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Chemical Reaction Engineering

7% of exam

Reaction rates and order, the Arrhenius rate constant, conversion, yield, and selectivity, series, parallel, homogeneous, heterogeneous, and biological reactions, reactor types (batch, semibatch, CSTR, plug flow, gas and liquid phase), and catalysis.

5 concepts

A. Reaction rates and order

Rate Laws, Integrated Forms and Arrhenius

Reaction order and rate laws, the zero/first/second-order integrated forms and half-life, and the Arrhenius temperature dependence — the kinetic backbone of every reactor problem.

Kinetics is where the FE Chemical exam separates the reactor topics from the equilibrium topics: thermodynamics tells you whether a reaction ·can· go and how far, but kinetics tells you how ·fast· — and the rate law is the bridge from a beaker measurement to a reactor you can size. Almost every reactor-design question in the Chemical Engineering chapter starts by handing you a rate

-r_A = kC_A^n

Fluid Mechanics and Dynamics

Thermodynamics

Material and Energy Balances

Heat Transfer

Mass Transfer and Separation

Solids Handling

Chemical Reaction Engineering

Engineering Economics

Process Design

Process Control

Safety, Health, and Environment

Ethics and Professional Practice

and asking what happens over time or down a tube, so fluency with the rate law, its integrated forms, and the temperature dependence of

k

is worth more raw points than any other single skill in this domain. This concept follows the FE Reference Handbook — Chemical Engineering (Chemical Reaction Engineering), where the rate definition, the Arrhenius equation, and the zero/first/second-order integrated forms are tabulated. The point most often lost here is a sign or a basis error: forgetting that

-r_A

is a ·disappearance· rate, mixing concentration with conversion, or putting a temperature in

^\circ\text{C}

where the Arrhenius exponent demands kelvin.

Defining the rate of reaction

The rate of reaction of a species is the moles of that species formed per unit time per unit volume. For a reactant

A

that disappears, the rate

r_A

is negative, so we work with

-r_A

, a positive number. In a closed, constant-volume system this reduces to a derivative of concentration, but never lose sight of the per-volume definition — it is what lets the same

-r_A

appear in a batch balance and in a flow-reactor balance. Keep the stoichiometric bookkeeping straight: for

aA \to cC

, the rates relate as

-r_A/a = r_C/c

.

-r_A = -\dfrac{1}{V}\dfrac{dN_A}{dt} \;\xrightarrow{V=\text{const}}\; -\dfrac{dC_A}{dt}

Rate law and reaction order

A power-law rate expresses

-r_A

as a rate constant times concentrations raised to empirical exponents. If

-r_A = kC_A^x C_B^y

, the reaction is

x

order in

A

,

y

order in

B

, and overall order

n = x + y

. Order is an experimental quantity, not the stoichiometric coefficient — only for a true elementary step do the two coincide. The units of

k

are set entirely by the overall order: a first-order

k

is in

\text{time}^{-1}

, a second-order

k

in

\text{volume}\cdot\text{mol}^{-1}\cdot\text{time}^{-1}

, and a zero-order

k

in

\text{mol}\cdot\text{volume}^{-1}\cdot\text{time}^{-1}

.

-r_A = kC_A^{x}C_B^{y}, \qquad n = x + y

Integrated forms: zero, first, second order

For a constant-volume batch reactor, separate

-dC_A/dt = kC_A^n

and integrate. Zero order gives a straight

C_A

decline,

C_A = C_{A0} - kt

. First order gives an exponential decay whose hallmark is that

\ln C_A

is linear in time. Second order gives a reciprocal that is linear in time,

1/C_A

versus

t

. The diagnostic value is huge: whichever plot is straight tells you the order. Convert any of these to conversion with

C_A = C_{A0}(1-X_A)

.

\begin{aligned} n=0:&\quad C_A = C_{A0}-kt \\ n=1:&\quad \ln\!\frac{C_{A0}}{C_A} = kt \\ n=2:&\quad \frac{1}{C_A}-\frac{1}{C_{A0}} = kt \end{aligned}

