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)15 min0.9

In a first-order reaction the concentration of A falls from 1.5 mol/L to 0.9 mol/L in 15 min. Find the rate constant k.

CA0 = 1.5 mol/LCA = 0.9 mol/Lt = 15 min
k = min⁻¹
2.Reaction kineticsHard
Tk30 °C0.00550 °C0.0952

The rate constant of a reaction is 0.005 min⁻¹ at 30 °C and 0.0952 min⁻¹ at 50 °C. Find the activation energy Ea.

k1 = 0.005 min⁻¹T1 = 30 °Ck2 = 0.0952 min⁻¹T2 = 50 °CR = 8.314 J/(mol·K)
Ea = kJ/mol
3.Reactor sizing (CSTR & PFR)Medium
QX = 80%same Q, k and X as the CSTR

A first-order liquid-phase reaction must reach 80% 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 = 80%first order, same k and Q
VCSTR/VPFR =
4.Reactor sizing (CSTR & PFR)Hard
V = 1000 LV = 1000 LQ = 50 L/minX = ?

A first-order liquid-phase reaction (k = 0.12 min⁻¹) runs in two equal CSTRs in series, each of 1000 L. The feed rate is 50 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 = 50 L/mink = 0.12 min⁻¹
X = %
5.Chemical equilibriumMedium
H2​I2​HII0.40.40C−x−x+2xE0.4 − x0.4 − x2x

0.4 mol/L of H₂ and 0.4 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 = 0.4 mol/LKc = 25
[H2] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.2500C−x+x+xE0.25 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.2 at the reaction temperature. A container starts with 0.25 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.25 mol/LKc = 0.2
[Cl2] = mol/L
7.Mass balanceMedium
MixerStream 11000 kg/h · 20% NaOHStream 2ṁ₂ = ? · 40% 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 = 1000 kg/hx1 = 20 %x2 = 40 %x3 = 25 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F3600 kg/h · 7% saltProduct PP = ? · 35%Vapour W (pure water)0% salt

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

F = 3600 kg/hxF = 7 %xP = 35 %
P = kg/h

3Solutions

1.Reaction kinetics

Answer: k = 0.03406 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(1.5/0.9)/15 = 0.5108/15 = 0.03406 min⁻¹

Answer

k = 0.03406 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 = 303.1 K, T2 = 323.1 K

Solve Arrhenius for Ea

Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.947/(-2.042e-4) = 119993 J/mol
Ea = 120 kJ/mol

Answer

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

Answer: VCSTR/VPFR = 2.485

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) = 0.8/0.2 = 4
PFR: k·τ = ln(1/(1 − X)) = ln(5) = 1.609

Ratio of volumes

VCSTR/VPFR = 4/1.609 = 2.485

Answer

VCSTR/VPFR = 2.485

The gap grows quickly at high conversion.

4.Reactor sizing (CSTR & PFR)

Answer: X = 91.35 %

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 = 1000/50 = 20 min, kτ = 0.12·20 = 2.4

Conversion

C1/CA0 = 1/(1 + kτ) = 1/(1 + 2.4) = 0.2941
C2/CA0 = 1/(1 + kτ)² = 0.08651
X = 1 − 0.08651 = 91.35%

Answer

X = 91.35%
5.Chemical equilibrium

Answer: [H2] = 0.1143 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/(0.4 − x)²

Take the square root

5 = 2x/(0.4 − x) → x = 0.4·5/(2 + 5) = 0.2857 mol/L

Equilibrium concentration

[H₂] = 0.4 − x = 0.1143 mol/L

Answer

[H2] = 0.1143 mol/L
6.Chemical equilibrium

Answer: [Cl2] = 0.1449 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

0.2 = x²/(0.25 − x) → x² + 0.2·x − 0.05 = 0

Solve the quadratic

x = (−0.2 + √(0.2² + 4·0.05))/2 = 0.1449 mol/L

Equilibrium concentration

[Cl₂] = x = 0.1449 mol/L

Answer

[Cl2] = 0.1449 mol/L
7.Mass balance

Answer: ṁ2 = 333.3 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) = 1000·(20 − 25)/(25 − 40) = 333.3 kg/h

Answer

ṁ2 = 333.3 kg/h
8.Mass balance

Answer: P = 720 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 = 3600·7/35 = 720 kg/h

Answer

P = 720 kg/h

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