GATE Chemical Engineering Chemical Reaction Engineering Syllabus

Chemical Reaction Engineering syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Chemical Reaction Engineering in GATE Chemical Engineering, not a summary of it.

1. 1

   Theories of Reaction Rates

   2 topics

   • Kinetics of Homogeneous Reactions
   • Interpretation of Kinetic Data
2. 2

   Single and Multiple Reactions in Ideal Reactors

   1 topic

   • Kinetics of Enzyme Reactions
     • Michaelis-Menten Model
     • Monod Model
3. 3

   Non-Ideal Reactors

   2 topics

   • Residence Time Distribution
     • Single Parameter Model
   • Non-Isothermal Reactors
4. 4

   Kinetics of Heterogeneous Catalytic Reactions

   2 topics

   • Diffusion Effects in Catalysis
   • Rate and Performance Equations for Catalyst Deactivation

Chemical Reaction Engineering flashcards for GATE Chemical Engineering

23 of 51 cards from the Chemical Reaction Engineering deck — real questions with worked answers.

1. Define the rate of reaction based on unit volume of reacting fluid.

   $-r_A = -\dfrac{1}{V}\dfrac{dN_A}{dt}$, the moles of $A$ disappearing per unit volume per unit time (e.g. $\mathrm{mol\,m^{-3}\,s^{-1}}$).

2. What is the difference between the molecularity and the order of a reaction?

   Molecularity is the number of molecules taking part in an elementary reaction step and is always a positive integer. Order is the empirical sum of the exponents in the experimental rate law and may be fractional, zero, or even negative.

3. Write the general power-law rate expression and define overall order.

   $-r_A = k\,C_A^{a} C_B^{b}\cdots$; the overall order is $n = a + b + \cdots$.

4. State the Arrhenius equation and the meaning of each term.

   $k = A\,e^{-E_a/RT}$, where $A$ is the frequency (pre-exponential) factor, $E_a$ the activation energy, $R$ the gas constant, and $T$ the absolute temperature.

5. How is activation energy obtained from rate-constant data?

   Plot $\ln k$ versus $1/T$; the line is straight with slope $-E_a/R$, so $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$.

6. What are the units of the rate constant $k$ for a reaction of overall order $n$ (concentration in $\mathrm{mol/L}$, time in $s$)?

   $k$ has units $\left(\dfrac{\mathrm{mol}}{\mathrm{L}}\right)^{1-n} s^{-1}$.

7. Give the integrated rate law for an irreversible first-order reaction $A \to$ products in a batch reactor.

   $-\ln\dfrac{C_A}{C_{A0}} = -\ln(1-X_A) = kt$.

8. Give the integrated rate law for an irreversible second-order reaction $2A \to$ products (or $A+B$ with $C_{A0}=C_{B0}$).

   $\dfrac{1}{C_A} - \dfrac{1}{C_{A0}} = kt$, equivalently $\dfrac{X_A}{C_{A0}(1-X_A)} = kt$.

9. Give the integrated form for a zero-order reaction and the condition where it fails.

   $C_{A0} - C_A = kt$ (i.e. $C_{A0}X_A = kt$), valid only while $C_A > 0$; the reaction stops when $A$ is exhausted at $t = C_{A0}/k$.

10. What is the half-life of a first-order reaction and why is it notable?

    $t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{k}$; it is independent of the initial concentration.

11. How does the half-life of an $n$-th order reaction depend on initial concentration?

    $t_{1/2} \propto C_{A0}^{\,1-n}$, specifically $t_{1/2} = \dfrac{(2^{n-1}-1)}{k(n-1)}C_{A0}^{\,1-n}$ for $n \neq 1$.

12. In the integral method of analyzing kinetic data, how is reaction order confirmed?

    A rate law is assumed and integrated; the predicted concentration function is plotted against time. If the data give a straight line, the assumed order is correct and $k$ comes from the slope.

13. In the differential method of analyzing kinetic data, how is order determined?

    Take $-r_A = kC_A^n$, so $\ln(-r_A) = \ln k + n\ln C_A$. Plot $\ln(-r_A)$ vs $\ln C_A$; the slope is the order $n$ and the intercept gives $k$.

14. What is the half-life method for finding reaction order?

    Since $\ln t_{1/2} = \ln\!\big[\tfrac{2^{n-1}-1}{k(n-1)}\big] + (1-n)\ln C_{A0}$, plotting $\ln t_{1/2}$ versus $\ln C_{A0}$ gives a slope of $(1-n)$, yielding the order $n$.

15. State the method of initial rates for kinetic analysis.

    Measure the initial rate $-r_{A0}$ for several different initial concentrations; from $\ln(-r_{A0}) = \ln k + n\ln C_{A0}$ the slope gives order $n$. It avoids interference from products.

16. What does the steady-state approximation assume for a reactive intermediate?

    The net rate of formation of the intermediate is approximately zero ($\dfrac{d[I]}{dt}\approx 0$) because it is consumed as fast as it is produced, allowing its concentration to be eliminated from the rate law.

17. Give the rate expression for a first-order reversible reaction $A \rightleftharpoons R$ and its approach to equilibrium.

    $-\ln\dfrac{C_A - C_{Ae}}{C_{A0} - C_{Ae}} = (k_1 + k_2)t$, where $C_{Ae}$ is the equilibrium concentration and $K_C = k_1/k_2$.

18. Write the Michaelis-Menten rate equation for enzyme kinetics.

    $-r_S = \dfrac{r_{max}\,C_S}{K_M + C_S}$ (often written $v = \dfrac{V_{max}[S]}{K_M + [S]}$).

19. What is the physical meaning of the Michaelis constant $K_M$?

    $K_M$ equals the substrate concentration at which the reaction rate is half of $V_{max}$. A small $K_M$ indicates high enzyme-substrate affinity.

20. Derive the relation between rate and $V_{max}$ when $C_S = K_M$.

    $-r_S = \dfrac{V_{max}K_M}{K_M + K_M} = \dfrac{V_{max}}{2}$; the rate is exactly half the maximum.

21. What is the order of the Michaelis-Menten rate law in the limits of very low and very high substrate concentration?

    For $C_S \ll K_M$: $-r_S \approx \dfrac{V_{max}}{K_M}C_S$ (first order). For $C_S \gg K_M$: $-r_S \approx V_{max}$ (zero order, enzyme saturated).

22. Write the Lineweaver-Burk (double-reciprocal) linearization of the Michaelis-Menten equation.

    $\dfrac{1}{-r_S} = \dfrac{1}{V_{max}} + \dfrac{K_M}{V_{max}}\cdot\dfrac{1}{C_S}$; a plot of $1/(-r_S)$ vs $1/C_S$ gives intercept $1/V_{max}$ and slope $K_M/V_{max}$.

23. What is the mechanism underlying the Michaelis-Menten model?

    $\ce{E + S <=> ES -> E + P}$: enzyme and substrate reversibly form a complex $ES$, which then breaks down irreversibly to product and free enzyme. The steady-state (or rapid-equilibrium) assumption on $ES$ yields the rate law.

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