 Free FE Practice Test

# PrepFE™

## Free FE Chemical Example Practice Problems

We've selected 10 diverse practice problems from our question bank that you can use to review for the Chemical engineering FE exam and give you an idea about some of the content we provide.

### 1) For the system shown in the figure, compute the peak value time: ## Solutions

### 1) For the system shown in the figure, compute the peak value time: A.$0.726 \si{s}$
B.$5.111 \si{s}$
C.$1.231 \si{s}$
D.$0.940 \si{s}$

### Explanation: Refer to the Control Systems section in the Instrumentation, Measurement, and Control chapter of the FE Reference Handbook.

The controller plant, $G(s)$, is $\frac{25}{s(s+5)}$.

This is a unity feedback controller. It can be deduced that this is a unity feedback controller from looking at the classical model of a negative feedback controller, which is shown in the FE Reference Handbook as depicted in the image and comparing it to our system (noting that $G_2(s) = 1$, $H(s) = 1$ and $L(s) = 0$).

A unity feedback model is also shown explicitly in the section on Control Systems.

Thus, $H(s) = 1$ and the closed loop transfer function, $T(s)$, is $\frac{Y(s)}{R(s)}$ (output over the input) which is, $$T(s) = \frac{G(s)}{1+G(s)}$$ Substituting in $G(s)$, $$T(s) = \frac{\frac{25}{s(s+5)}}{1+\frac{25}{s(s+5)}}$$ $$T(s) = \frac{25}{s^2 + 5s + 25}$$ This is a second-order control system of the form shown in the handbook, $$\frac{Y(s)}{R(s)} = \frac{K\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$$ Where, $$\omega_n = \sqrt{25} = 5$$ $$2\zeta\omega_n = 5$$ Substituting $\omega_n$ and solving for $\zeta$ yields, $$\zeta = 0.5$$ Using the values for $\omega_n$ and $\zeta$, calculate the peak time from, $$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}}$$ $$t_p = \frac{\pi}{5\sqrt{1-0.5^2}} = 0.726 \si{s}$$

### 2) In the absorption of ammonia, $\si{NH_3}$, into water from an air-ammonia mixture in an absorption tower at $60^\circ \si F$ and $3.0 \si{atm}$ the individual film coefficients were estimated to be, $$k_L = 1.10 \si{lb\cdot mole\cdot NH_3/(hr\cdot ft^2 mole\cdot NH_3/ft^3)}$$ and $$k_G = 0.25 \si {lb\cdot mole \cdot NH_3/hr \cdot ft^2 \cdot atm}$$ The equilibrium partial pressure of ammonia over-dilute solutions of ammonia in water is given by, $$p_{A_i} = 0.25c_{A_i}$$ Where $p_{A_i}$ is in $\si {atm}$ and $c_{A_i}$ in $\si{mole\cdot NH_3/ft^3}$ of solution. Determine the mass transfer coefficient $K_L$.

A.$0.0592 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta P_A}}$
B.$0.015 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta C_A}}$
C.$1.223 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta C_A}}$
D.$3.765 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta C_A}}$

### Explanation:

Refer to the Convection section in the Chemical Engineering chapter of the FE Reference Handbook.

Apply the equation that relates the individual given coefficients together. $$\frac{1}{K_L}= \frac{1}{Hk_G} + \frac{1}{k_L}$$ The value $H$ is Henry's Law constant where $p_{A_i} = HC_{AL}$ which is given as $0.25$. Plugging in these values yields, $$\frac{1}{K_L}= \frac{1}{(0.25)(0.25)} + \frac{1}{1.10} = 16.91$$ Taking the reciprocal to find $K_L$, $$K_L = 0.0592 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta P_A}}$$

### 3) A continuous stirred-tank reactor (CSTR) is used to treat industrial waste according to a chemical reaction which destroys the waste according to the first-order kinetic equation: $$\frac{dC}{dt}|_{r} = -kC$$ Where $k= 0.2 \si {day}^{-1}$ The reactor volume is $500 \si m^3$, the volumetric flow of the singular intake and outtake is $50 \frac{\si {m^3}}{\si {day}}$, and the inlet waste concentration is $100 \frac{\si{mg}}{\si{L}}$. Determine most nearly the outlet concentration.

A.$33\frac{\si{mg}}{\si{L}}$
B.$45\frac{\si{mg}}{\si{L}}$
C.$100\frac{\si{mg}}{\si{L}}$
D.$21\frac{\si{mg}}{\si{L}}$

### Explanation:

Refer to the Fate and Transport section in the Environmental Engineering chapter of the FE Reference Handbook.

