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Chapter 8: Problem 96

If \(60 \%\) of a first order reaction was completed in 60 minutes, \(50 \%\) of the same reaction would be completed in approximately a. 50 minutes b. 45 minutes c. 60 minutes d. 40 minutes \((\log 4=0.60, \log 5=0.69)\)

Short Answer

Expert verified
The 50% reaction completion takes approximately 45 minutes.

Step by step solution

01

Understand the Problem

We know that 60% of a first-order reaction is completed in 60 minutes. We need to determine how long it takes for 50% of the reaction to complete. We are given the logarithmic values to help with the calculations.
02

Apply First Order Reaction Formula

For a first order reaction, the amount of substance at time \( t \) is given by the equation: \[[A_t] = [A_0]e^{-kt}\]Where \([A_0]\) is the initial concentration, \([A_t]\) is the concentration at time \(t\), and \(k\) is the rate constant.
03

Derive Rate Constant from Given Data

Using the fact that 60% of the reaction is complete, 40% remains. So, \[ rac{[A_t]}{[A_0]} = 0.40 = e^{-k imes 60} \]Take the natural log of both sides: \[ rac{-k imes 60}{2.303} = ext{log}(0.40) \]Using \( ext{log}(0.40) = -0.40 \), we solve for \(k\): \[ k = \frac{-2.303 imes (-0.40)}{60} \]
04

Calculate Rate Constant \(k\)

Compute \(k\): \[ k = \frac{2.303 imes 0.40}{60} = 0.01534 ext{ min}^{-1} \]
05

Use Rate Constant to Find Time for 50% Completion

For 50% completion, \[ rac{[A_t]}{[A_0]} = 0.50 = e^{-kt} \]Take the natural log of both sides: \[ -kt = 2.303 imes ext{log}(0.50) \]Using \( ext{log}(0.50) = -0.3010 \), we have: \[ t = \frac{2.303 imes 0.3010}{0.01534} \]
06

Perform the Calculation for Time

Calculate \(t\): \[ t = \frac{2.303 imes 0.3010}{0.01534} \approx 45 ext{ minutes} \]
07

Choose the Correct Answer

From our calculation, we found that approximately 45 minutes is required for 50% of the reaction to complete. Thus, the correct choice is (b) 45 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Constant
In the world of chemical kinetics, the rate constant, often represented by the symbol \(k\), is a crucial factor. It provides insight into the speed of a reaction at a given temperature. For first-order reactions, the rate constant is unique because it remains consistent across varying concentrations of the reactants.

  • The rate constant is determined experimentally by studying how the concentration of the reactant changes over time.
  • It has units of inverse time, such as \(\text{min}^{-1}\), indicating how frequently the reaction proceeds.
  • Knowing \(k\) allows us to predict how long a given percentage of a reactant will take to be consumed.
For example, from our problem's solution, the rate constant was found to be approximately 0.01534 \(\text{min}^{-1}\). With this, you can calculate how long it takes for the reaction to reach various levels of completion.
Reaction Completion Time
Calculating the time required for a reaction to reach a certain degree of completion is essential, especially for first-order reactions. The beauty of first-order kinetics is that the 'half-life,' or the time taken for the concentration to reduce by half, is constant for such reactions. This half-life is directly related to the rate constant.

In practice, the completion time can be derived using the equation \[-kt = \ln\left(\frac{[A_t]}{[A_0]}\right)\]where \([A_0]\) is the initial concentration and \([A_t]\) is the concentration at the desired time \(t\).

The above equation allows us to solve for \(t\) knowing \(k\) and the fraction of reaction completed. For instance, in the given exercise, to determine how long it takes for 50% of the reaction to complete once \(k\) is known, you substitute \(\frac{[A_t]}{[A_0]} = 0.50\) into the equation.
  • This illustrates that for a first-order reaction, the time to reach 50% completion is approximately 45 minutes.
Natural Logarithm in Kinetics
The natural logarithm is an essential mathematical tool in the study of kinetics. It helps in transforming exponential decay equations into linear forms, making them easier to analyze and solve.

