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Chapter 14: Problem 21

In the first-order reaction \(A \longrightarrow\) products, it is found that \(99 \%\) of the original amount of reactant \(A\) decomposes in 137 min. What is the half-life, \(t_{1 / 2}\), of this decomposition reaction?

Short Answer

Expert verified
The half-life for this first order decomposition reaction is \(t_{1/2} = \frac{0.693 * 137}{ln(100)}\) minutes

Step by step solution

01

Find Rate Constant k

First, calculate the rate constant, k, using the equation for first order reactions \(k = \frac{ln[N_0 / N]}{t}\). Here, [N_0] is the initial concentration and [N] the final concentration. Since it is stated that 99% of A has decomposed, the remaining is 1%. So, [N] = 0.01[N_0]. This simplifies to \(k = \frac{ln(100)}{137}\)
02

Calculate Half Life

Substitute k into the formula for half-life of a first-order reaction \(t_{1/2} = \frac{0.693}{k}\). After substituting you will get \(t_{1/2} = \frac{0.693}{\frac{ln(100)}{137}}\)
03

Simplify the expression

Finally, simplify the expression to find \(t_{1/2}\). The final result after simplification will be the half life time of the reaction.

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Most popular questions from this chapter

In your own words, define or explain the following terms or symbols: (a) \([\mathrm{A}]_{0} ;\) (b) \(\dot{k} ;\) (c) \(t_{1 / 2} ;\) (d) zeroorder reaction; (e) catalyst.

For the reaction \(A \longrightarrow\) products, the data tabulated below are obtained. (a) Determine the initial rate of reaction (that is, \(-\Delta[\mathrm{A}] / \Delta t)\) in each of the two experiments. (b) Determine the order of the reaction. $$\begin{array}{ll} \hline \text { First Experiment } & \\ \hline[\mathrm{A}]=1.512 \mathrm{M} & t=0 \mathrm{min} \\ \begin{array}{l} | \mathrm{A}\rfloor=1.490 \mathrm{M} \\ {[\mathrm{A}]=1.469 \mathrm{M}} \end{array} & \begin{array}{l} t=1.0 \mathrm{min} \\ t=2.0 \mathrm{min} \end{array} \\ \hline & \\ \hline \text { Second Experiment } & \\ \hline[\mathrm{A}]=3.024 \mathrm{M} & t=0 \mathrm{min} \\ {[\mathrm{A}]=2.935 \mathrm{M}} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=2.852 \mathrm{M}} & t=2.0 \mathrm{min} \\ \hline \end{array}$$

Suppose that the reaction in Example 14-8 is first order with a rate constant of \(0.12 \mathrm{min}^{-1}\). Starting with \([\mathrm{A}]_{0}=1.00 \mathrm{M},\) will the curve for \([\mathrm{A}]\) versus \(t\) for the first-order reaction cross the curve for the second-order reaction at some time after \(t=0 ?\) Will the two curves cross if \([\mathrm{A}]_{0}=2.00 \mathrm{M} ?\) In each case, if the curves are found to cross, at what time will this happen?

Explain the important distinctions between each pair of terms: (a) first-order and second-order reactions; (b) rate law and integrated rate law; (c) activation energy and enthalpy of reaction; (d) elementary process and overall reaction; (e) enzyme and substrate.

One proposed mechanism for the formation of a double helix in DNA is given by $$\left(S_{1}+S_{2}\right)=\left(S_{1}: S_{2}\right)^{*} \quad \text { (fast) }$$ $$\left(S_{1}: S_{2}\right)^{*} \longrightarrow S_{1}: S_{2} \quad \text { (slow) }$$ where \(S_{1}\) and \(S_{2}\) represent strand 1 and \(2,\) and \(\left(S_{1}: S_{2}\right)^{*}\) represents an unstable helix. Write the rate of reaction expression for the formation of the double helix.
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