FE Chemical Exam: 10 Practice Problems with Detailed Solutions

Answer: C

This is a straightforward stoichiometry problem involving excess air.

Step 1: Determine the theoretical (stoichiometric) oxygen requirement. From the balanced equation, 1 mol CH\(_4\) requires 2 mol O\(_2\):

$$\text{Theoretical O}_2 = 100 \times 2 = 200 \text{ mol}$$

Step 2: Apply the 20% excess. “Percent excess” means the percentage above the theoretical requirement:

$$\text{Actual O}_2 = 200 \times (1 + 0.20) = 200 \times 1.20$$

$$\text{Actual O}_2 = \mathbf{240 \text{ mol}}$$

With 20% excess air, 20% more oxygen is supplied than what is stoichiometrically required. The excess oxygen passes through unreacted and appears in the flue gas. Choice A (200 mol) is the theoretical amount without excess. Choice B (220 mol) would correspond to 10% excess. Always multiply the theoretical requirement by \((1 + \text{fraction excess})\).

Answer: B

Apply a steady-state material balance on the sugar (the non-volatile component).

Step 1: Define the system. Let \(F\) = feed rate, \(P\) = product rate, and \(V\) = vapor (water evaporated) rate.

Step 2: Write the sugar balance. Since no sugar leaves in the vapor:

$$F \cdot x_F = P \cdot x_P$$

$$1000 \times 0.10 = P \times 0.50$$

$$100 = 0.50P$$

$$P = \frac{100}{0.50} = \mathbf{200 \text{ kg/h}}$$

Step 3: Verify with the overall balance:

$$V = F - P = 1000 - 200 = 800 \text{ kg/h}$$

This confirms we evaporate 800 kg/h of water, concentrating the sugar from 10% to 50%. The key insight is to write the balance on the component that does not leave in the vapor stream. Choice D (800 kg/h) is the vapor rate, not the product rate — a common trap if you confuse the two exit streams.

Answer: C

For an adiabatic reactor (\(Q = 0\)), all the heat of reaction goes into raising the temperature of the process stream.

Step 1: Calculate the rate of heat generation from the reaction:

$$\dot{Q}_{\text{rxn}} = \dot{n}_A \times (-\Delta H_{\text{rxn}}) = 2 \text{ mol/s} \times 50{,}000 \text{ J/mol}$$

$$\dot{Q}_{\text{rxn}} = 100{,}000 \text{ J/s}$$

Step 2: Apply the energy balance. For an adiabatic system, the heat generated equals the sensible heat absorbed by the stream:

$$\dot{Q}_{\text{rxn}} = \sum \dot{n}_i C_{p,i} \cdot \Delta T$$

$$100{,}000 = 500 \times \Delta T$$

$$\Delta T = \frac{100{,}000}{500} = \mathbf{200 \text{ K}}$$

In an adiabatic exothermic reactor, the temperature always increases. The magnitude of the rise depends on the ratio of heat released to the heat capacity of the stream. Choice B (100 K) results from forgetting to convert kJ to J. This type of energy balance calculation is fundamental to non-isothermal reactor design and is commonly tested on the FE Chemical exam.

Answer: B

Raoult’s law states that the partial pressure of each component equals its liquid mole fraction multiplied by its saturation pressure:

$$p_i = x_i \cdot P^{\text{sat}}_i$$

Step 1: Calculate the partial pressure of each component:

$$p_{\text{benzene}} = x_{\text{B}} \cdot P^{\text{sat}}_{\text{B}} = 0.40 \times 150 = 60 \text{ kPa}$$

$$p_{\text{toluene}} = x_{\text{T}} \cdot P^{\text{sat}}_{\text{T}} = 0.60 \times 60 = 36 \text{ kPa}$$

Step 2: The total pressure at the bubble point is the sum of partial pressures:

$$P = p_{\text{benzene}} + p_{\text{toluene}} = 60 + 36 = \mathbf{96 \text{ kPa}}$$

Step 3: As a check, calculate the vapor composition:

$$y_{\text{benzene}} = \frac{p_{\text{benzene}}}{P} = \frac{60}{96} = 0.625$$

This confirms that benzene is enriched in the vapor phase (62.5% vs. 40% in the liquid), which makes physical sense because benzene is the more volatile component. Raoult’s law VLE calculations are among the most commonly tested thermodynamics problems on the FE Chemical exam. The benzene-toluene system is a classic example of a nearly ideal binary mixture.

