# Assignment 1 - 2013

 Due date(s): 21 January 2013 (PDF) Assignment questions (PDF) Assignment solutions

Assignment objectives: math refresher; chemistry refresher; review mol balances

• Always state assumptions in this assignment, midterms and exams.
• Never use an equation by just writing it down; state its origin and all simplifying assumptions. For example: using the general mol balance in a batch reactor, under the assumption of a well-mixed and constant volume system, we have: $$\dfrac{dN_j}{dt} = r_jV$$

Question 1 [10]

1. $$\displaystyle \int{ \frac{1}{x} \,dx} =$$
2. $$\displaystyle \int{ \frac{1}{x^2} \,dx} =$$
3. $$\displaystyle \int{ \frac{1}{ax+b} \,dx} =$$
4. $$\displaystyle \int{ \frac{1}{\sqrt{x}} \,dx} =$$
5. When do we require an integration constant; and when do we not require it?

Solution

1. $$\displaystyle \int{ \frac{1}{x} \,dx} = \ln{x} + C$$
2. $$\displaystyle \int{ \frac{1}{x^2} \,dx} = -\frac{1}{x} + C$$
3. $$\displaystyle \int{ \frac{1}{ax+b} \,dx} = \frac{1}{a}\ln(ax+b) + C$$
4. $$\displaystyle \int{ \frac{1}{\sqrt{x}} \,dx} = 2\sqrt{x} + C$$
5. We require an integration constant when the limits of integration are unspecified (indefinite integrals); otherwise we don't require the constant, since the integral is uniquely defined.

Question 2 [10]

1. A vessel contains a gas of concentration $$20\,\text{mol.m}^{-3}$$. The gas is stored at 375°C. Assuming this is an ideal gas, what is the pressure in the vessel measured in kPa? What assumption are you making (apart from the ideal-gas law)?
2. A constant volume batch reactor operates at 14.7 psi and 1340°F. The reactor volume is $$290\,\text{ft}^3$$. How many mols are in the system, assuming an ideal gas?

Solution

Assumptions: Pure species in tank, at constant volume and temperature.

1. $$T = 648\,\text{K}$$, so from the ideal gas law: $$P = CRT = (20\,\text{mol.m}^{-3})(8.314\,\text{J.mol}^{-1}.K^{-1})(648\,\text{K}) = 107.7\,\text{kPa}$$.
2. At $$P = 14.7\,\text{psi} = 1\,\text{atm}$$, with $$V = 290\,\text{ft}^3 = 8.205\,\text{m}^3$$ and $$T = 1000\,\text{K}$$, then $$R = 8.205746 \times 10^{-5}\, \text{m}^3\text{.atm.K}^{-1}\text{.mol}^{-1}$$, giving $$n = 100\,\text{mol}$$.

Question 3 [10]

Milk is pasteurized if it is heated to 63°C for 30 min, but if it is heated to 74°C it only needs 15 seconds for the same result. Find the activation energy of this sterilization process.

Recall the activation energy for a chemical reaction is the $$E$$ term, and the rate constant in is given by $$k = k_0 e^{\frac{-E}{RT}}$$.

Hint: assume pasteurization proceeds via first-order kinetics; what is the "reactant"?

Solution

To ask for the activation energy of a process means assuming an Arrhenius temperature dependency for the process. Here we are told that

• $$t_1 = 1800$$ seconds are required at temperature $$T_1$$ = 336 K
• $$t_2 = 15$$ seconds are required at temperature $$T_2$$ = 347 K

As told, assuming first order kinetics, $$-r_A = k_A C_A = k_0 e^{\frac{-E}{RT}} C_A$$.

For a batch system, making the regular assumption of constant volume and well-mixed (both are suitable for pasteurization):

$t = \int_{C_{A0}}^{C_A}{\frac{dC_A}{-r_A}} = \int_{C_{A0}}^{C_A}{\frac{dC_A}{-k_A C_A}} = \frac{1}{k_A}\ln\left(\frac{C_A}{C_{A0}} \right)$

So for the 2 systems we have that $$C_{A0}$$ and $$C_{A}$$ are the same, so we can write the ratio:

$\dfrac{t_1}{t_2} = \frac{1800}{15} = 120 = \frac{k_{A,2}}{k_{A,1}} = \frac{k_{0} e^{\frac{-E}{RT_2}}}{k_{0} e^{\frac{-E}{RT_1}}}$

Simplifying, recognizing that $$k_0$$, and $$R$$ are the same for both systems:

$\begin{split}120 &= \exp\left(-\frac{E}{RT_2} + \frac{E}{RT_1} \right) \\ \ln(120) &= -\frac{E}{R}\left(\frac{1}{347} - \frac{1}{336}\right)\\ -E &= \frac{\ln(120)(8.314)}{-9.43\times 10^{-5}}\\ E &= 421 885\,\text{J.mol}^{-1}\end{split}$

Question 4 [13]

The fermentation of an active ingredient $$A$$ is to be carried out in a reactor. The reaction kinetics are given by:

$A \longrightarrow R$$-r_A = \frac{0.1 C_A}{1+0.5 C_A } \left[\dfrac{\text{mol}}{\text{L.min}}\right]$
1. Consider a batch reactor filled with 750 L of reactant at $$C_{A,0} = 2\,\text{mol.L}^{-1}$$. How long must the reactor be operated to achieve an exit concentration of A of $$0.1\,\text{mol.L}^{-1}$$?

