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) has k = 0.5 L/(mol·min) and an initial concentration of 0.8 mol/L. Find its first half-life t½.
k = 0.5 L/(mol·min)CA0 = 0.8 mol/L
t½ = min
2.Reaction kineticsHard
A reaction has an activation energy of 70 kJ/mol. At 20 °C it is too slow. To what temperature T2 (in °C) must it be heated to make it 5 times faster?
Ea = 70 kJ/molT1 = 20 °Ck2/k1 = 5R = 8.314 J/(mol·K)
T2 = °C
3.Reactor sizing (CSTR & PFR)Medium
A first-order liquid-phase reaction A → B with k = 0.08 min⁻¹ is carried out in a plug-flow reactor (PFR) fed at 40 L/min. The density is constant and the reactor is at steady state. Find the reactor volume V in litres for 85% conversion of A.
Q = 40 L/mink = 0.08 min⁻¹X = 85%
V = L
4.Reactor sizing (CSTR & PFR)Hard
A first-order liquid-phase reaction (k = 0.05 min⁻¹) runs in an 800 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 = 800 LQ = 25 L/mink = 0.05 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 = 64. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 64
[H2] = mol/L
6.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 at the reaction temperature. A container starts with 0.1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.1 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 30% sugar? Find ṁ2.
ṁ1 = 700 kg/hx1 = 15 %x2 = 35 %x3 = 30 %
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 = 3100 kg/hxF = 6 %xP = 55 %
W = kg/h
3Solutions
1.Reaction kinetics
Answer: t½ = 2.5 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/(0.5·0.8) = 2.5 min
Answer
t½ = 2.5 min
Each later half-life is twice as long, because C₀ has halved.
2.Reaction kinetics
Answer: T2 = 37.4 °C
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 temperature
T1 = 293.1 K
Solve Arrhenius for T₂
1/T2 = 1/293.1 − 8.314·ln(5)/70000 = 3.411 × 10⁻³ − 1.912 × 10⁻⁴ = 3.220 × 10⁻³ K⁻¹
T2 = 310.6 K = 37.4 °C
Answer
T2 = 37.4 °C
3.Reactor sizing (CSTR & PFR)
Answer: V = 948.6 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 − 0.85))/0.08 = 1.897/0.08 = 23.71 min
Volume
V = Q·τ = 40·23.71 = 948.6 L
Answer
V = 948.6 L
4.Reactor sizing (CSTR & PFR)
Answer: X = 79.81 %
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 = 800/25 = 32 min, kτ = 0.05·32 = 1.6
Conversion
X = 1 − e−kτ = 1 − e−1.6 = 79.81%
Answer
X = 79.81%
5.Chemical equilibrium
Answer: [H2] = 0.08 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
64 = (2x)²/(0.4 − x)²
Take the square root
8 = 2x/(0.4 − x) → x = 0.4·8/(2 + 8) = 0.32 mol/L
Equilibrium concentration
[H₂] = 0.4 − x = 0.08 mol/L
Answer
[H2] = 0.08 mol/L
6.Chemical equilibrium
Answer: [Cl2] = 0.05 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/(0.1 − x) → x² + 0.05·x − 0.005 = 0
Solve the quadratic
x = (−0.05 + √(0.05² + 4·0.005))/2 = 0.05 mol/L
Equilibrium concentration
[Cl₂] = x = 0.05 mol/L
Answer
[Cl2] = 0.05 mol/L
7.Mass balance
Answer: ṁ2 = 2100 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) = 700·(15 − 30)/(30 − 35) = 2100 kg/h
Answer
ṁ2 = 2100 kg/h
8.Mass balance
Answer: W = 2762 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 = 3100·6/55 = 338.2 kg/h
Total balance
W = F − P = 3100 − 338.2 = 2762 kg/h
Answer
W = 2762 kg/h
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