Answer: B

The Reynolds number determines whether pipe flow is laminar or turbulent.

Step 1: Calculate the cross-sectional area of the pipe:

$$A = \frac{\pi D^2}{4} = \frac{\pi (0.050)^2}{4} = 1.963 \times 10^{-3} \text{ m}^2$$

Step 2: Calculate the average velocity:

$$v = \frac{Q}{A} = \frac{2.0 \times 10^{-3}}{1.963 \times 10^{-3}} = 1.019 \text{ m/s}$$

Step 3: Calculate the Reynolds number:

$$\text{Re} = \frac{\rho v D}{\mu} = \frac{1000 \times 1.019 \times 0.050}{1.0 \times 10^{-3}}$$

$$\text{Re} = \frac{50.93}{1.0 \times 10^{-3}} = \mathbf{50{,}930}$$

Step 4: Determine the flow regime. Since Re > 4000, the flow is fully turbulent.

For pipe flow: Re < 2300 is laminar, 2300 < Re < 4000 is transitional, and Re > 4000 is turbulent. Choice A would result from using the diameter in cm instead of m in the velocity calculation. Choice D would correspond to a much lower flow rate. Always convert all units to a consistent system (SI) before calculating.

Answer: B

For a counterflow heat exchanger, the temperature differences at each end are found by pairing the hot inlet with the cold outlet, and the hot outlet with the cold inlet.

Step 1: Identify the terminal temperature differences:

$$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 150 - 75 = 75\text{°C}$$

$$\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 60 - 25 = 35\text{°C}$$

Step 2: Apply the LMTD formula:

$$\text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{75 - 35}{\ln(75 / 35)}$$

$$\text{LMTD} = \frac{40}{\ln(2.143)} = \frac{40}{0.7621}$$

$$\text{LMTD} = \mathbf{52.5\text{°C}}$$

Note that the LMTD (52.5°C) is less than the arithmetic mean of the two temperature differences, which would be \((75 + 35)/2 = 55\)°C (choice C). The LMTD is always less than or equal to the arithmetic mean — they are equal only when \(\Delta T_1 = \Delta T_2\). Choice D (75°C) is simply the larger terminal temperature difference. On the FE exam, always verify whether the problem specifies parallel flow or counterflow, as this changes which temperatures are paired at each end.

Answer: B

For a first-order reaction in a CSTR, the design equation gives a direct relationship between conversion and the Damköhler number.

Step 1: Write the CSTR design equation for a first-order reaction. The CSTR mole balance is:

$$\tau = \frac{C_{A0} \cdot X}{(-r_A)}$$

For a first-order reaction, \(-r_A = kC_A = kC_{A0}(1 - X)\). Substituting:

$$\tau = \frac{X}{k(1 - X)}$$

Step 2: Calculate the Damköhler number:

$$Da = k\tau = 0.5 \times 4 = 2.0$$

Step 3: Solve for conversion:

$$k\tau = \frac{X}{1 - X}$$

$$2.0 = \frac{X}{1 - X}$$

$$2.0(1 - X) = X$$

$$2.0 = 3.0X$$

$$X = \frac{2.0}{3.0} = \mathbf{0.667 = 66.7\%}$$

For comparison, a PFR with the same space time and rate constant would achieve \(X = 1 - e^{-k\tau} = 1 - e^{-2} = 86.5\%\) (choice D). This illustrates the well-known result that a PFR always requires less volume than a CSTR for the same conversion of a positive-order reaction. Choice D is a common trap if you accidentally use the PFR equation instead of the CSTR equation.