Half-life

t_{1/2}

is the time for

C_A

to fall to

C_{A0}/2

, and its dependence on initial concentration is itself a fingerprint of order. For first order it is constant,

t_{1/2}=\ln 2/k

, independent of

C_{A0}

— the signature of exponential decay and the reason radioactive-style problems are always first order. For zero order

t_{1/2}=C_{A0}/2k

(grows with

C_{A0}

); for second order

t_{1/2}=1/(kC_{A0})

(shrinks with

C_{A0}

). If doubling the starting concentration leaves the half-life unchanged, you are looking at a first-order process.

t_{1/2}^{(0)}=\frac{C_{A0}}{2k},\quad t_{1/2}^{(1)}=\frac{\ln 2}{k},\quad t_{1/2}^{(2)}=\frac{1}{kC_{A0}}

The Arrhenius equation

The rate constant climbs with temperature through the Arrhenius law,

k = A\,e^{-E_a/RT}

, where

A

is the pre-exponential (frequency) factor,

E_a

is the activation energy, and

T

is in kelvin. A plot of

\ln k

against

1/T

is a straight line of slope

-E_a/R

— the standard way to extract

E_a

from data. With

k

measured at two temperatures you can solve for

E_a

directly, eliminating

A

. The intuition:

E_a

is the energy barrier reacting molecules must clear, so a larger

E_a

makes the rate far more temperature-sensitive.

k = A\,e^{-E_a/RT}, \qquad \ln\!\frac{k_2}{k_1} = \frac{E_a}{R}\,\frac{T_2-T_1}{T_1 T_2}

Finding order from data

Two routes dominate the exam. The ·integral· method assumes an order, plots the corresponding linearized form (table above), and accepts the order whose plot is straight. The ·differential· method takes

\ln(-r_A) = \ln k + n\ln C_A

and reads the order as the slope of

\ln(-r_A)

versus

\ln C_A

. A faster shortcut on multiple-choice items is the half-life test, or the method of initial rates: if doubling

C_{A0}

doubles the initial rate, the reaction is first order; if it quadruples the rate, second order.

\ln(-r_A) = \ln k + n\,\ln C_A

Exam strategy

Read the rate law and immediately classify the order — it fixes the units of

k

, the integrated form, and the half-life behavior in one stroke. Always carry temperatures in kelvin inside any Arrhenius exponent or ratio; the most common silent error is leaving

T

in

^\circ\text{C}

. When a problem gives two

(k,T)

pairs and asks for

E_a

, use the two-point form so

A

cancels; when it gives data and asks for order, reach for the linear plot whose slope you can identify. Keep conversion and concentration interchangeable through

C_A=C_{A0}(1-X_A)

, and remember

-r_A

is positive for a disappearing reactant.

Key equations

Rate of reaction (constant volume)

-r_A = -\dfrac{1}{V}\dfrac{dN_A}{dt} = -\dfrac{dC_A}{dt}

Moles of $A$ consumed per volume per time; positive for a disappearing reactant. $C_A$ in $\text{mol/L}$ , $t$ in $\text{s}$ or $\text{min}$

Power-law rate and order

-r_A = kC_A^{x}C_B^{y}, \quad n = x+y

Fractional conversion

C_A = C_{A0}(1-X_A)

