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)20 min0.32

In a first-order reaction the concentration of A falls from 0.8 mol/L to 0.32 mol/L in 20 min. Find the rate constant k.

CA0 = 0.8 mol/LCA = 0.32 mol/Lt = 20 min
k = min⁻¹
2.Reaction kineticsHard
Tk30 °C0.0240 °C?

A reaction has a rate constant of 0.02 min⁻¹ at 30 °C and an activation energy of 80 kJ/mol. Find the rate constant k2 at 40 °C.

k1 = 0.02 min⁻¹T1 = 30 °CT2 = 40 °CEa = 80 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹
3.Reactor sizing (CSTR & PFR)Medium
Q = 20 L/minCA0​ = 2.5 mol/LCA​ = ?V = 150 L, k = 0.08 min⁻¹

A liquid-phase first-order reaction A → B (k = 0.08 min⁻¹) runs in a 150 L plug-flow reactor (PFR). The feed enters at 20 L/min with CA0 = 2.5 mol/L. The density is constant and the reactor is at steady state. Find the outlet concentration CA.

V = 150 LQ = 20 L/mink = 0.08 min⁻¹CA0 = 2.5 mol/L
CA = mol/L
4.Reactor sizing (CSTR & PFR)Hard
V = 500 LQ = 25 L/minX = ?

A first-order liquid-phase reaction (k = 0.08 min⁻¹) runs in a 500 L CSTR fed at 25 L/min. The density is constant and the reactor is at steady state. Find the conversion X in percent.

V = 500 LQ = 25 L/mink = 0.08 min⁻¹
X = %
5.Chemical equilibriumMedium
H2​I2​HII0.80.80C−x−x+2xE0.8 − x0.8 − x2x

0.8 mol/L of H₂ and 0.8 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 = 0.8 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 = 0.04 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 = 0.04
[PCl5] = mol/L
7.Mass balanceMedium
MixerStream 1500 kg/h · 40% NaOHStream 2ṁ₂ = ? · 15% NaOHProduct20% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 20% NaOH? Find ṁ2.

ṁ1 = 500 kg/hx1 = 40 %x2 = 15 %x3 = 20 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F1300 kg/h · 7% sugarProduct PP = ? · 60%Vapour W (pure water)0% sugar

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

F = 1300 kg/hxF = 7 %xP = 60 %
P = kg/h

3Solutions

1.Reaction kinetics

Answer: k = 0.04581 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(0.8/0.32)/20 = 0.9163/20 = 0.04581 min⁻¹

Answer

k = 0.04581 min⁻¹
2.Reaction kinetics

Answer: k2 = 0.05511 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 = 303.1 K, T2 = 313.1 K

Arrhenius ratio

ln(k2/k1) = −(80000/8.314)·(1/313.1 − 1/303.1) = 1.014
k2 = 0.02·e1.014 = 0.05511 min⁻¹

Answer

k2 = 0.05511 min⁻¹
3.Reactor sizing (CSTR & PFR)

Answer: CA = 1.372 mol/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 = V/Q = 150/20 = 7.5 min

Outlet concentration

CA = 2.5·e−0.08·7.5 = 1.372 mol/L

Answer

CA = 1.372 mol/L

Conversion: X = 45.12%.

4.Reactor sizing (CSTR & PFR)

Answer: X = 61.54 %

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 = 500/25 = 20 min, kτ = 0.08·20 = 1.6

Conversion

X = kτ/(1 + kτ) = 1.6/(1 + 1.6) = 61.54%

Answer

X = 61.54%
5.Chemical equilibrium

Answer: [H2] = 0.1778 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/(0.8 − x)²

Take the square root

7 = 2x/(0.8 − x) → x = 0.8·7/(2 + 7) = 0.6222 mol/L

Equilibrium concentration

[H₂] = 0.8 − x = 0.1778 mol/L

Answer

[H2] = 0.1778 mol/L
6.Chemical equilibrium

Answer: [PCl5] = 0.2919 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/(0.4 − x) → x² + 0.04·x − 0.016 = 0

Solve the quadratic

x = (−0.04 + √(0.04² + 4·0.016))/2 = 0.1081 mol/L

Equilibrium concentration

[PCl₅] = 0.4 − 0.1081 = 0.2919 mol/L

Answer

[PCl5] = 0.2919 mol/L
7.Mass balance

Answer: ṁ2 = 2000 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·(40 − 20)/(20 − 15) = 2000 kg/h

Answer

ṁ2 = 2000 kg/h
8.Mass balance

Answer: P = 151.7 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 = 1300·7/60 = 151.7 kg/h

Answer

P = 151.7 kg/h

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