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Chapter 14: Problem 47

The reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude and units of the first order rate constant: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { (atm) } \\ \hline 0 & 1.000 \\ 2,500 & 0.947 \\ 5,000 & 0.895 \\ 7,500 & 0.848 \\ 10,000 & 0.803 \\ \hline \end{array} $$

Short Answer

Expert verified
The magnitude of the first-order rate constant k for the given reaction is \(2.2314 \times 10^{-5}\) and its units are \(s^{-1}\) (per second).

Step by step solution

01

Write the first-order rate law for the reaction

The rate law for a first-order reaction is given by: \(Rate = k[SO_2Cl_2]\), where Rate is the reaction rate, k is the rate constant, and [SO_2Cl_2] is the concentration of SO2Cl2.
02

Express the first-order rate law in terms of pressure instead of concentration

The pressure of SO2Cl2 is proportional to its concentration and, thus, the same will hold for the rate expression: \(Rate = kP_{SO_2Cl_2}\), where \(P_{SO_2Cl_2}\) is the pressure of SO2Cl2.
03

Define the integrated rate law and its variables

To find k, we must use the integrated rate law for a first-order reaction. The integrated rate law is given by: \(ln\frac{P_{SO_{2}Cl_{2, initial}}}{P_{SO_{2}Cl_{2}}} = kt\), where \(P_{SO_2Cl_{2, initial}}\) is the initial pressure at time t = 0, \(P_{SO_2Cl_2}\) is the pressure at any time t, and t is the elapsed time in seconds.
04

Choose data points to substitute into the integrated rate law equation

We can choose any two points from the given data to find the rate constant. For simplicity, let's choose the initial point (Time = 0 s and Pressure = 1.000 atm) and the final point (Time = 10,000 s and Pressure = 0.803 atm).
05

Plug the data points into the integrated rate law equation

Now we can substitute the chosen values into the equation: \(ln\frac{1.000}{0.803} = k \times 10,000\)
06

Solve the equation for the rate constant k

First, we will calculate the value of the natural logarithm and then rearrange the equation to isolate k: \(ln(1.25) = 0.22314 = 10,000k\) \(k = \frac{0.22314}{10,000}\)
07

Calculate the rate constant k

Now we can find the rate constant by dividing 0.22314 by 10,000: \(k = 2.2314 \times 10^{-5}\) Hence, the magnitude of the first-order rate constant k is \(2.2314 \times 10^{-5}\).
08

Determine the units of the rate constant k

Since this is a first-order reaction, and the pressures are given in atm and time in seconds, the units of the rate constant k will be \(s^{-1}\), which means the rate constant is expressed in terms of "per second". In conclusion, the magnitude of the first-order rate constant k is \(2.2314 \times 10^{-5}\) and its units are \(s^{-1}\).

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Most popular questions from this chapter

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M}\) ?

There are literally thousands of enzymes at work in complex living systems such as human beings. What properties of enzymes give rise to their ability to distinguish one substrate from another?

Sketch a graph for the generic first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) that has concentration of \(\mathrm{A}\) on the vertical axis and time on the horizontal axis. (a) Is this graph linear? Explain. (b) Indicate on your graph the half-life for the reaction.

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{NH}_{3}(g)\)

The isomerization of methyl isonitrile \(\left(\mathrm{CH}_{3} \mathrm{NC}\right)\) to acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) was studied in the gas phase at \(215^{\circ} \mathrm{C},\) and the following data were obtained: $$ \begin{array}{rl} \hline \text { Time (s) } & {\left[\mathrm{CH}_{3} \mathrm{NC}\right](\boldsymbol{M})} \\ \hline 0 & 0.0165 \\ 2,000 & 0.0110 \\ 5,000 & 0.00591 \\ 8,000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (a) Calculate the average rate of reaction, in \(M / s\), for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from \(t=0\) to \(t=15,000 \mathrm{~s}\). (c) Graph [CH \(\left._{3} \mathrm{NC}\right]\) versus time and determine the instantaneous rates in \(M /\) s at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).
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