If the feed rate is continuously fed at $$25\,\text{L.min}^{-1}$$, with $$C_{A,0} = 2\,\text{mol.L}^{-1}$$. Determine the volume required for a

1. CSTR
2. PFR

to achieve an exit concentration of A of $$0.1\,\text{mol.L}^{-1}$$.

1. Which of the CSTR or PFR require a smaller volume?

Solution

1. For a well-mixed batch reactor, the design equation is:

$\begin{split}\dfrac{dC_A}{dt} &= r_A \\ -\int_{C_{A0}}^{C_A}{\dfrac{1+0.5C_A}{0.1C_A}dC_A} &= \int_{t=0}^{t}{dt}\\ -10\ln\left(\dfrac{0.1}{2.0}\right) - 5.0\left(0.1 - 2.0\right) &= t\\ t &= 29.95 + 9.5 = 39.45\,\text{minutes}\end{split}$
2. Using the design equation for CSTRs (assuming it to be well-mixed and operating at steady state, with a constant volumetric flow rate of $$25\,\text{L.min}^{-1}$$):

$\begin{split}V &= \dfrac{F_{A0}-F_A}{-r_A} \\ &= \dfrac{50-2.5}{\frac{(0.1)(0.1)}{1+(0.5)(0.1) }} \\ &= 4987.5 \approx 5000\,\text{m}^3\end{split}$
3. Use the mole balance equation for PFRs, making the usual assumption of steady state, and well-mixed in the radial direction, and using that $$(q)(dC_A) = dF_A$$, assuming constant volumetric flow throughout the PFR:

$\begin{split}\dfrac{dF_A}{dV} &= r_A \\ dV &= q\dfrac{dC_A}{r_A} \\ V &= q \int_{C_{A0}}^{C_A}{-\frac{1.0 + 0.5C_A}{0.1C_A}dC_A}\\ V &= -(25)(10) \ln\left(\dfrac{0.1}{2.0}\right) -(25)(5)(0.1 - 2.0)\\ V &= 986\,\text{L}\end{split}$

Question 5 [7]

The gas phase reaction:

$A \longrightarrow B + C$

is carried out at 100°C in a 20 L constant-volume, sealed batch reactor, at atmospheric pressure. The reaction is second order: $$-r_A = k C_A^2$$ where $$k = 2\,\text{L.mol}^{-1}\text{.min}^{-1}$$.

One mole of pure A is initially placed in the reactor, which is well mixed (is this a reasonable assumption?). Determine:

1. the partial pressure due to A in the reactor
2. the concentration of A in the reactor after 5 minutes have elapsed
3. the partial pressure due to A in the reactor after 5 minutes have elapsed.

Solution

There was an error in the question, so as long as you attempted the question you got full grade. The problem was over-specified, so you get a partial pressure of A that exceeds atmospheric pressure.

1. $$p_a V = n_A RT$$, or solving for $$p_A = \dfrac{(1)(8.314)(373)}{20 \times 10^{-3}} = 155056\,\text{Pa}$$.

2. After 5 minutes in a batch reactor that is assumed to be well-mixed:

$\begin{split}\frac{dN_A}{dt} &= r_AV \\ \frac{dC_A}{dt} &= r_A \\ \int_{t=0}^{t}{dt} &= \int_{C_{A0}}^{C_A}{\dfrac{dC_A}{-kC_A^2}} \\ t &= \dfrac{1}{k_A}\left(\dfrac{1}{C_A} - \frac{1}{C_{A0}}\right)\\ C_A &= 0.033\,\text{mol.L}^{-1}\end{split}$
3. The partial pressure due to A after 5 minutes have elapsed is $$p_A = C_A RT = (33.0)(8.314)(373) =102337\,\text{Pa}$$

Question 6 [10]

Consider a municipal water treatment plant for a smallish community. Waste water at $$32,000\,\text{m}^3\text{.day}^{-1}$$, flows through the treatment plant with a mean residence time of 8 hours. Air is bubbled through the tanks, and microbes in the tank attack and break down the organic material:

$\text{Organic waste} + \text{O}_2 \xrightarrow{\text{microbes}} \text{CO}_2 + \text{H}_2\text{O}$

A typical entering feed has a BOD (biological oxygen demand) of $$200\,(\text{mg O}_2)\text{.L}^{-1}$$, the effluent has a negligible BOD. Find the average rate of reaction, or decrease in BOD, in the treatment tanks.

Solution

The rate of reaction is defined as $$-r_{\text{O}_2} = \dfrac{\text{mol O_2 used}}{(\text{day})(\text{m}^3)}$$.

We can obtain an estimate of the tank volume from the residence time equation: $$\tau = \dfrac{V}{q}$$ or that $$V = \dfrac{32000}{3} = 10667 \,\text{m}^3$$.

To obtain a value for the numerator, recognize that 1 mole of waste requires 1 mole of oxygen to remove its biological oxygen demand. We require 200 mg of oxygen though per litre of waste, or $$\dfrac{0.2\,\text{g.L}^{-1}}{32\,\text{g.mol}^{-1}}\times 1000\,\text{L.m}^{-3} = 6.25$$ mol of oxygen required per $$\text{m}^3$$ of waste. There will be $$32 000\,\text{m}^3\text{.day}^{-1}$$ of waste water to treat, so this corresponds then to an average reaction rate of $$\dfrac{6.25 \times 32 000}{10667} = 18.75\,\dfrac{\text{mol O_2 used}}{(\text{day})(\text{m}^3)}$$.