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The first order reaction, \(\mathrm{SO}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{Cl}_{2}\), has a half life of \(8.75\) hours at \(593 \mathrm{~K}\). How long will it take for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to fall to \(12.5 \%\) of its initial value? a. \(6.24\) hours b. \(26.2\) hours c. \(16.2\) hours d. \(22.6\) hours

Short Answer

Expert verified
It takes approximately 26.2 hours for the concentration to fall to 12.5%.

Step by step solution

01

Understand the Half-Life Formula for First Order Reactions

For first order reactions, the relationship between the half-life \(t_{1/2}\) and the rate constant \(k\) is given by the formula \(t_{1/2} = \frac{0.693}{k}\). We will use this formula to find \(k\).
02

Calculate the Rate Constant k

Given the half-life \(t_{1/2} = 8.75\) hours, substitute it into the half-life formula: \[8.75 = \frac{0.693}{k}\].Solve for \(k\):\[k = \frac{0.693}{8.75} \approx 0.0792 \, \text{hr}^{-1}\].
03

Use the Integrated Rate Law for First Order Reactions

The integrated rate law for a first order reaction is \( [A] = [A]_0 e^{-kt} \), where \([A]\) is the concentration at time \(t\), and \([A]_0\) is the initial concentration. We want to find \(t\) when \([A] = 0.125[A]_0\) (i.e., 12.5% of the initial concentration).
04

Set Up the Equation to Solve for Time t

Use the relationship from Step 3:\[0.125[A]_0 = [A]_0 e^{-kt}\]Cancel \([A]_0\) from both sides:\[0.125 = e^{-kt}\].
05

Apply the Natural Logarithm to Solve for Time t

Take the natural logarithm on both sides to solve for \(t\):\[\ln(0.125) = -kt\]\[-2.079 = -0.0792t\].
06

Solve the Equation for Time t

Finally, solve for \(t\):\[t = \frac{2.079}{0.0792} \approx 26.2\, \text{hours}\].

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Key Concepts

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

half-life calculation
In a first order reaction, half-life is an essential concept that helps understand how long it takes for a reactant to decrease to half its original amount. For these reactions, an important feature is that the half-life is constant and does not depend on the initial concentration. It's determined using the simple formula:
\( t_{1/2} = \frac{0.693}{k} \)
where \( t_{1/2} \) is the half-life, and \( k \) is the rate constant.
  • This relationship is specifically useful because, unlike zero-order or second-order reactions, knowing just the rate constant allows you to determine how long the reactant takes to reach half its original concentration regardless of starting amounts.
  • In practical applications, the stability and predictability of the half-life measurement allow chemists to predict reaction progress over time.
rate constant determination
The rate constant \( k \) is a critical factor in first order reactions, defining the speed at which the reaction proceeds. Calculating it is straightforward with the half-life data:
First, use the known formula of half-life:
\[ 8.75 = \frac{0.693}{k} \]From this equation, solve for \( k \) to find:
\[ k = \frac{0.693}{8.75} \approx 0.0792 \text{ hr}^{-1} \]
  • The value of \( k \) tells us that each hour, about 7.92% of the reactant decomposes, which is a direct translation of the reaction's speed.
  • This constant remains the same throughout the reaction and is an important factor in calculating how the concentration of reactants will change over time.
integrated rate law
The integrated rate law for first order reactions is crucial when determining the concentration of a reactant at any point in time. It is expressed as:
\[ [A] = [A]_0 e^{-kt} \] where:
  • \([A]\) is the concentration at time \( t \).
  • \([A]_0\) is the initial concentration.
    • \(e\) is the base of the natural logarithm.
    This formula helps assess the progress and control of chemical reactions by connecting concentration, reaction time, and the rate constant.To find the time when a concentration drops to a specific percentage, like 12.5%, the equation is rearranged:
    \[0.125[A]_0 = [A]_0 e^{-kt}\]Canceling \([A]_0\), you're left with:
    \[0.125 = e^{-kt}\]This lays the foundation of solving for \( t \) by isolating it effectively.
natural logarithm application
Natural logarithms can simplify the equations involved in reaction kinetics, specifically for first order processes. To find the time \( t \) when the concentration is a certain fraction of the original, apply the natural logarithm:
Taking the natural log of both sides of our equation from the integrated rate law gives us:
\[ \ln(0.125) = -kt \]Simplifying this by applying our value for \( k \):
\[ -2.079 = -0.0792t \]Finally, solving for time \( t \):
\[ t = \frac{2.079}{0.0792} \approx 26.2\, \text{hours} \] The use of logarithms here allows you to easily solve equations involving exponential concepts. This simplification is valuable in kinetic studies for determining time frames to reach specific concentration levels.

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Most popular questions from this chapter

A mechanism for a naturally occurring reaction that destroys ozone is: Step I: \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{HO}(\mathrm{g}) \rightarrow \mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) Step II: \(\mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{HO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) Which species is a catalyst? a. \(\mathrm{O}\) b. \(\mathrm{O}_{3}\) c. \(\mathrm{HO}_{2}\) d. \(\mathrm{HO}\)

In which of the following ways does an activated complex differ from an ordinary molecule? a. \(\Delta \mathrm{H}^{\circ}\) is probably positive. b. It is quite unstable and has no independent existence c. The system has no vibrational character d. The system has a greater vibrational character

Match the following: (Here \(\mathrm{a}=\) Initial concentration of the reactant, \(\mathrm{p}=\) Initial pressure of the reactant) List I List II A. \(t \frac{1}{2}=\) constant (p) Zero order B. \(\mathrm{t} \frac{1}{2} \alpha \mathrm{a}\) (q) First order C. \(\mathrm{t} 1 / 2 \alpha \mathrm{l} / \mathrm{a}\) (r) Second order D. \(t^{1 / 2} \alpha p^{-1}\) (s) Pseudo first order

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 rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-5} \mathrm{~s}^{-1}\). If the rate is \(2.40 \times 10^{-5}\) mol litre \(^{-1}\) \(\mathrm{s}^{-1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in mol litre \(^{-1}\) ) is a. \(1.4\) b. \(1.2\) c. \(0.04\) d. \(0.8\)

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