Chemical Reaction Engineering: cram pack

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

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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 = ?90%

A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.2 L/(mol·min) and starts at 0.5 mol/L. How long does it take to reach 90% conversion? Find t.

k = 0.2 L/(mol·min)CA0 = 0.5 mol/LX = 90 %
t = min
2.Reaction kineticsHard
Tk25 °C0.00245 °C?

A reaction has a rate constant of 0.002 min⁻¹ at 25 °C and an activation energy of 80 kJ/mol. Find the rate constant k2 at 45 °C.

k1 = 0.002 min⁻¹T1 = 25 °CT2 = 45 °CEa = 80 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹
3.Reactor sizing (CSTR & PFR)Medium
Q = 100 L/minX = 80%V = ?, k = 0.2 min⁻¹

A first-order liquid-phase reaction A → B with k = 0.2 min⁻¹ is carried out in a plug-flow reactor (PFR) fed at 100 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 = 100 L/mink = 0.2 min⁻¹X = 80%
V = L
4.Reactor sizing (CSTR & PFR)Hard
V = 400 LV = 400 LQ = 100 L/minX = ?

A first-order liquid-phase reaction (k = 0.4 min⁻¹) runs in two equal CSTRs in series, each of 400 L. The feed rate is 100 L/min, the density is constant and the reactors are at steady state. Find the overall conversion X in percent.

V = 400 L (×2)Q = 100 L/mink = 0.4 min⁻¹
X = %
5.Chemical equilibriumMedium
H2​I2​HII0.20.20C−x−x+2xE0.2 − x0.2 − x2x

0.2 mol/L of H₂ and 0.2 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 = 0.2 mol/LKc = 49
[H2] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.400C−x+x+xE0.4 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 at the reaction temperature. A container starts with 0.4 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].

[PCl₅]₀ = 0.4 mol/LKc = 0.05
[PCl5] = mol/L
7.Mass balanceMedium
MixerStream 1450 kg/h · 40% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct25% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 25% NaOH? Find ṁ2.

ṁ1 = 450 kg/hx1 = 40 %x2 = 15 %x3 = 25 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F1600 kg/h · 20% NaOHProduct P60% NaOHVapour W (pure water)W = ? kg/h

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

F = 1600 kg/hxF = 20 %xP = 60 %
W = kg/h

3Solutions

1.Reaction kinetics

Answer: t = 90 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 − 0.9) = 0.05 mol/L

Integrated second-order law

t = (1/CA − 1/CA0)/k = (20 − 2)/0.2 = 90 min

Answer

t = 90 min
2.Reaction kinetics

Answer: k2 = 0.01521 min⁻¹

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 temperatures

T1 = 298.1 K, T2 = 318.1 K

Arrhenius ratio

ln(k2/k1) = −(80000/8.314)·(1/318.1 − 1/298.1) = 2.029
k2 = 0.002·e2.029 = 0.01521 min⁻¹

Answer

k2 = 0.01521 min⁻¹
3.Reactor sizing (CSTR & PFR)

Answer: V = 804.7 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))/0.2 = 1.609/0.2 = 8.047 min

Volume

V = Q·τ = 100·8.047 = 804.7 L

Answer

V = 804.7 L
4.Reactor sizing (CSTR & PFR)

Answer: X = 85.21 %

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 = 400/100 = 4 min, kτ = 0.4·4 = 1.6

Conversion

C1/CA0 = 1/(1 + kτ) = 1/(1 + 1.6) = 0.3846
C2/CA0 = 1/(1 + kτ)² = 0.1479
X = 1 − 0.1479 = 85.21%

Answer

X = 85.21%
5.Chemical equilibrium

Answer: [H2] = 0.04444 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/(0.2 − x)²

Take the square root

7 = 2x/(0.2 − x) → x = 0.2·7/(2 + 7) = 0.1556 mol/L

Equilibrium concentration

[H₂] = 0.2 − x = 0.04444 mol/L

Answer

[H2] = 0.04444 mol/L
6.Chemical equilibrium

Answer: [PCl5] = 0.2814 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

0.05 = x²/(0.4 − x) → x² + 0.05·x − 0.02 = 0

Solve the quadratic

x = (−0.05 + √(0.05² + 4·0.02))/2 = 0.1186 mol/L

Equilibrium concentration

[PCl₅] = 0.4 − 0.1186 = 0.2814 mol/L

Answer

[PCl5] = 0.2814 mol/L
7.Mass balance

Answer: ṁ2 = 675 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 450·(40 − 25)/(25 − 15) = 675 kg/h

Answer

ṁ2 = 675 kg/h
8.Mass balance

Answer: W = 1067 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 = 1600·20/60 = 533.3 kg/h

Total balance

W = F − P = 1600 − 533.3 = 1067 kg/h

Answer

W = 1067 kg/h

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