/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 The decomposition of \(\mathrm{S... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 27

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$ \mathrm{sO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ is first-order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), and the reaction has a halflife of 245 minutes at 600 K. If you begin with \(3.6 \times 10^{-3}\) mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a \(1.0-\mathrm{L}\) flask, how long will it take for the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(2.00 \times 10^{-4}\) mol?

Short Answer

Expert verified
It will take approximately 1250.43 minutes.

Step by step solution

01

Identify the Reaction Order and Relevant Formula

The reaction is given as first-order in \(\text{SO}_{2}\text{Cl}_{2}\). For a first-order reaction, the formula to use is the exponential decay formula: \( [A] = [A]_0 e^{-kt} \), where \([A]\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
02

Calculate the Rate Constant

Use the half-life formula for a first-order reaction to find \(k\). The half-life \(t_{1/2}\) for a first-order reaction is given by \( t_{1/2} = \frac{0.693}{k} \). Substituting \(t_{1/2} = 245\) minutes: \[ k = \frac{0.693}{245} = 0.00283\, \text{min}^{-1} \]
03

Set Up the First-Order Decay Equation

We want to find the time \(t\) when the concentration of \(\text{SO}_{2}\text{Cl}_{2}\) decreases from \(3.6 \times 10^{-3}\, \text{mol} \) to \(2.00 \times 10^{-4}\, \text{mol}\). Substitute these into the first-order decay formula: \[ 2.00 \times 10^{-4} = 3.6 \times 10^{-3} e^{-0.00283t} \]
04

Solve for Time \(t\)

Isolate \(t\) by first dividing both sides of the equation by the initial concentration \(3.6 \times 10^{-3}\): \[ \frac{2.00 \times 10^{-4}}{3.6 \times 10^{-3}} = e^{-0.00283t} \] Take the natural logarithm of both sides: \[ \ln \left( \frac{2.00 \times 10^{-4}}{3.6 \times 10^{-3}} \right) = -0.00283t \] Solving for \(t\):\[ t = \frac{\ln(\frac{2.00 \times 10^{-4}}{3.6 \times 10^{-3}})}{-0.00283} = 1250.43\, \text{minutes} \]
05

Conclusion

It will take approximately 1250.43 minutes for the amount of \(\text{SO}_{2}\text{Cl}_{2}\) to decrease to \(2.00 \times 10^{-4}\) mol in the reaction flask.

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Key Concepts

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

Rate Constant Calculation
In order to understand how fast a reaction will proceed, chemists often calculate a key parameter known as the rate constant, represented by the symbol \( k \). For a first-order reaction, the rate constant is related to the half-life of the reaction, which is a fixed time it takes for half of the reactant to be consumed. To find the rate constant for first-order reactions, we use the formula:
  • \( t_{1/2} = \frac{0.693}{k} \)
To calculate \( k \) 鈥� the rate constant 鈥� we rearrange this formula to solve for \( k \):
  • \( k = \frac{0.693}{t_{1/2}} \)
In the given problem, the reaction has a half-life of 245 minutes. Substituting the value into the equation \[ k = \frac{0.693}{245} = 0.00283\, \text{min}^{-1} \] shows that the rate constant \( k \) is 0.00283 \( \text{min}^{-1} \). This figure is essential for predicting how long it takes for certain concentrations of reactants to convert into products.
Exponential Decay Formula
The exponential decay formula is crucial in understanding first-order reactions, which involve a constant fractional loss of reactant per unit time regardless of the concentration. This formula captures the essence of how a starting reactant amount decreases exponentially over time. The simplified exponential decay expression is:
  • \( [A] = [A]_0 e^{-kt} \)
Where:
  • \([A]\) is the concentration of the reactant at time \( t \)
  • \([A]_0\) is the initial concentration
  • \( k \) is the rate constant
In the original problem, the goal is to find how long it takes for \(3.6 \times 10^{-3}\) mol to reduce to \(2.00 \times 10^{-4}\) mol. Using the exponential decay formula helps in calculating the accurate time \( t \) by isolating \( t \) in the equation. Once plugged in with the known values, solving this formula provides the specific time required for the concentration to drop to a desired level.
Half-Life of a Reaction
The half-life of a reaction is a crucial concept in the study of reaction kinetics, particularly for first-order reactions. It is defined as the time required for the concentration of a reactant to decrease by half of its initial value. For first-order reactions, the half-life is unique because it is independent of the initial concentration.
  • First-order half-life formula: \( t_{1/2} = \frac{0.693}{k} \)
This formula allows us to determine that the half-life stays constant, making calculations straightforward and predictions about the time frame of reactions reliable.The half-life concept becomes very handy in predicting the temporal dynamics of reactions. It doesn't vary with how much substance you start with, which simplifies many calculations regarding decay and duration.In the exercise, the half-life of 245 minutes allowed for the rate constant calculation which, in turn, led to determining the exact time required for the reaction's progression to a specified concentration, showing the practical utility of this concept.

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Most popular questions from this chapter

Hydrogenation reactions, processes wherein \(\mathrm{H}_{2}\) is added to a molecule, are usually catalyzed. An excellent catalyst is a very finely divided metal suspended in the reaction solvent. Explain why finely divided rhodium, for example, is a much more efficient catalyst than a small block of the metal.

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g})\) (b) \(2 \mathrm{HOF}(\mathrm{g}) \rightarrow 2 \mathrm{HF}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\)

Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct. (a) The rate-determining elementary step in a reaction is the slowest step in a mechanism. (b) It is possible to change the rate constant by changing the temperature. (c) As a reaction proceeds at constant temperature, the rate remains constant. (d) A reaction that is third-order overall must involve more than one step.

The decomposition of dinitrogen pentaoxide $$ \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ has the following rate equation: Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) It has been found experimentally that the decomposition is \(20.5 \%\) complete in 13.0 hours at 298 K. Calculate the rate constant and the half-life at 298 K.

The reaction cyclopropane \(\rightarrow\) propene occurs on a platinum metal surface at \(200^{\circ} \mathrm{C}\). (The platinum is a catalyst.) The reaction is first-order in cyclopropane. Indicate how the following quantities change (increase, decrease, or no change) as this reaction progresses, assuming constant temperature. (a) [cyclopropane] (b) Ipropene] (c) [catalyst] (d) the rate constant, \(k\) (e) the order of the reaction (f) the half-life of cyclopropane
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