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)t½​ = ?C0​/2

A second-order reaction (−rA = k·CA²) has k = 0.02 L/(mol·min) and an initial concentration of 0.5 mol/L. Find its first half-life t½.

k = 0.02 L/(mol·min)CA0 = 0.5 mol/L
t½ = min
2.Reaction kineticsHard
CA0​ (mol/L)r (mol/(L·s))run 10.20.002run 20.40.008n = ?

Two experiments measure the initial rate of A → products. With CA0 = 0.2 mol/L the rate is 0.002 mol/(L·s); with CA0 = 0.4 mol/L it is 0.008 mol/(L·s). Find the reaction order n with respect to A.

CA0,1 = 0.2 mol/Lr1 = 0.002 mol/(L·s)CA0,2 = 0.4 mol/Lr2 = 0.008 mol/(L·s)
n =
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
Q = 50 L/minX = ?V = 50 L

A first-order liquid-phase reaction (k = 0.5 min⁻¹) runs in a 50 L plug-flow reactor (PFR) fed at 50 L/min. The density is constant and the reactor is at steady state. Find the conversion X in percent.

V = 50 LQ = 50 L/mink = 0.5 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 = 50. Find the equilibrium concentration [HI].

[H₂]₀ = [I₂]₀ = 0.8 mol/LKc = 50
[HI] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I0.800C−x+x+xE0.8 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.02 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 0.8 mol/LKc = 0.02
[Cl2] = mol/L
7.Mass balanceMedium
MixerStream 1250 kg/h · 40% sugarStream 2ṁ₂ = ? · 10% 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 = 250 kg/hx1 = 40 %x2 = 10 %x3 = 30 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F4800 kg/h · 6% NaOHProduct PP = ? · 45%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 = 4800 kg/hxF = 6 %xP = 45 %
P = kg/h

3Solutions

1.Reaction kinetics

Answer: t½ = 100 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).

Second-order half-life

t½ = 1/(k·CA0) = 1/(0.02·0.5) = 100 min

Answer

t½ = 100 min

Each later half-life is twice as long, because C₀ has halved.

2.Reaction kinetics

Answer: n = 2

Method of initial rates: for r = k·Cⁿ, comparing two runs that differ only in the starting concentration removes k: r₂/r₁ = (C₂/C₁)ⁿ. Doubling C doubles the rate for first order and quadruples it for second order.

Ratios

r2/r1 = 0.008/0.002 = 4
C2/C1 = 0.4/0.2 = 2

Order

n = ln(4)/ln(2) = 2

Answer

n = 2
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 = 39.35 %

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 = 50/50 = 1 min, kτ = 0.5·1 = 0.5

Conversion

X = 1 − e−kτ = 1 − e−0.5 = 39.35%

Answer

X = 39.35%
5.Chemical equilibrium

Answer: [HI] = 1.247 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/(0.8 − x)²

Take the square root

7.071 = 2x/(0.8 − x) → x = 0.8·7.071/(2 + 7.071) = 0.6236 mol/L

Equilibrium concentration

[HI] = 2x = 1.247 mol/L

Answer

[HI] = 1.247 mol/L
6.Chemical equilibrium

Answer: [Cl2] = 0.1169 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

0.02 = x²/(0.8 − x) → x² + 0.02·x − 0.016 = 0

Solve the quadratic

x = (−0.02 + √(0.02² + 4·0.016))/2 = 0.1169 mol/L

Equilibrium concentration

[Cl₂] = x = 0.1169 mol/L

Answer

[Cl2] = 0.1169 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
sugar: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 250·(40 − 30)/(30 − 10) = 125 kg/h

Answer

ṁ2 = 125 kg/h
8.Mass balance

Answer: P = 640 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 = 4800·6/45 = 640 kg/h

Answer

P = 640 kg/h

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