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)
Chemical Reaction Engineering: cram pack
Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.
1Formulas and methods
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
- Write Kc with products over reactants, each raised to the power of its coefficient.
- Set up an ICE table: Initial, Change (in steps of x times the coefficients), Equilibrium.
- Put the equilibrium expressions into Kc and solve for x.
- 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
A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.4 L/(mol·min) and starts at 1.5 mol/L. How long does it take to reach 95% conversion? Find t.
k = 0.4 L/(mol·min)CA0 = 1.5 mol/LX = 95 %
t = min
2.Reaction kineticsHard
A reaction has a rate constant of 0.02 min⁻¹ at 30 °C and an activation energy of 120 kJ/mol. Find the rate constant k2 at 55 °C.
k1 = 0.02 min⁻¹T1 = 30 °CT2 = 55 °CEa = 120 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹
3.Reactor sizing (CSTR & PFR)Medium
A first-order liquid-phase reaction A → B with k = 0.5 min⁻¹ is carried out in a plug-flow reactor (PFR) fed at 5 L/min. The density is constant and the reactor is at steady state. Find the reactor volume V in litres for 75% conversion of A.
Q = 5 L/mink = 0.5 min⁻¹X = 75%
V = L
4.Reactor sizing (CSTR & PFR)Hard
A first-order liquid-phase reaction (k = 0.15 min⁻¹) runs in a 200 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 = 200 LQ = 25 L/mink = 0.15 min⁻¹
X = %
5.Chemical equilibriumMedium
0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 16. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 16
[HI] = mol/L
6.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.01 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 0.2 mol/LKc = 0.01
[PCl5] = mol/L
7.Mass balanceMedium
How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% sugar? Find ṁ2.
ṁ1 = 750 kg/hx1 = 10 %x2 = 40 %x3 = 30 %
ṁ2 = kg/h
8.Mass balanceHard
An evaporator concentrates a sugar solution by boiling off pure water. Find the water evaporated W in kg/h.
F = 4900 kg/hxF = 7 %xP = 60 %
W = kg/h
3Solutions
1.Reaction kinetics
Answer: t = 31.67 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).
Concentration at the end
CA = 1.5·(1 − 0.95) = 0.075 mol/L
Integrated second-order law
t = (1/CA − 1/CA0)/k = (13.33 − 0.6667)/0.4 = 31.67 min
Answer
t = 31.67 min
2.Reaction kinetics
Answer: k2 = 0.7522 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 = 328.1 K
Arrhenius ratio
ln(k2/k1) = −(120000/8.314)·(1/328.1 − 1/303.1) = 3.627
k2 = 0.02·e3.627 = 0.7522 min⁻¹
Answer
k2 = 0.7522 min⁻¹
3.Reactor sizing (CSTR & PFR)
Answer: V = 13.86 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 = ln(1/(1 − 0.75))/0.5 = 1.386/0.5 = 2.773 min
Volume
V = Q·τ = 5·2.773 = 13.86 L
Answer
V = 13.86 L
4.Reactor sizing (CSTR & PFR)
Answer: X = 69.88 %
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 = 200/25 = 8 min, kτ = 0.15·8 = 1.2
Conversion
X = 1 − e−kτ = 1 − e−1.2 = 69.88%
Answer
X = 69.88%
5.Chemical equilibrium
Answer: [HI] = 0.3333 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
16 = (2x)²/(0.25 − x)²
Take the square root
4 = 2x/(0.25 − x) → x = 0.25·4/(2 + 4) = 0.1667 mol/L
Equilibrium concentration
[HI] = 2x = 0.3333 mol/L
Answer
[HI] = 0.3333 mol/L
6.Chemical equilibrium
Answer: [PCl5] = 0.16 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
0.01 = x²/(0.2 − x) → x² + 0.01·x − 0.002 = 0
Solve the quadratic
x = (−0.01 + √(0.01² + 4·0.002))/2 = 0.04 mol/L
Equilibrium concentration
[PCl₅] = 0.2 − 0.04 = 0.16 mol/L
Answer
[PCl5] = 0.16 mol/L
7.Mass balance
Answer: ṁ2 = 1500 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) = 750·(10 − 30)/(30 − 40) = 1500 kg/h
Answer
ṁ2 = 1500 kg/h
8.Mass balance
Answer: W = 4328 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 = 4900·7/60 = 571.7 kg/h
Total balance
W = F − P = 4900 − 571.7 = 4328 kg/h
Answer
W = 4328 kg/h
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