Zero-order integrated form

C_A = C_{A0}-kt, \quad t_{1/2}=\dfrac{C_{A0}}{2k}

First-order integrated form

\ln\dfrac{C_{A0}}{C_A}=kt,\quad C_A=C_{A0}e^{-kt}

First-order half-life

t_{1/2}=\dfrac{\ln 2}{k}

Independent of $C_{A0}$

Second-order integrated form

\dfrac{1}{C_A}-\dfrac{1}{C_{A0}}=kt,\quad t_{1/2}=\dfrac{1}{kC_{A0}}

Arrhenius equation

k = A\,e^{-E_a/RT}

$A$ = pre-exponential factor (same units as $k$

Two-temperature Arrhenius

\ln\dfrac{k_2}{k_1}=\dfrac{E_a}{R}\dfrac{T_2-T_1}{T_1 T_2}

Arrhenius linear form

\ln k = \ln A - \dfrac{E_a}{R}\dfrac{1}{T}

Differential method for order

\ln(-r_A)=\ln k + n\ln C_A

Slope of $\ln(-r_A)$

Worked examples

First-order decomposition: half-life, time to 90%, and residual

Problem. A liquid-phase decomposition $A \to$ products is first order with $k = 0.0231\ \text{min}^{-1}$ . Find the half-life, the time to reach $90\%$ conversion, and the fraction of $A$ remaining after $60\ \text{min}$ .

Solution. Half-life (independent of

C_{A0}

):

t_{1/2} = \ln 2/k = 0.6931/0.0231 = 30.0\ \text{min}

t = \dfrac{-\ln(1-0.90)}{0.0231} \approx 99.7\ \text{min}

Second-order batch reactor

Problem. A reaction $A \to$ products follows $-r_A = kC_A^2$ with $k = 0.50\ \text{L·mol}^{-1}\text{min}^{-1}$

Activation energy from two temperatures

Problem. A rate constant doubles when temperature rises from $300\ \text{K}$ to $310\ \text{K}$ . Find the activation energy, then estimate by what factor $k$ increases from $300\ \text{K}$ to $320\ \text{K}$ .

Common pitfalls

•Leaving temperature in $^\circ\text{C}$ inside an Arrhenius exponent or ratio. The Boltzmann factor $e^{-E_a/RT}$ demands absolute temperature — always convert to kelvin first.
•Confusing reaction order with stoichiometric coefficient. Order is experimental; the two match only for a genuinely elementary step, never assume it.
•Sign and basis slips: $-r_A$ is positive for a disappearing reactant, and $C_A = C_{A0}(1-X_A)$
•Carrying the wrong units on $k$ . A first-order $k$ is $\text{time}^{-1}$ , second-order is $\text{L·mol}^{-1}\text{time}^{-1}$
•Assuming half-life is always constant. Only first-order half-life is independent of $C_{A0}$ ; zero-order grows and second-order shrinks with initial concentration.
•Forgetting $R$ must match $E_a$ units: use $R = 8.314\ \text{J/(mol·K)}$ with $E_a$

References

NCEES FE Reference Handbook — Chemical Engineering
NCEES FE Reference Handbook — Chemistry and Biology — Reaction rates and equilibrium background.
Fogler, Elements of Chemical Reaction Engineering — Rate laws, order determination, and Arrhenius analysis.

C. Conversion, yield, and selectivity

Conversion, Yield and Selectivity

Fractional conversion, overall and instantaneous yield, selectivity for parallel and series networks, and how concentration and temperature steer a process toward the desired product.

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D. Type of reactions

Reaction Types: Series, Parallel, Reversible and Biological

Series and parallel networks, reversible reactions and equilibrium-limited conversion, homogeneous versus heterogeneous, and enzyme kinetics via Michaelis-Menten.

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E. Reactor types

Reactor Design: Batch, CSTR and PFR

The three ideal-reactor design equations, space-time and space-velocity, CSTR-vs-PFR sizing for a target conversion, and reactors in series and parallel.

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F. Catalysis

Catalysis and Heterogeneous Reactions

How a catalyst lowers activation energy, homogeneous versus heterogeneous catalysis, surface-catalyzed kinetics with a rate-limiting step, and pore-diffusion effectiveness basics.

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.

Order $x,y$ are empirical; overall order $n$ sets the units of $k$ .

Links concentration and conversion at constant $V$ ; $X_A$ = mol $A$ reacted per mol $A$ fed.

$k$ in $\text{mol·L}^{-1}\text{time}^{-1}$ . $C_A$ linear in $t$ ; half-life grows with $C_{A0}$ .