The control volume is the CSTR itself and there is a single intake and outtake. The question asks for the concentration of the outlet only and all values are constants. Thus, it is a steady-state problem and we can begin by writing an expression for the flow, $$\frac{dm}{dt} = \frac{dm_{in}}{dt} - \frac{dm_{out}}{dt} + \frac{dm_{r}}{dt}$$ $$0 = QC_{in} - QC - kCV$$ Solving for the unknown $C$, $$C = C_{in} \cdot \frac{Q}{Q +kV}$$ Substituting in the known values, $$C = 100 \frac{\si{mg}}{\si{L}} \cdot \frac{50 \frac{\si {m^3}}{\si {day}}} {50 \frac{\si {m^3}}{\si {day}}+(0.2 \si {day}^{-1})(500 \si m^3)} = 33.3\frac{\si{mg}}{\si{L}} \approx 33 \frac{\si{mg}}{\si{L}}$$

### 4) Consider the following reaction and the accompanying data for three sets of experiments: $$\si{2NO} + \si{2H_2} \rightarrow \si{N_2}+2\si{H_2O}$$ $$\begin{array}{c|l c|lc |cc } & \text{Exp 1} & & \text{Exp 2} & & \text{Exp 3} \\ \text{Time (Hr)} & \si{[NO]} & \si{[H_2]} & \si{[NO]} & \si{[H_2]} & \si{[NO]} & \si{[H_2]} \\ \hline 0 & 1.00 & 1.00 &2.00 &1.00 &1.00 &2.00 \\ 1.0 & 0.80 & 0.80 &1.20& 0.20 &0.60 &1.60 \\ \end{array}$$ Determine the rate equation for this reaction.

A.$r = k \si{[NO]^2}$
B.$r = k\si{[NO][H_2]}$
C.$r = k \si{[NO]^2[H_2]}$
D.$r = k\si{[H_2]^2}$

### Explanation:

Refer to the Chemical Reaction Engineering section in the Chemical Engineering chapter of the FE Reference Handbook.

The reaction rate equation will have the form, $$r = kC_A^xC_B^y$$ Where $C_A$ and $C_B$ are the reactants and $x$ and $y$ are the respective orders. Thus, we know the expression needs to be of the form, $$r = k \si{[NO]}^x\si{[H_2]}^y$$ Next, we can use the data from the table to determine the reaction order. In experiments 1 and 2, the concentration of $\si{H_2}$ is maintained constant. Thus we can write the ratio, $$\frac{r_1}{r_2} = \frac{k\si{[NO]_1^x[H_2]_1^y}}{k\si{[NO]_2^x [H_2]_2^y}}$$ The rates of reaction for $\si {NO}$ can be deduced from the table as, $$r_1 = \frac{\Delta A_1}{\Delta t} = \frac{1-0.8}{1 \si {hr}} = 0.2 \si{M/hr} \\ r_2 = \frac{\Delta A_2}{\Delta t} = \frac{2-1.2}{1 \si {hr}} = 0.8 \si{M/hr}$$ Plugging in the values (noting that $k$ and the concentration of $\si H_2$ are cancelled out), $$\frac{0.2}{0.8} = \frac{1.0^x}{2.0^x}$$ Solve for $x$, $$\ln (\frac{0.2}{0.8})= x \ln \frac{1}{2} \\ x = 2$$ Apply the same logic to solve for $y$, this time using experiments 1 and 3 where the initial concentrations of $\si {NO}$ are held constant. $$r_1 = \frac{\Delta B_1}{\Delta t} = \frac{1-0.8}{1 \si {hr}} = 0.2 \si{M/hr} \\ r_2 = \frac{\Delta B_2}{\Delta t} = \frac{2-1.6}{1 \si {hr}} = 0.4 \si{M/hr}$$ Using the ratio, $$\frac{0.2}{0.4} = \frac{1.00^y}{2.00^y}$$ Solve for $y$, $$\ln(\frac{0.2}{0.4}) = y\ln\frac{1}{2} \\ y = 1$$ Thus, the final rate equation for this reaction is therefore, $$r = k \si{[NO]^2[H_2]}$$

### 5) Consider the balanced molecular equation expressing the decomposition of nitrogen dioxide into nitric oxide and molecular oxygen. $$2\si{NO_2} \rightarrow 2\si{NO} + \si{O_2}$$ Assume that you are able to follow the rate of decomposition of $\si{NO_2}$ at a certain temperature by the increase in the concentration of $\si{NO}$ (in moles/liter) with the passage of time, $t$. After $1.0 \si {min}$ elapses, the concentration of $\si{NO}$ is $0.10 \si{M}$; and after $5.0\si{min}$ the concentration is $0.70\si{M}$. What is the average rate of formation of $\si{NO}$?

