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 = 0.2 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 5 min.
k = 0.2 L/(mol·min)CA0 = 0.5 mol/Lt = 5 min
CA = mol/L
2.Reaction kineticsHard
The rate constant of a reaction is 0.05 min⁻¹ at 20 °C and 0.549 min⁻¹ at 35 °C. Find the activation energy Ea.
k1 = 0.05 min⁻¹T1 = 20 °Ck2 = 0.549 min⁻¹T2 = 35 °CR = 8.314 J/(mol·K)
Ea = kJ/mol
3.Reactor sizing (CSTR & PFR)Medium
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
A first-order liquid-phase reaction (k = 0.4 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 = 0.4 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 = 36. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 36
[H2] = mol/L
6.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.25 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 0.25 mol/LKc = 0.08
[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 45% salt? Find ṁ2.
ṁ1 = 350 kg/hx1 = 50 %x2 = 20 %x3 = 45 %
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 = 1700 kg/hxF = 10 %xP = 40 %
W = kg/h
3Solutions
1.Reaction kinetics
Answer: CA = 0.3333 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 = 2 + 0.2·5 = 3 L/mol
CA = 1/3 = 0.3333 mol/L
Answer
CA = 0.3333 mol/L
2.Reaction kinetics
Answer: Ea = 120 kJ/mol
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 = 293.1 K, T2 = 308.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.396/(-1.661e-4) = 119970 J/mol
Ea = 120 kJ/mol
Answer
Ea = 120 kJ/mol
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 = 32.97 %
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 = 0.4·1 = 0.4
Conversion
X = 1 − e−kτ = 1 − e−0.4 = 32.97%
Answer
X = 32.97%
5.Chemical equilibrium
Answer: [H2] = 0.1 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
36 = (2x)²/(0.4 − x)²
Take the square root
6 = 2x/(0.4 − x) → x = 0.4·6/(2 + 6) = 0.3 mol/L
Equilibrium concentration
[H₂] = 0.4 − x = 0.1 mol/L
Answer
[H2] = 0.1 mol/L
6.Chemical equilibrium
Answer: [PCl5] = 0.143 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/(0.25 − x) → x² + 0.08·x − 0.02 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.02))/2 = 0.107 mol/L
Equilibrium concentration
[PCl₅] = 0.25 − 0.107 = 0.143 mol/L
Answer
[PCl5] = 0.143 mol/L
7.Mass balance
Answer: ṁ2 = 70 kg/h
Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.
Balances
Total: ṁ3 = ṁ1 + ṁ2
salt: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3
Solve for ṁ2
ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 350·(50 − 45)/(45 − 20) = 70 kg/h
Answer
ṁ2 = 70 kg/h
8.Mass balance
Answer: W = 1275 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 = 1700·10/40 = 425 kg/h
Total balance
W = F − P = 1700 − 425 = 1275 kg/h
Answer
W = 1275 kg/h
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