Reaction Engineering - Video Lectures
Topic outline
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Arrhenius Model of Rate Constant Page
Which is the correct statement from the following statements on the Arrhenius model of the rate constant \(k = Ae^{-E/RT}\)?
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\(A\) is always dimensionless
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For two reactions 1 and 2, if \(A_1 = A_2\) and \(E_1 > E_2\), then \(k_1(T) > k_2(T)\)
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For a given reaction, the % change of \(k\) with respect to temperature is higher at lower temperatures
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The % change of \(k\) with respect to temperature is higher for higher \(A\)
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Molecularity of Reaction Page
Molecularity of an elementary reaction \(P +Q \rightarrow R + S\) is
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1
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2
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3
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4
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Order of Reaction and Units of Rate Constant Page
Overall order of reaction for which the rate constant has units of (mol/litre)\(^{-3/2}\).s\(^{-1}\) is
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\(-3/2\)
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1/2
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3/2
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5/2
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Constant Volume Reactions in Batch Reactor PageLiquid \(A\) decomposes in a batch reactor, and 50% of \(A\) is converted in a 5-minute run. How much additional time would it take to reach 75% conversion, with:Enrol me in this course
- first order reaction _______ min
- second order reaction _______ min
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Order of Reaction from Conversion Data PageIn a homogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 mol/liter and also for 0.8 mol/liter.Enrol me in this course
- What is the order (\(n\)) of reaction? (Choose from: 0, 0.5, 1, 2)
- What is the value of \(k\) in mol, liter, min units? _________ \(\times 10^{-3} \text{ (mol/liter)$^{1-n}$.min$^{-1}$}\) where \(n\) is the order of reaction.
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A 10-minute experimental run in a batch reactor shows that 75% of liquid reactant is converted to product by a \(1/2\)-order rate. What would be the fraction converted in a half-hour run? ___________Enrol me in this course
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Rate Expression from Conversion Data PageAfter 8 minutes in a batch reactor, reactant (\(C_{A0} = 1\) mol/liter) is 80% converted; after 18 minutes, conversion is 90%.Enrol me in this course
- What is the order (\(n\)) of reaction? (Choose from: 0, 0.5, 1, 2)
- What is the value of \(k\) in mol, liter, min units. _______ \(\text{ (mol/liter)$^{1-n}$.min$^{-1}$}\) where \(n\) is the order of reaction.
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Upper Limit of Reaction Rate Page
For a non-catalytic homogeneous reaction \(A \rightarrow B\), the rate expression at 300 K is \[ -r_A \quad (\text {mol.m$^{-3}$.s$^{-1}$})=\frac {10C_A}{1+5C_A} \] where \(C_A\) is the concentration of \(A\) (in mol/m3). Theoretically, the upper limit for the magnitude of the reaction rate (\(-r_A\) in mol.m\(^{-3}\).s-1, rounded off to the first decimal place) at 300 K is ____________
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First Order Reversible Reaction PageThe first-order reversible liquid reaction \[ A \rightleftharpoons R, \qquad \qquad C_{A0} = 0.5 \text{ mol/liter}, \qquad \qquad C_{R0} = 0\] takes place in a batch reactor. After 8 minutes, conversion of \(A\) is 33.3% while equilibrium conversion is 66.7%. Find the constants (\(k_1 \&\ k_2\) round off to 4 decimal places) of this rate equation given as \(-r_A = k_1C_A - k_2C_R\).Enrol me in this course
- \(k_1 = \)__________ min-1
- \(k_2 = \)__________ min-1
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Time Constant of First Order Reversible Reaction PageA first order reversible reaction \(A\xrightleftharpoons [k_2]{k_1} B\) occurs in a batch reactor. The exponential decay of the concentration of \(A\) has the time constantEnrol me in this course
- \(\dfrac {1}{k_1}\)
- \(\dfrac {1}{k_2}\)
- \(\dfrac {1}{k_1-k_2}\)
- \(\dfrac {1}{k_1+k_2}\)
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Maximum Conversion of Reversible Reaction Page
The rate of the liquid phase reversible reaction \(A \rightleftharpoons 2B\) in (kmol.m-3.min-1) at 298 K, is \[ -r_A = 0.02 C_A - 0.01 C_B \] where the concentrations \(C_A\) and \(C_B\) are expressed in (kmol.m-3). What is the maximum limiting conversion of \(A\) achievable in an isothermal CSTR at 298 K, assuming pure \(A\) is fed at the inlet?
