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 = 0.1 L/(mol·min) and an initial concentration of 0.2 mol/L. Find its first half-life t½.

k = 0.1 L/(mol·min)CA0 = 0.2 mol/L
t½ = min
2.Reaction kineticsHard
CA0​ (mol/L)r (mol/(L·s))run 10.20.008run 20.80.032n = ?

Two experiments measure the initial rate of A → products. With CA0 = 0.2 mol/L the rate is 0.008 mol/(L·s); with CA0 = 0.8 mol/L it is 0.032 mol/(L·s). Find the reaction order n with respect to A.

CA0,1 = 0.2 mol/Lr1 = 0.008 mol/(L·s)CA0,2 = 0.8 mol/Lr2 = 0.032 mol/(L·s)
n =
3.Reactor sizing (CSTR & PFR)Medium
QX = 95%same Q, k and X as the CSTR

A first-order liquid-phase reaction must reach 95% 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 = 95%first order, same k and Q
VCSTR/VPFR =
4.Reactor sizing (CSTR & PFR)Hard
V = 1500 LQ = 20 L/minX = ?

A first-order liquid-phase reaction (k = 0.12 min⁻¹) runs in a 1500 L CSTR fed at 20 L/min. The density is constant and the reactor is at steady state. Find the conversion X in percent.

V = 1500 LQ = 20 L/mink = 0.12 min⁻¹
X = %
5.Chemical equilibriumMedium
H2​I2​HII0.20.20C−x−x+2xE0.2 − x0.2 − x2x

0.2 mol/L of H₂ and 0.2 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 = 0.2 mol/LKc = 49
[H2] = mol/L
6.Chemical equilibriumHard
PCl5​PCl3​Cl2​I1.500C−x+x+xE1.5 − xxx

PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 1.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].

[PCl₅]₀ = 1.5 mol/LKc = 0.08
[Cl2] = mol/L
7.Mass balanceMedium
MixerStream 11000 kg/h · 15% NaOHStream 2ṁ₂ = ? · 45% NaOHProduct40% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 40% NaOH? Find ṁ2.

ṁ1 = 1000 kg/hx1 = 15 %x2 = 45 %x3 = 40 %
ṁ2 = kg/h
8.Mass balanceHard
EvaporatorFeed F2800 kg/h · 5% NaOHProduct PP = ? · 40%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 = 2800 kg/hxF = 5 %xP = 40 %
P = kg/h

3Solutions

1.Reaction kinetics

Answer: t½ = 50 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/(0.1·0.2) = 50 min

Answer

t½ = 50 min

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

2.Reaction kinetics

Answer: n = 1

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 = 0.032/0.008 = 4
C2/C1 = 0.8/0.2 = 4

Order

n = ln(4)/ln(4) = 1

Answer

n = 1
3.Reactor sizing (CSTR & PFR)

Answer: VCSTR/VPFR = 6.342

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) = 0.95/0.05 = 19
PFR: k·τ = ln(1/(1 − X)) = ln(20) = 2.996

Ratio of volumes

VCSTR/VPFR = 19/2.996 = 6.342

Answer

VCSTR/VPFR = 6.342

The gap grows quickly at high conversion.

4.Reactor sizing (CSTR & PFR)

Answer: X = 90 %

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 = 1500/20 = 75 min, kτ = 0.12·75 = 9

Conversion

X = kτ/(1 + kτ) = 9/(1 + 9) = 90%

Answer

X = 90%
5.Chemical equilibrium

Answer: [H2] = 0.04444 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/(0.2 − x)²

Take the square root

7 = 2x/(0.2 − x) → x = 0.2·7/(2 + 7) = 0.1556 mol/L

Equilibrium concentration

[H₂] = 0.2 − x = 0.04444 mol/L

Answer

[H2] = 0.04444 mol/L
6.Chemical equilibrium

Answer: [Cl2] = 0.3087 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

0.08 = x²/(1.5 − x) → x² + 0.08·x − 0.12 = 0

Solve the quadratic

x = (−0.08 + √(0.08² + 4·0.12))/2 = 0.3087 mol/L

Equilibrium concentration

[Cl₂] = x = 0.3087 mol/L

Answer

[Cl2] = 0.3087 mol/L
7.Mass balance

Answer: ṁ2 = 5000 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·(15 − 40)/(40 − 45) = 5000 kg/h

Answer

ṁ2 = 5000 kg/h
8.Mass balance

Answer: P = 350 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 = 2800·5/40 = 350 kg/h

Answer

P = 350 kg/h

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