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 first-order reaction in a batch reactor has k = 0.01 min⁻¹. How long does it take to reach 50% conversion? Find t.
k = 0.01 min⁻¹X = 50 %
t = min
2.Reaction kineticsHard
A reaction has a rate constant of 0.05 min⁻¹ at 40 °C and an activation energy of 40 kJ/mol. Find the rate constant k2 at 50 °C.
k1 = 0.05 min⁻¹T1 = 40 °CT2 = 50 °CEa = 40 kJ/molR = 8.314 J/(mol·K)
k2 = min⁻¹
3.Reactor sizing (CSTR & PFR)Medium
A first-order liquid-phase reaction must reach 90% 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 = 90%first order, same k and Q
VCSTR/VPFR =
4.Reactor sizing (CSTR & PFR)Hard
A first-order liquid-phase reaction (k = 0.15 min⁻¹) runs in two equal CSTRs in series, each of 200 L. The feed rate is 20 L/min, the density is constant and the reactors are at steady state. Find the overall conversion X in percent.
V = 200 L (×2)Q = 20 L/mink = 0.15 min⁻¹
X = %
5.Chemical equilibriumMedium
1 mol/L of H₂ and 1 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 = 1 mol/LKc = 49
[H2] = mol/L
6.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 1.5 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 1.5 mol/LKc = 0.04
[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% NaOH? Find ṁ2.
ṁ1 = 500 kg/hx1 = 40 %x2 = 20 %x3 = 30 %
ṁ2 = kg/h
8.Mass balanceHard
An evaporator concentrates a sugar solution by boiling off pure water. Find the product flow P in kg/h.
F = 4900 kg/hxF = 7 %xP = 30 %
P = kg/h
3Solutions
1.Reaction kinetics
Answer: t = 69.31 min
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).
Conversion and concentration
C/C0 = 1 − X = 0.5
Solve for the time
t = −ln(1 − X)/k = −ln(0.5)/0.01 = 69.31 min
Answer
t = 69.31 min
2.Reaction kinetics
Answer: k2 = 0.08044 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 = 313.1 K, T2 = 323.1 K
Arrhenius ratio
ln(k2/k1) = −(40000/8.314)·(1/323.1 − 1/313.1) = 0.4754
k2 = 0.05·e0.4754 = 0.08044 min⁻¹
Answer
k2 = 0.08044 min⁻¹
3.Reactor sizing (CSTR & PFR)
Answer: VCSTR/VPFR = 3.909
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) = 0.9/0.1 = 9
PFR: k·τ = ln(1/(1 − X)) = ln(10) = 2.303
Ratio of volumes
VCSTR/VPFR = 9/2.303 = 3.909
Answer
VCSTR/VPFR = 3.909
The gap grows quickly at high conversion.
4.Reactor sizing (CSTR & PFR)
Answer: X = 84 %
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 = 200/20 = 10 min, kτ = 0.15·10 = 1.5
Conversion
C1/CA0 = 1/(1 + kτ) = 1/(1 + 1.5) = 0.4
C2/CA0 = 1/(1 + kτ)² = 0.16
X = 1 − 0.16 = 84%
Answer
X = 84%
5.Chemical equilibrium
Answer: [H2] = 0.2222 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/(1 − x)²
Take the square root
7 = 2x/(1 − x) → x = 1·7/(2 + 7) = 0.7778 mol/L
Equilibrium concentration
[H₂] = 1 − x = 0.2222 mol/L
Answer
[H2] = 0.2222 mol/L
6.Chemical equilibrium
Answer: [PCl5] = 1.274 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
0.04 = x²/(1.5 − x) → x² + 0.04·x − 0.06 = 0
Solve the quadratic
x = (−0.04 + √(0.04² + 4·0.06))/2 = 0.2258 mol/L
Equilibrium concentration
[PCl₅] = 1.5 − 0.2258 = 1.274 mol/L
Answer
[PCl5] = 1.274 mol/L
7.Mass balance
Answer: ṁ2 = 500 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) = 500·(40 − 30)/(30 − 20) = 500 kg/h
Answer
ṁ2 = 500 kg/h
8.Mass balance
Answer: P = 1143 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 = 4900·7/30 = 1143 kg/h
Answer
P = 1143 kg/h
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