Chemical Reaction Engineering: cram pack

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

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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)40 min0.15

In a first-order reaction the concentration of A falls from 0.5 mol/L to 0.15 mol/L in 40 min. Find the rate constant k.

CA0 = 0.5 mol/LCA = 0.15 mol/Lt = 40 min
k = min⁻¹
2.Reaction kineticsHard
Tk35 °C0.00250 °C0.0176

The rate constant of a reaction is 0.002 min⁻¹ at 35 °C and 0.0176 min⁻¹ at 50 °C. Find the activation energy Ea.

k1 = 0.002 min⁻¹T1 = 35 °Ck2 = 0.0176 min⁻¹T2 = 50 °CR = 8.314 J/(mol·K)
Ea = kJ/mol
3.Reactor sizing (CSTR & PFR)Medium
QX = 50%same Q, k and X as the CSTR

A first-order liquid-phase reaction must reach 50% 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 = 50%first order, same k and Q
VCSTR/VPFR =
4.Reactor sizing (CSTR & PFR)Hard
Q = 25 L/minX = ?V = 1000 L

A first-order liquid-phase reaction (k = 0.05 min⁻¹) runs in a 1000 L plug-flow reactor (PFR) fed at 25 L/min. The density is constant and the reactor is at steady state. Find the conversion X in percent.

V = 1000 LQ = 25 L/mink = 0.05 min⁻¹
X = %
5.Chemical equilibriumMedium
H2​I2​HII0.50.50C−x−x+2xE0.5 − x0.5 − x2x

0.5 mol/L of H₂ and 0.5 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 50. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 50
[HI] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.100C−x+x+xE0.1 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 0.1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].

[PCl₅]₀ = 0.1 mol/LKc = 0.5
[PCl5] = mol/L
7.Mass balanceMedium
MixerStream 1600 kg/h · 35% sugarStream 2ṁ₂ = ? · 20% sugarProduct30% sugar

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

ṁ1 = 600 kg/hx1 = 35 %x2 = 20 %x3 = 30 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F1300 kg/h · 5% NaOHProduct PP = ? · 65%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 = 1300 kg/hxF = 5 %xP = 65 %
P = kg/h

3Solutions

1.Reaction kinetics

Answer: k = 0.0301 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(0.5/0.15)/40 = 1.204/40 = 0.0301 min⁻¹

Answer

k = 0.0301 min⁻¹
2.Reaction kinetics

Answer: Ea = 120 kJ/mol

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 = 308.1 K, T2 = 323.1 K

Solve Arrhenius for Ea

Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.175/(-1.506e-4) = 120031 J/mol
Ea = 120 kJ/mol

Answer

Ea = 120 kJ/mol
3.Reactor sizing (CSTR & PFR)

Answer: VCSTR/VPFR = 1.443

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) = 0.5/0.5 = 1
PFR: k·τ = ln(1/(1 − X)) = ln(2) = 0.6931

Ratio of volumes

VCSTR/VPFR = 1/0.6931 = 1.443

Answer

VCSTR/VPFR = 1.443

The gap grows quickly at high conversion.

4.Reactor sizing (CSTR & PFR)

Answer: X = 86.47 %

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 = 1000/25 = 40 min, kτ = 0.05·40 = 2

Conversion

X = 1 − e−kτ = 1 − e−2 = 86.47%

Answer

X = 86.47%
5.Chemical equilibrium

Answer: [HI] = 0.7795 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

50 = (2x)²/(0.5 − x)²

Take the square root

7.071 = 2x/(0.5 − x) → x = 0.5·7.071/(2 + 7.071) = 0.3898 mol/L

Equilibrium concentration

[HI] = 2x = 0.7795 mol/L

Answer

[HI] = 0.7795 mol/L
6.Chemical equilibrium

Answer: [PCl5] = 0.01459 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

0.5 = x²/(0.1 − x) → x² + 0.5·x − 0.05 = 0

Solve the quadratic

x = (−0.5 + √(0.5² + 4·0.05))/2 = 0.08541 mol/L

Equilibrium concentration

[PCl₅] = 0.1 − 0.08541 = 0.01459 mol/L

Answer

[PCl5] = 0.01459 mol/L
7.Mass balance

Answer: ṁ2 = 300 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) = 600·(35 − 30)/(30 − 20) = 300 kg/h

Answer

ṁ2 = 300 kg/h
8.Mass balance

Answer: P = 100 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 = 1300·5/65 = 100 kg/h

Answer

P = 100 kg/h

ExamProblems · examproblems.com/courses/chemical-reaction-engineering