A.$1.725 \text{ mole} \si{liter^{-1}}\si{min}^{-1}$
B.$1.725 \text{ mole} \si{min}^{-1}$
C.$0.15 \text{ mole} \si{liter^{-1}}\si{min}^{-1}$
D.$0.15 \text{ mole} \si{min}^{-1}$

### Explanation:

Refer to the Chemical Reaction Engineering section in the Chemical Engineering chapter of the FE Reference Handbook.

Use the expression to determine a general equation for the rate, $$r = \frac{\si{[NO]_2} - \si{[NO]_1}}{t_2-t_1} = \frac{\Delta[NO]}{\Delta t}$$ Now plug in the given values for the concentrations at times $t_2$ and $t_1$. $$r = \frac{(0.70-0.10) \si {M/liter}}{(5.0 - 1.0) \si{min}} =0.15 \text{ mole} \si{liter^{-1}}\si{min}^{-1}$$

### 6) Calculate the activation energy of a chemical reaction if the pre-exponential factor is $15 \si {M^{-1}s^{-1}}$, and the rate constant is $12\si {M^{-1}s^{-1}}$ at temperature $22\si K$.

A.$29.22\si{J/mol}$
B.$36.47\si{J/mol}$
C.$40.81\si{J/mol}$
D.$53.90\si{J/mol}$

### Explanation:

Refer to the Chemical Reaction Engineering section in the Chemical Engineering chapter of the FE Reference Handbook.

The Arrhenius equation can be applied directly from the givens, $$k = Ae^{\frac{-E_a}{\overline R T}}$$ where \begin{align} A & = \text{pre-exponential factor} \\ E_a &= \text{activation energy} (\si{J/mol, cal/mol)} \\ T &= \text{temperature } ({\si K}) \\ \overline R &= \text{gas law constant} = 8.314 \si {J/(mol \cdot K)} \end{align} Rearrange the equation to solve for $E_a$, $$E_a =- \ln\frac{k}{A} (\overline RT)$$ Plug in the values, $$E_a = \ln \left ( \frac{12}{15} \right ) (8.314 \times 22) = 40.81 \si{J/mol}$$

A.12 kg
B.10 kg
C.14 kg
D.7 kg

### Explanation:

Refer to the Ideal Gas section in the Thermodynamics chapter of the FE Reference Handbook.

Because we are given the pressure, volume, and temperature and wish to find mass, we will use the form of ideal gas law, $$PV = mRT$$ where we must find the gas constant for this particular ideal gas. It is $$R = \frac{\bar{R}}{mol. wt} = \frac{\SI{8.314}{\kilo\pascal\meter\cubed\per\kilo\mole\per\kelvin}}{\SI{24}{\kilo\gram\per\kilo\mole}} = \SI{0.346}{\kilo\pascal\meter\cubed\per\kilo\gram\per\kelvin}$$ Thus the mass is $$m = \frac{PV}{RT} = \frac{(\SI{2050}{\kilo\pascal})\SI{0.6}{\meter\cubed}}{(\SI{0.346}{\kilo\pascal\meter\cubed\per\kilo\gram\per\kelvin})\cdot\SI{ (273+25)}{\kelvin}}=\SI{11.9}{\kilo\gram}$$

### 8) In a rigid tank is a mixture of oxygen, O$_2$, and carbon dioxide, CO$_2$ at $\SI{15}{\celsius}$. The partial pressure of oxygen is $P_{O_2} = \SI{80}{\kilo\pascal}$, the partial pressure of carbon dioxide is $P_{CO_2} = \SI{20}{\kilo\pascal}$. What is the total internal energy of this gas mixture? Assume the internal energies of the two gasses are: $$u_{O_2} = \SI{188.5}{\kilo\joule\per\kilo\gram}\\ u_{CO_2} = \SI{151.1}{\kilo\joule\per\kilo\gram}\\$$

A.$\SI{178.9}{\kilo\joule\per\kilo\gram}$
B.$\SI{169.8}{\kilo\joule\per\kilo\gram}$
C.$\SI{181.0}{\kilo\joule\per\kilo\gram}$
D.$\SI{201.1}{\kilo\joule\per\kilo\gram}$

### Explanation:

Refer to the Ideal Gas Mixtures section in the Thermodynamics chapter of the FE Reference Handbook.

