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 reaction with rate law −rA = k·CA² has k = 0.05 L/(mol·min) and starts at 2 mol/L. Find the concentration CA after 20 min.
k = 0.05 L/(mol·min)CA0 = 2 mol/Lt = 20 min
CA = mol/L
2.Reaction kineticsHard
A first-order reaction has k = 0.002 min⁻¹ at 30 °C and an activation energy of 60 kJ/mol. The batch reactor is now run at 55 °C for 60 min. Find the conversion X reached.
k1 = 0.002 min⁻¹T1 = 30 °CEa = 60 kJ/molT2 = 55 °Ct = 60 minR = 8.314 J/(mol·K)
X = %
3.Reactor sizing (CSTR & PFR)Medium
A first-order liquid-phase reaction must reach 70% 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 = 70%first order, same k and Q
VCSTR/VPFR =
4.Reactor sizing (CSTR & PFR)Hard
A first-order liquid-phase reaction (k = 0.4 min⁻¹) runs in a 500 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 = 500 LQ = 50 L/mink = 0.4 min⁻¹
X = %
5.Chemical equilibriumMedium
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 = 50. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 50
[H2] = mol/L
6.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.5 mol/LKc = 0.05
[Cl2] = mol/L
7.Mass balanceMedium
How much of stream 2 must be added each hour to stream 1 so that the blended product contains 15% NaOH? Find ṁ2.
ṁ1 = 850 kg/hx1 = 30 %x2 = 10 %x3 = 15 %
ṁ2 = kg/h
8.Mass balanceHard
An evaporator concentrates a salt solution by boiling off pure water. Find the water evaporated W in kg/h.
F = 4000 kg/hxF = 17 %xP = 30 %
W = kg/h
3Solutions
1.Reaction kinetics
Answer: CA = 0.6667 mol/L
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).
Integrated second-order law
1/CA = 1/CA0 + kt = 0.5 + 0.05·20 = 1.5 L/mol
CA = 1/1.5 = 0.6667 mol/L
Answer
CA = 0.6667 mol/L
2.Reaction kinetics
Answer: X = 52.09 %
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).
Rate constant at the new temperature
T1 = 303.1 K, T2 = 328.1 K
k2 = 0.002·e−(60000/8.314)·(1/328.1 − 1/303.1) = 0.01227 min⁻¹
Conversion
X = 1 − e−0.01227·60 = 1 − 0.4791 = 52.09 %
Answer
X = 52.09 %
3.Reactor sizing (CSTR & PFR)
Answer: VCSTR/VPFR = 1.938
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) = 0.7/0.3 = 2.333
PFR: k·τ = ln(1/(1 − X)) = ln(3.333) = 1.204
Ratio of volumes
VCSTR/VPFR = 2.333/1.204 = 1.938
Answer
VCSTR/VPFR = 1.938
The gap grows quickly at high conversion.
4.Reactor sizing (CSTR & PFR)
Answer: X = 98.17 %
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 = 500/50 = 10 min, kτ = 0.4·10 = 4
Conversion
X = 1 − e−kτ = 1 − e−4 = 98.17%
Answer
X = 98.17%
5.Chemical equilibrium
Answer: [H2] = 0.08819 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/(0.4 − x)²
Take the square root
7.071 = 2x/(0.4 − x) → x = 0.4·7.071/(2 + 7.071) = 0.3118 mol/L
Equilibrium concentration
[H₂] = 0.4 − x = 0.08819 mol/L
Answer
[H2] = 0.08819 mol/L
6.Chemical equilibrium
Answer: [Cl2] = 0.1351 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
0.05 = x²/(0.5 − x) → x² + 0.05·x − 0.025 = 0
Solve the quadratic
x = (−0.05 + √(0.05² + 4·0.025))/2 = 0.1351 mol/L
Equilibrium concentration
[Cl₂] = x = 0.1351 mol/L
Answer
[Cl2] = 0.1351 mol/L
7.Mass balance
Answer: ṁ2 = 2550 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) = 850·(30 − 15)/(15 − 10) = 2550 kg/h
Answer
ṁ2 = 2550 kg/h
8.Mass balance
Answer: W = 1733 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 = 4000·17/30 = 2267 kg/h
Total balance
W = F − P = 4000 − 2267 = 1733 kg/h
Answer
W = 1733 kg/h
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