Answer: B

The Fenske equation gives the minimum number of equilibrium stages at total reflux:

$$N_{\min} = \frac{\ln\left[\left(\frac{x_D}{1 - x_D}\right)\left(\frac{1 - x_B}{x_B}\right)\right]}{\ln(\alpha)}$$

Step 1: Calculate the argument of the numerator logarithm:

$$\frac{x_D}{1 - x_D} = \frac{0.95}{0.05} = 19$$

$$\frac{1 - x_B}{x_B} = \frac{0.95}{0.05} = 19$$

$$\text{Product} = 19 \times 19 = 361$$

Step 2: Calculate the minimum stages:

$$N_{\min} = \frac{\ln(361)}{\ln(2.5)} = \frac{5.889}{0.9163}$$

$$N_{\min} = 6.43$$

Step 3: Since we need whole equilibrium stages, round up:

$$N_{\min} = \mathbf{7 \text{ stages}}$$

The Fenske equation is found in the FE Reference Handbook and is one of the most important distillation equations on the exam. It gives the absolute minimum number of stages needed for a given separation at total reflux (infinite reflux ratio). The actual number of stages at finite reflux will always be higher. Note that higher relative volatility means easier separation (fewer stages), and that the equation uses mole fractions of the light key component in both distillate and bottoms.

Answer: A

The standard form of a first-order transfer function is:

$$G(s) = \frac{K}{\tau s + 1}$$

where \(K\) is the steady-state gain and \(\tau\) is the time constant.

Step 1: Compare the given transfer function to the standard form:

$$G(s) = \frac{3}{5s + 1}$$

By direct comparison: \(K = 3\) and \(\tau = 5\).

Step 2: For a unit step input, the steady-state output equals the gain \(K\) multiplied by the step magnitude:

$$y(\infty) = K \times 1 = 3$$

Alternatively, apply the final value theorem:

$$y(\infty) = \lim_{s \to 0} s \cdot G(s) \cdot \frac{1}{s} = \lim_{s \to 0} \frac{3}{5s + 1} = \frac{3}{1} = \mathbf{3}$$

The time constant \(\tau = 5\) means the output reaches 63.2% of its final value after one time constant, and approximately 95% after \(3\tau = 15\) time units. Choice B reverses the gain and time constant. Choice D (15) is the product \(K \times \tau\), not a meaningful quantity. First-order transfer functions are the most commonly tested process control topic on the FE Chemical exam. Know how to read off \(K\) and \(\tau\) from the standard form and how to apply the final value theorem.

Answer: B

The flammable range of a gas lies between the Lower Explosive Limit (LEL) and the Upper Explosive Limit (UEL). A mixture can only ignite if its concentration falls within this range.

Step 1: Compare the measured concentration to the flammable range:

$$\text{Measured: } 2.0\% \quad \text{vs.} \quad \text{LEL: } 5.0\% \quad \text{UEL: } 15.0\%$$

Since 2.0% < 5.0% (the LEL), the mixture is below the lower explosive limit.

Step 2: Calculate the percentage of LEL:

$$\%\text{LEL} = \frac{2.0}{5.0} \times 100 = \mathbf{40\% \text{ of LEL}}$$

A concentration below the LEL means there is not enough fuel in the mixture to sustain combustion — the mixture is “too lean.” However, even though the mixture cannot currently ignite, 40% of the LEL is still a significant concentration that warrants attention and monitoring. Standard industrial practice typically triggers alarms at 10–25% of the LEL.

Choice D is incorrect because while the 40% LEL value is correct, the conclusion (“requires immediate evacuation”) is overstated — the mixture is not in the flammable range. The correct action is to investigate the source of the leak and increase ventilation, not necessarily evacuate. On the FE exam, safety questions test whether you understand the meaning of flammability limits, not just the numerical values.