Free FE Practice Test# PrepFEâ„¢

## 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.

◯ A.

$r = k \si{[NO]^2}$

◯ B.

$r = k\si{[NO][H_2]}$

◯ C.

$r = k \si{[NO]^2[H_2]}$

◯ D.

$r = k\si{[H_2]^2}$

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.

$0.15 \text{ mole} \si{min}^{-1}$

◯ A.

Excellent/Very Free Flow

◯ B.

Good/Free Flow

◯ C.

Fair

◯ D.

Poor Flow

◯ E.

Very Poor Flow

◯ F.

Approximately No Flow

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.

$3.765 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta C_A}}$

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.

$\SI{201.1}{\kilo\joule\per\kilo\gram}$

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.

$21\frac{\si{mg}}{\si{L}}$

◯ A.

$29.22\si{J/mol}$

◯ B.

$36.47\si{J/mol}$

◯ C.

$40.81\si{J/mol}$

◯ D.

$53.90\si{J/mol}$

◯ A.

12 kg

◯ B.

10 kg

◯ C.

14 kg

◯ D.

7 kg

◯ A.

$0.726 \si{s}$

◯ B.

$5.111 \si{s}$

◯ C.

$1.231 \si{s}$

◯ D.

$0.940 \si{s}$

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

A.$r = k \si{[NO]^2}$

B.$r = k\si{[NO][H_2]}$

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

D.$r = k\si{[H_2]^2}$

The correct answer is C.

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, $$r = k \si{[NO]^2[H_2]}$$

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.$0.15 \text{ mole} \si{min}^{-1}$

The correct answer is C.

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 $t_1$. $$r = \frac{(0.70-0.10) \si {M/liter}}{(5.0 - 1.0) \si{min}} =0.15 \text{ mole} \si{liter^{-1}}\si{min}^{-1}$$

A.Excellent/Very Free Flow

B.Good/Free Flow

C.Fair

D.Poor Flow

E.Very Poor Flow

F.Approximately No Flow

The correct answer is A.

The pile of solids is formed in the shape of a cone by the depositing funnel. To calculate the angle of repose, we can drop a perpendicular from the top of the cone to the base to show the height and draw a bisector across the base to show the diameter.

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.

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.$3.765 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta C_A}}$

The correct answer is A.

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 $K_L$, $$K_L = 0.0592 \frac{\si{lb\cdot mole}}{\si{hr\cdot ft^2 \cdot \Delta P_A}}$$

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.$\SI{201.1}{\kilo\joule\per\kilo\gram}$

The correct answer is A.

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

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

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.$21\frac{\si{mg}}{\si{L}}$

The correct answer is A.

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, $$C = 100 \frac{\si{mg}}{\si{L}} \cdot \frac{50 \frac{\si {m^3}}{\si {day}}} {50 \frac{\si {m^3}}{\si {day}}+(0.2 \si {day}^{-1})(500 \si m^3)} = 33.3\frac{\si{mg}}{\si{L}} \approx 33 \frac{\si{mg}}{\si{L}} $$

A.$29.22\si{J/mol}$

B.$36.47\si{J/mol}$

C.$40.81\si{J/mol}$

D.$53.90\si{J/mol}$

The correct answer is C.

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

The correct answer is A.

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 $$ m = \frac{PV}{RT} = \frac{(\SI{2050}{\kilo\pascal})\SI{0.6}{\meter\cubed}}{(\SI{0.346}{\kilo\pascal\meter\cubed\per\kilo\gram\per\kelvin})\cdot\SI{ (273+25)}{\kelvin}}=\SI{11.9}{\kilo\gram} $$

A.$0.726 \si{s}$

B.$5.111 \si{s}$

C.$1.231 \si{s}$

D.$0.940 \si{s}$

The correct answer is A.

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, $$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}}$$ $$t_p = \frac{\pi}{5\sqrt{1-0.5^2}} = 0.726 \si{s}$$

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

The correct answer is A.

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)