Chemical Reaction Engineering: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Reaction kinetics

First order
ln(CA/CA0) = −kt
Half-life (first order)
t½ = ln 2/k
Second order
1/CA = 1/CA0 + kt
Arrhenius
k = A·e−Ea/(RT)
Two temperatures
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2)

Reactor sizing (CSTR & PFR)

Space time
τ = V/Q
CSTR, first order
τ = X/(k·(1 − X)), CA = CA0/(1 + kτ)
PFR, first order
τ = ln(1/(1 − X))/k, CA = CA0·e−kτ
Conversion from volume
CSTR: X = kτ/(1 + kτ), PFR: X = 1 − e−kτ
N equal CSTRs in series
X = 1 − 1/(1 + kτi)N

Chemical equilibrium

Equilibrium constant
Kc = [C]c[D]d/([A]a[B]b)
ICE table
Initial, Change (×coefficient), Equilibrium
Quadratic
x = (−b + √(b² − 4ac))/(2a)

Method

  1. Write Kc with products over reactants, each raised to the power of its coefficient.
  2. Set up an ICE table: Initial, Change (in steps of x times the coefficients), Equilibrium.
  3. Put the equilibrium expressions into Kc and solve for x.
  4. Check whether x is small enough for an approximation; otherwise solve the quadratic.

Mass balance

Steady state
in = out
Total balance
ṁ1 + ṁ2 = ṁ3
Component balance
ṁ1x1 + ṁ2x2 = ṁ3x3
Evaporator
F·xF = P·xP, W = F − P

2Practice problems

1.Reaction kineticsMedium
t (min)X (%)t = ?75%

A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.05 L/(mol·min) and starts at 0.5 mol/L. How long does it take to reach 75% conversion? Find t.

k = 0.05 L/(mol·min)CA0 = 0.5 mol/LX = 75 %
t = min
2.Reaction kineticsHard
Tk50 °Ck1​?10·k1​

A reaction has an activation energy of 55 kJ/mol. At 50 °C it is too slow. To what temperature T2 (in °C) must it be heated to make it 10 times faster?

Ea = 55 kJ/molT1 = 50 °Ck2/k1 = 10R = 8.314 J/(mol·K)
T2 = °C
3.Reactor sizing (CSTR & PFR)Medium
Q = 10 L/minX = 80%V = ?, k = 0.5 min⁻¹

A first-order liquid-phase reaction A → B with k = 0.5 min⁻¹ is carried out in a plug-flow reactor (PFR) fed at 10 L/min. The density is constant and the reactor is at steady state. Find the reactor volume V in litres for 80% conversion of A.

Q = 10 L/mink = 0.5 min⁻¹X = 80%
V = L
4.Reactor sizing (CSTR & PFR)Hard
V = 250 LV = 250 LQ = 25 L/minX = ?

A first-order liquid-phase reaction (k = 0.15 min⁻¹) runs in two equal CSTRs in series, each of 250 L. The feed rate is 25 L/min, the density is constant and the reactors are at steady state. Find the overall conversion X in percent.

V = 250 L (×2)Q = 25 L/mink = 0.15 min⁻¹
X = %
5.Chemical equilibriumMedium
H2​I2​HII0.50.50C−x−x+2xE0.5 − x0.5 − x2x

0.5 mol/L of H₂ and 0.5 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 49. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 49
[H2] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.800C−x+x+xE0.8 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.8 mol/LKc = 0.04
[Cl2] = mol/L
7.Mass balanceMedium
MixerStream 1550 kg/h · 10% saltStream 2ṁ₂ = ? · 50% saltProduct40% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 40% salt? Find ṁ2.

ṁ1 = 550 kg/hx1 = 10 %x2 = 50 %x3 = 40 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F5000 kg/h · 16% saltProduct P35% saltVapour W (pure water)W = ? kg/h

An evaporator concentrates a salt solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 5000 kg/hxF = 16 %xP = 35 %
W = kg/h

3Solutions

1.Reaction kinetics

Answer: t = 120 min

Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).