$k$ in $\text{time}^{-1}$ . Equivalent: $-\ln(1-X_A)=kt$ . $\ln C_A$ linear in $t$ .

— the diagnostic signature of first-order kinetics.

$k$ in $\text{L·mol}^{-1}\text{time}^{-1}$ . $1/C_A$ linear in $t$ ; half-life shrinks with $C_{A0}$ .

),

E_a

= activation energy (

\text{J/mol}

),

T

in K,

R=8.314\ \text{J/(mol·K)}

.

Solve for $E_a$ from two $(k,T)$ pairs; $A$ cancels. $T$ in K.

Slope $-E_a/R$ , intercept $\ln A$ on a $\ln k$ vs $1/T$ plot.

vs

\ln C_A

gives the order

n

.

. Time to

X_A = 0.90

:

-\ln(1-X_A)=kt \Rightarrow t = -\ln(0.10)/0.0231 = 2.3026/0.0231 = 99.7\ \text{min}

. Residual at

60\ \text{min}

:

C_A/C_{A0} = e^{-kt} = e^{-0.0231(60)} = e^{-1.386} = 0.250

, i.e.

25.0\%

remains. Sanity check:

60\ \text{min}

is two half-lives (

2\times 30

), and two halvings give

(1/2)^2 = 0.25

— consistent. Answers:

t_{1/2}=30.0\ \text{min}

,

t_{90}=99.7\ \text{min}

, residual

=25.0\%

.

and

C_{A0} = 2.0\ \text{mol/L}

. How long to reach

C_A = 0.50\ \text{mol/L}

, and what conversion is that?

Solution. Second-order integrated form:

\dfrac{1}{C_A}-\dfrac{1}{C_{A0}} = kt

.

\dfrac{1}{0.50}-\dfrac{1}{2.0} = 2.0 - 0.50 = 1.5\ \text{L/mol}

.

t = 1.5/0.50 = 3.00\ \text{min}

. Conversion:

X_A = (C_{A0}-C_A)/C_{A0} = (2.0-0.50)/2.0 = 0.750

. Sanity check: units of

kt = (0.50\ \text{L·mol}^{-1}\text{min}^{-1})(3.00\ \text{min}) = 1.5\ \text{L/mol}

, matching the reciprocal-concentration difference. Answers:

t = 3.00\ \text{min}

,

X_A = 75.0\%

.

t = \dfrac{1}{k}\left(\dfrac{1}{C_A}-\dfrac{1}{C_{A0}}\right) = \dfrac{1.5}{0.50}=3.00\ \text{min}

Solution.

Two-point Arrhenius:

\ln(k_2/k_1) = \dfrac{E_a}{R}\dfrac{T_2-T_1}{T_1 T_2}

.

E_a = R\ln 2 \cdot \dfrac{T_1 T_2}{T_2-T_1} = 8.314(0.6931)\dfrac{(300)(310)}{10} = 8.314(0.6931)(9300) = 5.36\times 10^4\ \text{J/mol} = 53.6\ \text{kJ/mol}

. For

300\to 320\ \text{K}

:

\ln(k/k_1) = \dfrac{53594}{8.314}\left(\dfrac{1}{300}-\dfrac{1}{320}\right) = 6446(2.083\times 10^{-4}) = 1.343

, so

k/k_1 = e^{1.343} = 3.83

. Sanity check: a

10\ \text{K}

rise doubled

k

; a further

10\ \text{K}

should roughly double again, and

2^2 = 4 \approx 3.83

(slightly less because the exponent is in

1/T

, not

T

). Answers:

E_a = 53.6\ \text{kJ/mol}

, factor

\approx 3.83

.

E_a = R\ln 2\,\dfrac{T_1 T_2}{T_2-T_1} \approx 53.6\ \text{kJ/mol}

— do not plug conversion where a concentration belongs.

— a mismatch is a giveaway you used the wrong integrated form.

in

\text{J/mol}

(or

1.987\ \text{cal/(mol·K)}

with

E_a

in

\text{cal/mol}

).