.1 seconds.

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Most popular questions from this chapter

The decomposition of ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( on an alumina \)\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)$ surface $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ was studied at 600 \(\mathrm{K}\) . Concentration versus time data were collected for this reaction, and a plot of [A] versus time resulted in a straight line with a slope of $-4.00 \times 10^{-5} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}$ . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. If the initial concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) was \(1.25 \times 10^{-2} M\) calculate the half-life for this reaction. c. How much time is required for all the \(1.25 \times 10^{-2} \mathrm{M}\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) to decompose?

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

For enzyme-catalyzed reactions that follow the mechanism $$ \begin{aligned} \mathrm{E}+\mathrm{S} & \rightleftharpoons \mathrm{E} \cdot \mathrm{S} \\ \mathrm{E} \cdot \mathrm{S} & \rightleftharpoons \mathrm{E}+\mathrm{P} \end{aligned} $$ a graph of the rate as a function of [S], the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.

For the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C},\) explain at least two ways in which the rate law could be zero order in chemical A.

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 \(\mathrm{h}\) ?
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