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)60 min0.6

In a first-order reaction the concentration of A falls from 1 mol/L to 0.6 mol/L in 60 min. Find the rate constant k.

CA0 = 1 mol/LCA = 0.6 mol/Lt = 60 min
k = min⁻¹
2.Reaction kineticsHard
1/Tln kT = 25 °Cln 0.002ln A = ?Ea​ = 70 kJ/mol

A first-order reaction has a rate constant of 0.002 s⁻¹ at 25 °C and an activation energy of 70 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.

k = 0.002 s⁻¹T = 25 °CEa = 70 kJ/molR = 8.314 J/(mol·K)
A = s⁻¹
3.Reactor sizing (CSTR & PFR)Medium
QX = 60%same Q, k and X as the CSTR

A first-order liquid-phase reaction must reach 60% conversion. How many times larger must a CSTR be than a PFR for the same feed rate, rate constant and conversion? Find the ratio VCSTR/VPFR.

X = 60%first order, same k and Q
VCSTR/VPFR =
4.Reactor sizing (CSTR & PFR)Hard
Q = 80 L/minX = ?V = 100 L

A first-order liquid-phase reaction (k = 0.25 min⁻¹) runs in a 100 L plug-flow reactor (PFR) fed at 80 L/min. The density is constant and the reactor is at steady state. Find the conversion X in percent.

V = 100 LQ = 80 L/mink = 0.25 min⁻¹
X = %
5.Chemical equilibriumMedium
H2​I2​HII0.250.250C−x−x+2xE0.25 − x0.25 − x2x

0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 25. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 25
[H2] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.500C−x+x+xE0.5 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.5 mol/LKc = 0.04
[Cl2] = mol/L
7.Mass balanceMedium
MixerStream 1500 kg/h · 20% NaOHStream 2ṁ₂ = ? · 45% 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 = 500 kg/hx1 = 20 %x2 = 45 %x3 = 25 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F2100 kg/h · 5% NaOHProduct P55% 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 = 2100 kg/hxF = 5 %xP = 55 %
W = kg/h

3Solutions

1.Reaction kinetics

Answer: k = 0.008514 min⁻¹

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).

Integrated first-order law

k = ln(CA0/CA)/t = ln(1/0.6)/60 = 0.5108/60 = 0.008514 min⁻¹

Answer

k = 0.008514 min⁻¹
2.Reaction kinetics

Answer: A = 3.674 × 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 = 25 + 273.15 = 298.1 K

Solve Arrhenius for A

Ea/(RT) = 70000/(8.314·298.1) = 28.24
A = k·e28.24 = 0.002·1.837 × 10¹² = 3.674 × 10⁹ s⁻¹

Answer

A = 3.674 × 10⁹ s⁻¹
3.Reactor sizing (CSTR & PFR)

Answer: VCSTR/VPFR = 1.637

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.

k·τ for each reactor

CSTR: k·τ = X/(1 − X) = 0.6/0.4 = 1.5
PFR: k·τ = ln(1/(1 − X)) = ln(2.5) = 0.9163

Ratio of volumes

VCSTR/VPFR = 1.5/0.9163 = 1.637

Answer

VCSTR/VPFR = 1.637

The gap grows quickly at high conversion.

4.Reactor sizing (CSTR & PFR)

Answer: X = 26.84 %

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 = 100/80 = 1.25 min, kτ = 0.25·1.25 = 0.3125

Conversion

X = 1 − e−kτ = 1 − e−0.3125 = 26.84%

Answer

X = 26.84%
5.Chemical equilibrium

Answer: [H2] = 0.07143 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

25 = (2x)²/(0.25 − x)²

Take the square root

5 = 2x/(0.25 − x) → x = 0.25·5/(2 + 5) = 0.1786 mol/L

Equilibrium concentration

[H₂] = 0.25 − x = 0.07143 mol/L

Answer

[H2] = 0.07143 mol/L
6.Chemical equilibrium

Answer: [Cl2] = 0.1228 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/(0.5 − x) → x² + 0.04·x − 0.02 = 0

Solve the quadratic

x = (−0.04 + √(0.04² + 4·0.02))/2 = 0.1228 mol/L

Equilibrium concentration

[Cl₂] = x = 0.1228 mol/L

Answer

[Cl2] = 0.1228 mol/L
7.Mass balance

Answer: ṁ2 = 125 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) = 500·(20 − 25)/(25 − 45) = 125 kg/h

Answer

ṁ2 = 125 kg/h
8.Mass balance

Answer: W = 1909 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 = 2100·5/55 = 190.9 kg/h

Total balance

W = F − P = 2100 − 190.9 = 1909 kg/h

Answer

W = 1909 kg/h

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