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Chapter 2: Problem 32

A concentration \(C(\mathrm{mol} / \mathrm{L})\) varies with time (min) according to the equation \(C=3.00 \exp (-2.00 t)\) (a) What are the implicit units of 3.00 and 2.00? (b) Suppose the concentration is measured at \(t=0\) and \(t=1\) min. Use two- point linear interpolation or extrapolation to estimate \(C(t=0.6 \mathrm{min})\) and \(t(C=0.10 \mathrm{mol} / \mathrm{L})\) from the measured values, and compare these results with the true values of these quantities. (c) Sketch a curve of \(C\) versus \(t,\) and show graphically the points you determined in Part (b).

Short Answer

Expert verified
The units of 3.00 and -2.00 are \(\mathrm{mol}/\mathrm{L}\) and \(\mathrm{L}/(\mathrm{mol}\cdot\mathrm{min})\) respectively. The points \(C(t=0.6 \mathrm{min})\) and \(t(C=0.10 \mathrm{mol}/\mathrm{L})\) can be estimated using two-point linear interpolation or extrapolation and compared with the true values. A sketch of the curve of \(C\) versus \(t\) helps to visualize these points.

Step by step solution

01

Identify the Units

First, let's examine the equation \(C=3.00 \exp (-2.00 t)\). In this equation, \(C\) represents concentration, which we know is measured in moles per liter (\(\mathrm{mol}/\mathrm{L}\)). Therefore, since the exponential factor must be dimensionless, we can say that \(t\) is in minutes, and hence -2.00 has units of \(\mathrm{L}/(\mathrm{mol.min})\). As \(C\) and \(3.00\) share the same units, \(3.00\) would have the units \(\mathrm{mol}/\mathrm{L}\).
02

Use Two-point Linear Interpolation or Extrapolation

Next, we derive the estimated concentration at \(t=0.6\) minutes using two-point linear interpolation. For \(t=0\), \(C=3.00 \exp (0)\), so \(C=3.00 \mathrm{mol}/\mathrm{L}\). For \(t=1\), \(C=3.00 \exp (-2.00 \times 1)\), so \(C=3.00 \exp (-2.00)=1.35 \mathrm{mol}/\mathrm{L}\). Using the interpolation formula \(C(t)=(C(t_b)-C(t_a))/(t_b-t_a) \times (t-t_a)+C(t_a)\), we can substitute \(t_a=0\), \(t_b=1\), \(C_a=3.00\), and \(C_b=1.35\) to find \(C(0.6)\). Similarly, we derive the time when \(C=0.10 \mathrm{mol}/\mathrm{L}\).
03

Visualize with Curve Sketching

Lastly, sketch a curve depicting the relationship between concentration \(C\) and time \(t\) based on the equation \(C=3.00 \exp (-2.00 t)\). Also, plot the points that have been determined in the previous part for \(C(t=0.6 \mathrm{min})\) and \(t(C=0.10 \mathrm{mol}/\mathrm{L})\).
04

Compare Results

Now, let's compare the estimated values we found earlier with the true values from the equation \(C=3.00 \exp (-2.00 t)\). This will allow us to assess the accuracy of our two-point linear interpolation.

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Key Concepts

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

Units of Measurement
In chemical kinetics equations, it's essential to understand implicit units of the constants involved. This is particularly important when dealing with expressions like \[C = 3.00 \exp(-2.00t).\]Here, \(C\) is concentration in moles per liter (\(\mathrm{mol}/\mathrm{L}\). Because this concentration equation involves an exponential function, the units must be consistent to result in a dimensionless exponent.

- For the equation \(\exp(-2.00t)\), the term \(-2.00t\) must be dimensionless. Clearly, the time \(t\) is given in minutes. Therefore, the factor \(-2.00\), which is the rate constant, possesses units of \(\mathrm{min}^{-1}\).

- The coefficient 3.00, being directly in front of the exponential, has the same units as \(C\), which are \(\mathrm{mol}/\mathrm{L}\).

