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)C (mol/L)t½​ = ?C0​/2

A second-order reaction (−rA = k·CA²) has k = 0.05 L/(mol·min) and an initial concentration of 2 mol/L. Find its first half-life t½.

k = 0.05 L/(mol·min)CA0 = 2 mol/L
t½ = min
2.Reaction kineticsHard
1/Tln kT = 100 °Cln 0.01ln A = ?Ea​ = 80 kJ/mol

A first-order reaction has a rate constant of 0.01 s⁻¹ at 100 °C and an activation energy of 80 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.01 s⁻¹T = 100 °CEa = 80 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹
3.Reactor sizing (CSTR & PFR)Medium
Q = 100 L/minX = 75%V = ?, k = 0.12 min⁻¹

A first-order liquid-phase reaction A → B with k = 0.12 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 75% conversion of A.

Q = 100 L/mink = 0.12 min⁻¹X = 75%
V = L
4.Reactor sizing (CSTR & PFR)Hard
V = 1000 LV = 1000 LQ = 40 L/minX = ?

A first-order liquid-phase reaction (k = 0.15 min⁻¹) runs in two equal CSTRs in series, each of 1000 L. The feed rate is 40 L/min, the density is constant and the reactors are at steady state. Find the overall conversion X in percent.

V = 1000 L (×2)Q = 40 L/mink = 0.15 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 = 16. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 16
[HI] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I100C−x+x+xE1 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 1 mol/LKc = 0.1
[Cl2] = mol/L
7.Mass balanceMedium
MixerStream 1650 kg/h · 20% saltStream 2ṁ₂ = ? · 35% saltProduct30% salt

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% salt? Find ṁ2.

ṁ1 = 650 kg/hx1 = 20 %x2 = 35 %x3 = 30 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F3300 kg/h · 19% NaOHProduct PP = ? · 45%Vapour W (pure water)0% NaOH

An evaporator concentrates a NaOH solution by boiling off pure water. Find the product flow P in kg/h.

F = 3300 kg/hxF = 19 %xP = 45 %
P = kg/h

3Solutions

1.Reaction kinetics

Answer: t½ = 10 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).

Second-order half-life

t½ = 1/(k·CA0) = 1/(0.05·2) = 10 min

Answer

t½ = 10 min

Each later half-life is twice as long, because C₀ has halved.

2.Reaction kinetics

Answer: A = 1.581 × 10⁹ s⁻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 temperature

T = 100 + 273.15 = 373.1 K

Solve Arrhenius for A

Ea/(RT) = 80000/(8.314·373.1) = 25.79
A = k·e25.79 = 0.01·1.581 × 10¹¹ = 1.581 × 10⁹ s⁻¹

Answer

A = 1.581 × 10⁹ s⁻¹
3.Reactor sizing (CSTR & PFR)

Answer: V = 1155 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 − 0.75))/0.12 = 1.386/0.12 = 11.55 min

Volume

V = Q·τ = 100·11.55 = 1155 L

Answer

V = 1155 L
4.Reactor sizing (CSTR & PFR)

Answer: X = 95.57 %

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 = 1000/40 = 25 min, kτ = 0.15·25 = 3.75

Conversion

C1/CA0 = 1/(1 + kτ) = 1/(1 + 3.75) = 0.2105
C2/CA0 = 1/(1 + kτ)² = 0.04432
X = 1 − 0.04432 = 95.57%

Answer

X = 95.57%
5.Chemical equilibrium

Answer: [HI] = 0.2667 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

16 = (2x)²/(0.2 − x)²

Take the square root

4 = 2x/(0.2 − x) → x = 0.2·4/(2 + 4) = 0.1333 mol/L

Equilibrium concentration

[HI] = 2x = 0.2667 mol/L

Answer

[HI] = 0.2667 mol/L
6.Chemical equilibrium

Answer: [Cl2] = 0.2702 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

0.1 = x²/(1 − x) → x² + 0.1·x − 0.1 = 0

Solve the quadratic

x = (−0.1 + √(0.1² + 4·0.1))/2 = 0.2702 mol/L

Equilibrium concentration

[Cl₂] = x = 0.2702 mol/L

Answer

[Cl2] = 0.2702 mol/L
7.Mass balance

Answer: ṁ2 = 1300 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) = 650·(20 − 30)/(30 − 35) = 1300 kg/h

Answer

ṁ2 = 1300 kg/h
8.Mass balance

Answer: P = 1393 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 = 3300·19/45 = 1393 kg/h

Answer

P = 1393 kg/h

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