The internal energy of a mixture of ideal gasses is just the mass-weighted sum of the internal energy of the constituent gasses, $$u = \sum{y_iu_i} = y_{O_2}u_{O_2} + y_{CO_2}u_{CO_2}$$ where $y_i$ is the mass fraction of constituent $i$. In the problem statement, we are only given the partial pressures which means we can find the mole fraction, $$x_i = \frac{P_i}{P}$$ or, $$x_{O_2} = \frac{P_{O_2}}{P_{O_2} + P_{CO_2}} = \frac{80}{80 + 20} = 0.8\\ x_{CO_2} = \frac{P_{CO_2}}{P_{O_2} + P_{CO_2}} = 0.2$$ Which we can then use to convert from mole fraction to mass fraction via the formula given in the FE Handbook, $$y_i = \frac{x_i M_i}{\sum{x_i M_i}}$$ which for oxygen is $$y_{O_2} = \frac{0.8 (\SI{32}{\kilo\gram\per\kilo\mole})}{\left[ 0.8 (\SI{32}{\kilo\gram\per\kilo\mole}) \right] + \left[ 0.2 (\SI{44}{\kilo\gram\per\kilo\mole}) \right]} = 0.744$$ where $M_{O_2}$ and $M_{CO_2}$ can be found in the back of the Thermodynamics chapter in the Thermal and Physical Property Table. At this point, we could use the same formula to calculate the mole fraction of CO$_2$ but we can also use a shortcut because we know the total sum of mass fraction must equal 1. In this case we have only 2 constituent gases so we can straight away calculate $y_{CO_2} = 1 - 0.744 = 0.256$.

Finally we can evaluate for the total internal energy, $$u = y_{O_2}u_{O_2} + y_{CO_2}u_{CO_2} = 0.744\cdot\SI{188.5}{\kilo\joule\per\kilo\gram} + 0.256\cdot\SI{151.1}{\kilo\joule\per\kilo\gram} \\ = \SI{178.9}{\kilo\joule\per\kilo\gram}$$

### 9) A piston-cylinder contains $\SI{1}{\kilo\gram}$ water at $\SI{20}{\celsius}$ with an absolute pressure of $\SI{101.3}{\kilo\pascal}$. The piston is pulled upward until the pressure in the piston is reduced to $\SI{50}{\kilo\pascal}$. The process is done slowly and the liquid and any vapor in the piston remain at $\SI{20}{\celsius}$. What can be said of the phase of water in the piston?

A.Compressed liquid
B.Liquid-vapor mixture
C.Superheated steam
D.Super-critical phase

### Explanation:

Refer to the Steam Tables section in the Thermodynamics chapter of the FE Reference Handbook.

In the steam tables, we can see for a liquid at $\SI{20}{\celsius}$ the saturate pressure is $\SI{2.339}{\kilo\pascal}$. This tells us that, at this temperature, water will remain a liquid until the pressure is reduced to $\SI{2.339}{\kilo\pascal}$. Our system pressure, $\SI{50}{\kilo\pascal}$, is clearly higher and thus we remain in the compressed liquid phase (this is also referred to by some as the sub-cooled liquid phase)

### 10) A funnel method is to be used to determine the flowability of a powder. The following table shows flowability expected for a given range of angles of repose. $$\begin{array} {|r|r|}\hline \textbf{Flowability Expected} & \textbf{Angle of Repose} \\ \hline \text{Excellent/Very Free Flow} & 25-30 \\ \hline \text{Good/Free Flow} & 31-35 \\ \hline \text{Fair (discharging aid may not be required)} & 36-40 \\ \hline \text{Passable} & 41-45 \\ \hline \text{Poor Flow/Cohesive (discharging aid required)}& 46-55 \\ \hline \text{Very Poor Flow/Very Cohesive} & 56-65 \\ \hline \text{Approximately No Flow} & >66 \\ \hline \end{array}$$ If the measured height of the pile after depositing is $100\si{cm}$ and the radius of the pile is $200\si{cm}$ then the flowability of the powder is:

A.Excellent/Very Free Flow
B.Good/Free Flow
C.Fair
D.Poor Flow
E.Very Poor Flow
F.Approximately No Flow
The angle of repose is now the angle of a right triangle which is adjacent to half the length of the diameter, $d$, and opposite the length of the height $h$. In other words, the trigonometric tangent function. Calling the angle of repose $\theta$, we have, $$\tan{\theta} = \frac{h}{\frac{1}{2}d}$$ The radius is half the diameter by definition, so, $$\tan{\theta} = \frac{h}{r}$$ Solving for $\theta$, $$\theta = \tan^{-1}\frac{h}{r}$$ Plug in the given values of $r$ and $h$, $$\theta = \tan^{-1}\frac{100\si{cm}}{200\si{cm}} = 26.56^\circ$$ Thus, according to the table, this powder has an Excellent/Very Free Flow. 