  • In first-order kinetics, the rate equation is exponential: \([A_t] = [A_0]e^{-kt}\).
  • To find the time or concentration, we use the natural logarithm to linearize the equation: \(-kt = \ln\left(\frac{[A_t]}{[A_0]}\right)\).
  • This conversion allows for straightforward calculations of unknowns like time or the rate constant, as seen in the problem solution where logarithmic values are used to solve for time when 50% of the reaction is complete.
By understanding and applying the natural logarithm, students can accurately analyze reaction kinetics and make meaningful predictions about reaction behavior.

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Most popular questions from this chapter

The following set of data was obtained by the method of initial rates for the reaction: $$ \begin{aligned} &\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow \\ &2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}-(\mathrm{aq}) \end{aligned} $$ What is the rate law for the reaction? $$ \begin{array}{lll} \hline\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right], \mathrm{M} & {[\mathrm{I}-], \mathrm{M}} & \text { Initial rate, } \mathrm{M} \mathrm{s}^{-1} \\ \hline 0.25 & 0.10 & 9.00 \times 10^{-3} \\ 0.10 & 0.10 & 3.60 \times 10^{-3} \\ 0.20 & 0.30 & 2.16 \times 10^{-2} \\ \hline \end{array} $$ a. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{2}\) b. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]^{2}\left[\mathrm{I}^{-}\right]\) c. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]\) d. Rate \(=\mathrm{k}\left[\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\right]\left[\mathrm{I}^{-}\right]^{5}\)

The aquation of tris-(1,10-phenanthroline) iron (II) in acid solution takes place according to the equation: $$ \begin{aligned} &\mathrm{Fe}(\mathrm{phen})_{3}^{2}+3 \mathrm{H}_{3} \mathrm{O}^{+}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow \\ &\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}+3 \text { (phen) } \mathrm{H}^{+} \end{aligned} $$ If the activation energy is \(126 \mathrm{~kJ} / \mathrm{mol}\) and frequency factor is \(8.62 \times 10^{17} \mathrm{~s}^{-1}\), at what temperature is the rate constant equal to \(3.63 \times 10^{-3} \mathrm{~s}^{-1}\) for the first order reaction? a. \(0^{\circ} \mathrm{C}\) b. \(50^{\circ} \mathrm{C}\) c. \(45^{\circ} \mathrm{C}\) d. \(90^{\circ} \mathrm{C}\)

In a hypothetical reaction given below $$ 2 \mathrm{XY}_{2}(\mathrm{aq})+2 \mathrm{Z}^{-}(\mathrm{aq}) \rightarrow $$ (Excess) $$ 2 \mathrm{XY}_{2}^{-}(\mathrm{aq})+\mathrm{Z}_{2}(\mathrm{aq}) $$ \(\mathrm{XY}_{2}\) oxidizes \(\mathrm{Z}\) - ion in aqueous solution to \(\mathrm{Z}_{2}\) and gets reduced to \(\mathrm{XY}_{2}-\) The order of the reaction with respect to \(\mathrm{XY}_{2}\) as concentration of \(Z\) - is essentially constant. Rate \(=\mathrm{k}\left[\mathrm{XY}_{2}\right]^{\mathrm{m}}\) Given below the time and concentration of \(\mathrm{XY}_{2}\) taken (s) Time \(\left(\mathrm{XY}_{2}\right) \mathrm{M}\) \(0.00\) \(4.75 \times 10^{-4}\) \(1.00\) \(4.30 \times 10^{-4}\) \(2.00\) \(3.83 \times 10^{-4}\) The half life of the reaction (in seconds) is a. \(2.39\) b. \(13.35\) c. \(6.93\) d. \(19.63\)

The basic theory behind Arrhenius's equation is that a. The activation energy and pre-exponential factor are always temperature- independent b. The rate constant is a function of temperature c. The number of effective collisions is proportional to the number of molecules above a certain threshold energy d. As the temperature increases, so does the number of molecules with energies exceeding the threshold energy.

In a first order reaction the concentration of reactant decreases from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times\) \(10^{4} \mathrm{sec}\). The rate constant of reaction in \(\mathrm{sec}^{-1}\) is a. \(2 \times 10^{4}\) b. \(3.45 \times 10^{-5}\) c. \(1.386 \times 10^{-4}\) d. \(2 \times 10^{-4}\)
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