Understanding these implicit units assists in correctly applying the equation and ensures that resulting calculations maintain the accurate dimensions throughout the process.
Exponential Decay
The mathematical principle of exponential decay plays a vital role in chemical kinetics. In the context of the given equation, \(C = 3.00 \exp(-2.00t)\), the concentration \(C\) changes over time according to the exponential term.

- **Exponential Function**: The term \(\exp(-2.00t)\) implies that the concentration decreases exponentially as time increases. This reflects a decaying process, akin to common phenomena such as radioactive decay or cooling processes.

- **Decay Rate**: The rate at which this process occurs is governed by the rate constant. In this equation, \(-2.00\) dictates this speed. A more negative rate constant would correspond to a faster decay.

- **Half-life Concept**: While not directly given here, the concept of a half-life鈥攖ime it takes for a concentration to reduce to half its original value鈥攃an be explored through this principle. In an exponential decay, this remains constant irrespective of the starting concentration.

Exponential decay provides a foundation for understanding how chemical reactions diminish over time.
Interpolation and Extrapolation
In many scientific fields, including chemical kinetics, interpolation and extrapolation are valuable techniques used to estimate unknown values from data points.

- **Two-Point Interpolation**: Given measurements of concentration at specific times, this technique allows you to estimate concentrations at times between these measurements. For example; using the values at \(t=0\,\mathrm{min}\,\) (\(C=3.00\,\mathrm{mol/L}\)) and \(t=1\,\mathrm{min}\,\) (\(C=1.35\,\mathrm{mol/L}\)), you can run: \[ C(t) = \left( \frac{1.35-3.00}{1-0} \right) \times (0.6-0) + 3.00 \]to find an approximate value at \(t=0.6\,\mathrm{min}\).

- **Extrapolation**: Conversely, when you need to guess the value of \(t\) for a known concentration value outside the tested interval, you rely on a similar logical extension. This method stressed in finding \(t(C=0.10 \mathrm{mol}/\mathrm{L})\).

Remember:
  • Interpolation provides estimates within the dataset range.
  • Extrapolation goes beyond the range, hence often less accurate.
Skillful application of these techniques aids in making educated estimations in kinetic studies.

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Most popular questions from this chapter

You arrive at your lab at 8 A.M. and add an indeterminate quantity of bacterial cells to a flask. At 11 A.M. you measure the number of cells using a spectrophotometer (the absorbance of light is directly related to the number of cells) and determine from a previous calibration that the flask contains 3850 cells, and at 5 P.M. the cell count has reached 36,530. (a) Fit each of the following formulas to the two given data points (that is, determine the values of the two constants in each formula): linear growth, \(C=C_{0}+k t ;\) exponential growth, \(C=C_{0} e^{k t} ;\) power-law growth, \(C=k t^{b} .\) In these expressions, \(C_{0}\) is the initial cell concentration and \(k\) and \(b\) are constants. (b) Select the most reasonable of the three formulas and justify your selection. (c) Estimate the initial number of cells present at 8 A.M. \((t=0)\). State any assumptions you make. (d) The culture needs to be split into two equal parts once the number of cells reaches 2 million. Estimate the time at which you would have to come back to perform this task. State any assumptions you make. If this is a routine operation that you must perform often, what does your result suggest about the scheduling of the experiment?

State what you would plot to get a straight line if experimental ( \(x, y\) ) data are to be correlated by the following relations, and what the slopes and intercepts would be in terms of the relation parameters. If you could equally well use two different kinds of plots (e.g., rectangular or semilog), state what you would plot in each case. [The solution to Part (a) is given as an example.] (a) \(y^{2}=a e^{-b / x}\). (b) \(y^{2}=m x^{3}-n\) (c) \(1 / \ln (y-3)=(1+a \sqrt{x}) / b\) (d) \((y+1)^{2}=\left[a(x-3)^{3}\right]^{-1}\) (e) \(y=\exp (a \sqrt{x}+b)\) (f) \(x y=10^{\left[a\left(x^{2}+y^{2}\right)+b\right]}\) (g) \(y=[a x+b / x]^{-1}\)