- 1
- 2/3
- 1/2
- 1/3
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Enzymatic Reaction PageEnzyme \(E\) catalyzes the transformation of reactant \(A\) to product \(R\) as follows: \[ A \stackrel{\text{enzyme}}{\longrightarrow} R, \qquad -r_A = \frac{200C_AC_{E0}}{2+C_A}\frac{\text{mol}}{\text{liter}\cdot\text{min}} \] If we introduce enzyme (\(C_{E0} = 0.001\) mol/liter) and reactant (\(C_{A0}=10\) mol/liter) into a batch reactor and let the reaction proceed, find the time needed for the concentration of the reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.Enrol me in this course
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First Order Variable Volume Reaction PageFind the first-order rate constant for the disappearance of \(A\) in the gas phase reaction \(2A \rightarrow R\) if, on holding the pressure constant, the volume of the reaction mixture, starting with 80% \(A\), decreases by 20% in 3 min.Enrol me in this course
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Rate of Formation or Consumption of Each Components PageFor the reaction \(2R + S \rightarrow T\), the rates of formation, \(r_R\), \(r_S\) and \(r_T\) of the substances \(R\), \(S\) and \(T\) respectively, are related byEnrol me in this course
- \(2  r_R = r_S = r_T\)
- \(2  r_R = r_S = -r_T\)
- \( r_R = 2  r_S = 2  r_T\)
- \( r_R = 2  r_S = -2 r_T\)
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Rate of Formation / Consumption of Components of Reaction Page
For the reaction \(A + B \rightarrow 2B + C\),
- \(r_{A} = r_{B}\)
- \(r_{A} = -r_{B}\)
- \(r_{A} = 2r_{B}\)
- \(r_{A} = r_{B}/2\)
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Order of Reaction from Rate vs Concentration PageConsider a reaction \(aG + bH \rightarrow \text{ Products}\). When concentration of both the reactants \(G\) and \(H\) is doubled, the rate increases by eight times. However, when concentration of \(G\) is doubled keeping the concentration of \(H\) fixed, the rate is doubled. The overall order of the reaction is ________ (Choose from: 0, 1, 2, 3)Enrol me in this course
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Order of Reaction from Half-life Data PageThe following half-life data are available for the irreversible liquid phase reaction, \[ A \rightarrow \text { Products} \]Enrol me in this course
Initial concentration
(kmol/m3)Half-life
(min)2 2 8 1 -
Reaction Order with Half-life Time Equaling Fifty Percent of Full-life Time Page
For which reaction order, the half-life of the reactant is half of the full lifetime (time for 100% conversion) of the reactant?
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Zero order
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Half order
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First order
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Second order
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Half-life vs Initial Concentration - Straight Line Plot Page
Consider the n\(^{th}\) order irreversible liquid phase reaction \(A \rightarrow B\). Which one of the following plots involving half-life of the reaction \((t_{1/2})\) and the initial reactant concentration \((C_{A0})\) gives a straight line plot?
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\(C_{A0}\) vs. \(t_{1/2}\)
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\(\ln C_{A0}\) vs. \(t_{1/2}\)
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\(C_{A0}\) vs. \(\ln t_{1/2}\)
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\(\ln C_{A0}\) vs. \(\ln t_{1/2}\)
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Inert Content for a Variable Volume Reaction PageAn irreversible gas phase reaction \(A \rightarrow 5B\) is conducted in an isothermal batch reactor at constant pressure in the presence of an inert. The feed contains no \(B\). If the volume of the gas at complete conversion must not exceed three times the initial volume, the minimum mole percent of the inert in the feed must be _______ (choose from: 0, 20, 33, 50)Enrol me in this course
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Decomposition of N2O5 - Non-elementary Reaction Page
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Decomposition of Azomethane - Steps Involved Page
The gas phase decomposition of azomethane to give ethane and nitrogen takes place according to the following sequence of elementary reactions.
\[ \begin {align*} (\text {CH}_3)_2\text {N}_2 + (\text {CH}_3)_2\text {N}_2 &\stackrel {k_1}{\longrightarrow } (\text {CH}_3)_2\text {N}_2 + [(\text {CH}_3)_2\text {N}_2]^* \\ [(\text {CH}_3)_2\text {N}_2]^* + (\text {CH}_3)_2\text {N}_2 &\stackrel {k_2}{\longrightarrow } (\text {CH}_3)_2\text {N}_2 + (\text {CH}_3)_2\text {N}_2 \\ [(\text {CH}_3)_2\text {N}_2]^* &\stackrel {k_3}{\longrightarrow } \text {C}_2\text {H}_6 + \text {N}_2 \end {align*} \]
Using the pseudo-steady-state-approximation for \([(\text {CH}_3)_2\text {N}_2]^*\), the order with respect to azo-methane in the rate expression for the formation of ethane, in the limit of high concentrations of azomethane, is ____________
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Mechanism of Non-elementary Reaction Page
Find a mechanism that is consistent with the rate equation and the reaction given below: \[ 2A + B \rightarrow A_2B \quad \quad \quad -r_A = kC_AC_B \]
\(A + B \rightleftharpoons AB; \quad \quad AB + A \rightarrow A_2B\)
\(A + B \rightarrow AB; \quad \quad AB + A \rightarrow A_2B\)
\(A + A \rightarrow AA; \quad \quad AA + B \rightarrow A_2B\)
\(A + A \rightleftharpoons AA; \quad \quad AA + B \rightarrow A_2B\)
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