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 = ?95%

A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.5 L/(mol·min) and starts at 1 mol/L. How long does it take to reach 95% conversion? Find t.

k = 0.5 L/(mol·min)CA0 = 1 mol/LX = 95 %
t = min
2.Reaction kineticsHard
Tk40 °C0.0570 °C?

A reaction has a rate constant of 0.05 min⁻¹ at 40 °C and an activation energy of 50 kJ/mol. Find the rate constant k2 at 70 °C.

k1 = 0.05 min⁻¹T1 = 40 °CT2 = 70 °CEa = 50 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹
3.Reactor sizing (CSTR & PFR)Medium
Q = 10 L/minX = 95%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 95% conversion of A.

Q = 10 L/mink = 0.5 min⁻¹X = 95%
V = L
4.Reactor sizing (CSTR & PFR)Hard
V = 200 LV = 200 LQ = 20 L/minX = ?

A first-order liquid-phase reaction (k = 0.05 min⁻¹) runs in two equal CSTRs in series, each of 200 L. The feed rate is 20 L/min, the density is constant and the reactors are at steady state. Find the overall conversion X in percent.

V = 200 L (×2)Q = 20 L/mink = 0.05 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 = 9. Find the equilibrium concentration [H2].

[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 9
[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 [Cl2].

[PCl₅]₀ = 0.4 mol/LKc = 0.05
[Cl2] = mol/L
7.Mass balanceMedium
MixerStream 1350 kg/h · 5% sugarStream 2ṁ₂ = ? · 50% sugarProduct45% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 45% sugar? Find ṁ2.

ṁ1 = 350 kg/hx1 = 5 %x2 = 50 %x3 = 45 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F3100 kg/h · 19% NaOHProduct P35% 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 = 3100 kg/hxF = 19 %xP = 35 %
W = kg/h

3Solutions

1.Reaction kinetics

Answer: t = 38 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 = 1·(1 − 0.95) = 0.05 mol/L

Integrated second-order law

t = (1/CA − 1/CA0)/k = (20 − 1)/0.5 = 38 min

Answer

t = 38 min
2.Reaction kinetics

Answer: k2 = 0.268 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 = 313.1 K, T2 = 343.1 K

Arrhenius ratio

ln(k2/k1) = −(50000/8.314)·(1/343.1 − 1/313.1) = 1.679
k2 = 0.05·e1.679 = 0.268 min⁻¹

Answer

k2 = 0.268 min⁻¹
3.Reactor sizing (CSTR & PFR)

Answer: V = 59.91 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 − 0.95))/0.5 = 2.996/0.5 = 5.991 min

Volume

V = Q·τ = 10·5.991 = 59.91 L

Answer

V = 59.91 L
4.Reactor sizing (CSTR & PFR)

Answer: X = 55.56 %

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 = 200/20 = 10 min, kτ = 0.05·10 = 0.5

Conversion

C1/CA0 = 1/(1 + kτ) = 1/(1 + 0.5) = 0.6667
C2/CA0 = 1/(1 + kτ)² = 0.4444
X = 1 − 0.4444 = 55.56%

Answer

X = 55.56%
5.Chemical equilibrium

Answer: [H2] = 0.1 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

9 = (2x)²/(0.25 − x)²

Take the square root

3 = 2x/(0.25 − x) → x = 0.25·3/(2 + 3) = 0.15 mol/L

Equilibrium concentration

[H₂] = 0.25 − x = 0.1 mol/L

Answer

[H2] = 0.1 mol/L
6.Chemical equilibrium

Answer: [Cl2] = 0.1186 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

[Cl₂] = x = 0.1186 mol/L

Answer

[Cl2] = 0.1186 mol/L
7.Mass balance

Answer: ṁ2 = 2800 kg/h

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

Balances

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

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 350·(5 − 45)/(45 − 50) = 2800 kg/h

Answer

ṁ2 = 2800 kg/h
8.Mass balance

Answer: W = 1417 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 = 3100·19/35 = 1683 kg/h

Total balance

W = F − P = 3100 − 1683 = 1417 kg/h

Answer

W = 1417 kg/h

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