The following empirical equation correlates the values of variables in a system in which solid particles are suspended in a flowing gas: $$\frac{k_{g} d_{p} y}{D}=2.00+0.600\left(\frac{\mu}{\rho D}\right)^{1 / 3}\left(\frac{d_{p} u \rho}{\mu}\right)^{1 / 2}$$ Both \((\mu / \rho D)\) and \(\left(d_{p} u \rho / \mu\right)\) are dimensionless groups; \(k_{g}\) is a coefficient that expresses the rate at which a particular species transfers from the gas to the solid particles; and the coefficients 2.00 and 0.600 are dimensionless constants obtained by fitting experimental data covering a wide range of values of the equation variables. The value of \(k_{g}\) is needed to design a catalytic reactor. since this coefficient is difficult to determine directly, values of the other variables are measured or estimated and \(k_{g}\) is calculated from the given correlation. The variable values are as follows: $$\begin{aligned}d_{p} &=5.00 \mathrm{mm} \\\y &=0.100 \quad(\text { dimensionless }) \\\D &=0.100 \mathrm{cm}^{2} / \mathrm{s} \\\\\mu &=1.00 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2} \\\\\rho &=1.00 \times 10^{-3} \mathrm{g} / \mathrm{cm}^{3} \\\u &=10.0 \mathrm{m} / \mathrm{s}\end{aligned}$$ (a) What is the estimated value of \(k_{g} ?\) (Give its value and units.) (b) Why might the true value of \(k_{g}\) in the reactor be significantly different from the value estimated in Part (a)? (Give several possible reasons.) (c) Create a spreadsheet in which up to five sets of values of the given variables ( \(d_{p}\) through \(u\) ) are entered in columns and the corresponding values of \(k_{g}\) are calculated. Test your program using the following variable sets: (i) the values given above; (ii) as above, only double the particle diameter \(d_{p}\) (making it \(10.00 \mathrm{mm}\) ); (iii) as above, only double the diffusivity \(D ;\) (iv) as above, only double the viscosity \(\mu ;(\mathrm{v})\) as above, only double the velocity \(u\). Report all five calculated values of \(k_{g}\).

In modeling the effect of an impurity on crystal growth, the following equation was derived: \(\frac{G-G_{\mathrm{L}}}{G_{0}-G}=\frac{1}{K_{\mathrm{L}} C^{m}}\) where \(C\) is impurity concentration, \(G_{\mathrm{L}}\) is a limiting growth rate, \(G_{0}\) is the growth rate of the crystal with no impurity present, and \(K_{\mathrm{L}}\) and \(m\) are model parameters. In a particular experiment, \(G_{0}=3.00 \times 10^{-3} \mathrm{mm} / \mathrm{min},\) and \(G_{\mathrm{L}}=1.80 \times 10^{-3} \mathrm{mm} / \mathrm{min} .\) Growth rates are measured for several impurity concentrations \(C\) (parts per million, or ppm), with the following results: $$\begin{array}{|c|c|c|c|c|c|}\hline C(\mathrm{ppm}) & 50.0 & 75.0 & 100.0 & 125.0 & 150.0 \\\\\hline G(\mathrm{mm} / \mathrm{min}) \times 10^{3} & 2.50 & 2.20 & 2.04 & 1.95 & 1.90 \\\\\hline\end{array}$$ (For example, when \(\left.C=50.0 \mathrm{ppm}, G=2.50 \times 10^{-3} \mathrm{mm} / \mathrm{min}\right)\). (a) Determine \(K_{\mathrm{L}}\) and \(m,\) giving both numerical values and units. (b) A solution is fed to a crystallizer in which the impurity concentration is 475 ppm. Estimate the expected crystal growth rate in (mm/min). Then state why you would be extremely skeptical about this result.

On a website devoted to answering engineering questions, viewers were invited to determine how much power a 100 -MW power plant generates annually. The answer declared to be best was submitted by a civil engineering student, who stated, "It produces \(100 \mathrm{MW} / \mathrm{hr}\) so over the year that's \(100^{*} 24^{*} 365.25 \&\) do the math." (a) Carry out the calculation, showing all the units. (b) What is wrong with the statement of the question? (c) Why was the student wrong in saying that the plant produces \(100 \mathrm{MW} / \mathrm{hr} ?\)
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