Concentration at the end

CA = 0.5·(1 − 0.75) = 0.125 mol/L

Integrated second-order law

t = (1/CA − 1/CA0)/k = (8 − 2)/0.05 = 120 min

Answer

t = 120 min
2.Reaction kinetics

Answer: T2 = 90.95 °C

Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).

Absolute temperature

T1 = 323.1 K

Solve Arrhenius for T₂

1/T2 = 1/323.1 − 8.314·ln(10)/55000 = 3.095 × 10⁻³ − 3.481 × 10⁻⁴ = 2.746 × 10⁻³ K⁻¹
T2 = 364.1 K = 90.95 °C

Answer

T2 = 90.95 °C
3.Reactor sizing (CSTR & PFR)

Answer: V = 32.19 L

Ideal reactors, first-order reaction (−rA = k·CA) at steady state, with space time τ = V/Q. A CSTR is perfectly mixed, so the whole volume reacts at the low outlet concentration: τ = X/(k·(1 − X)) and CA = CA0/(1 + k·τ). In a PFR each slice of fluid reacts like a small batch reactor: τ = ln(1/(1 − X))/k and CA = CA0·e−k·τ. For the same conversion a CSTR always needs more volume than a PFR.

Space time

τ = ln(1/(1 − 0.8))/0.5 = 1.609/0.5 = 3.219 min

Volume

V = Q·τ = 10·3.219 = 32.19 L

Answer

V = 32.19 L
4.Reactor sizing (CSTR & PFR)

Answer: X = 84 %

Ideal reactors, first-order reaction (−rA = k·CA) at steady state, with space time τ = V/Q. A CSTR is perfectly mixed, so the whole volume reacts at the low outlet concentration: τ = X/(k·(1 − X)) and CA = CA0/(1 + k·τ). In a PFR each slice of fluid reacts like a small batch reactor: τ = ln(1/(1 − X))/k and CA = CA0·e−k·τ. For the same conversion a CSTR always needs more volume than a PFR.

Space time

τ = V/Q = 250/25 = 10 min, kτ = 0.15·10 = 1.5

Conversion

C1/CA0 = 1/(1 + kτ) = 1/(1 + 1.5) = 0.4
C2/CA0 = 1/(1 + kτ)² = 0.16
X = 1 − 0.16 = 84%

Answer

X = 84%
5.Chemical equilibrium

Answer: [H2] = 0.1111 mol/L

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

49 = (2x)²/(0.5 − x)²

Take the square root

7 = 2x/(0.5 − x) → x = 0.5·7/(2 + 7) = 0.3889 mol/L

Equilibrium concentration

[H₂] = 0.5 − x = 0.1111 mol/L

Answer

[H2] = 0.1111 mol/L
6.Chemical equilibrium

Answer: [Cl2] = 0.16 mol/L

Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.

Equilibrium expression

0.04 = x²/(0.8 − x) → x² + 0.04·x − 0.032 = 0

Solve the quadratic

x = (−0.04 + √(0.04² + 4·0.032))/2 = 0.16 mol/L

Equilibrium concentration

[Cl₂] = x = 0.16 mol/L

Answer

[Cl2] = 0.16 mol/L
7.Mass balance

Answer: ṁ2 = 1650 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
salt: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 550·(10 − 40)/(40 − 50) = 1650 kg/h

Answer

ṁ2 = 1650 kg/h
8.Mass balance

Answer: W = 2714 kg/h

Tie component: a component that passes through unchanged in one stream (here the solute) links the flows directly. Then the total balance gives the rest.

Solute balance

F·xF = P·xP → P = 5000·16/35 = 2286 kg/h

Total balance

W = F − P = 5000 − 2286 = 2714 kg/h

Answer

W = 2714 kg/h

ExamProblems · examproblems.com/courses/chemical-reaction-engineering