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The reaction described by the equation $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is first order with a rate constant of \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\) at \(320^{\circ} \mathrm{C}\). What fraction of a sample of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) will remain if it is heated for \(5.0\) hours at \(320^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
67.3% of the sample remains.

Step by step solution

01

Convert time into seconds

Since the rate constant is given in terms of seconds, we need to convert the time from hours into seconds. There are 3600 seconds in one hour, so: \[ t = 5.0 \text{ hours} \times 3600 \text{ seconds/hour} = 18000 \text{ seconds} \]
02

Use the first-order kinetics formula

For a first-order reaction, the relationship between concentration and time is given by:\[ \ln(\frac{[A]_t}{[A]_0}) = -kt \]where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time the reaction has proceeded.
03

Calculate the fraction of the sample remaining

We need to find \(\frac{[A]_t}{[A]_0}\), which represents the fraction of the initial sample that remains after time \(t\). Rewriting the formula:\[ \ln(\frac{[A]_t}{[A]_0}) = - (2.2 \times 10^{-5} \text{ s}^{-1}) \times 18000 \text{ s} \]\[ \ln(\frac{[A]_t}{[A]_0}) = -0.396 \]Solve for \(\frac{[A]_t}{[A]_0}\):\[ \frac{[A]_t}{[A]_0} = e^{-0.396} \]\[ \frac{[A]_t}{[A]_0} \approx 0.673 \]
04

Interpret the result

The fraction 0.673 means that approximately 67.3% of the original \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(g)\) sample remains after being heated for 5.0 hours at 320°C.

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Key Concepts

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

Reaction Kinetics
Reaction kinetics is an area of chemistry that studies the speed or rate at which chemical reactions occur. It seeks to understand the steps involved in the transformation of reactants into products. The rate of a reaction is crucial, as it tells us how fast a reactant is converted to a product over time. Understanding kinetics can help in industrial processes where controlling the rate of a reaction is vital.

Factors such as temperature, concentration of reactants, and presence of a catalyst can influence the rate of reaction. Reaction kinetics is especially important in determining conditions under which a chemical reaction can proceed safely and efficiently. In the case of first-order reactions, the rate depends on the concentration of a single reactant. This simple dependency on concentration allows us to derive straightforward mathematical expressions to describe the rate.
Rate Constant
The rate constant is a key parameter in reaction kinetics that quantifies the speed of a reaction. It is represented by the symbol "k". For a first-order reaction, the unit of the rate constant is reciprocal time, such as s\(^{-1}\). This value is unique for each reaction and can vary with changes in temperature or pressure.

In the given exercise, the rate constant is recorded as \(2.2 \times 10^{-5} \text{ s}^{-1}\). This value tells us how quickly the concentration of SO\(_2\)Cl\(_2\) decreases per second at 320°C. If you have the rate constant and the time elapsed, you can predict how much of a substance will remain after a certain period using the first-order kinetics formula. The rate constant is essential for calculating the fraction of reactants remaining or turned into products over time.
Exponential Decay
Exponential decay describes the continuous and consistent decrease of a quantity over time. In chemistry, this often applies to the concentration of a reactant in a first-order reaction. It follows an exponential pattern, where the rate of decay is proportional to the current amount present.

This concept is highlighted by the formula: \[ \ln\left(\frac{[A]_t}{[A]_0}\right) = -kt \] where \([A]_t\) and \([A]_0\) are the concentrations at time \(t\) and initially, \(k\) is the rate constant, and \(e\) is the base of natural logarithms.

Using this equation, we see that with time, the concentration of SO\(_2\)Cl\(_2\) decreases exponentially, following an e\(^{-0.396}\) pattern in the exercise. This mathematical relationship helps to describe how quantities diminish in processes from radioactive decay to chemical reactions.
Half-Life
The half-life of a reaction is the period required for half of the reactant to be consumed. In first-order reactions, the half-life remains constant, irrespective of the initial concentration of the reactant. This is a unique feature of first-order kinetics, making it relatively easy to predict how long it will take until a certain amount of reactant remains.

The half-life is calculated using the equation:\[ t_{1/2} = \frac{0.693}{k}\]where \(t_{1/2}\) is the half-life and \(k\) is the rate constant.

In our problem, with a rate constant of \(2.2 \times 10^{-5} \text{ s}^{-1}\), the half-life can be determined and allows us to understand better how quickly the reaction progresses. Knowing the half-life is critical in fields such as pharmacology or nuclear physics, where the timing of reactions greatly influences outcomes.

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Most popular questions from this chapter

Cesium-137 is produced in nuclear reactors. If this isotope has a half-life of \(30.2\) years, how many years will it take for it to decay to one tenth of a percent of its initial amount?

The rate law for the reaction described by the equation $$ 2 \mathrm{~N}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) $$ is second order in \(\left[\mathrm{N}_{2} \mathrm{O}\right]\). The reaction was carried out at \(900 \mathrm{~K}\) with an initial concentration of \(\mathrm{N}_{2} \mathrm{O}(g)\) of \(2.0 \times 10^{-2}\) M. It took 4500 seconds for \(\left[\mathrm{N}_{2} \mathrm{O}\right]\) to fall to half its initial value. Determine the value of the rate constant for this reaction.

Express the rate of the following reaction equation in terms of the rate of concentration change for each of the three species involved: $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) $$

Phosphine, \(\mathrm{PH}_{3}(g)\) decomposes according to the equation $$ 4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The kinetics of the decomposition of phosphine at. \(950 \mathrm{~K}\) was followed by measuring the total pressure in the system as a function of time. The following data were obtained in a run where the reaction chamber contained only pure phosphine at the start of the reaction: \begin{tabular}{cc} \hline\(t / \min\) & \(P_{\text {total }} /\) Torr \\ \hline 0 & 100 \\ 40 & 150 \\ 80 & 167 \\ 100 & 172 \\ \hline \end{tabular} Determine the reaction rate law and calculate the value of the rate constant.

The table below gives the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) as a function of time for the reaction described by the equation \begin{tabular}{lc} \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) \\ \hline\(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right] / \mathrm{M}\) & \(t / \mathrm{min}\) \\ \hline \(0.0345\) & 0 \\ \(0.0245\) & \(3.8\) \\ \(0.0212\) & \(5.6\) \\ \(0.0154\) & \(9.8\) \\ \(0.0103\) & \(14.0\) \\ \hline \end{tabular} Verify that this reaction is a first-order reaction by plotting \(\ln \left(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=\mathrm{M}\right)\) versus time and determine the value of the rate constant.

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