Chemical Reaction Engineering: Constant Volume Batch Reactions, Lecture notes of Chemical Kinetics

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CONSTANT VOLUME BATCH

REACTIONS

ChE 2115: Chemical Reaction Engineering

Basil James S Santos

Definition of Rate

For a constant volume system

Recall ChE 2115: Chemical

Rate Law

𝑖 𝑛

n – order of the reaction

n = 1, first-order reaction

n = 2, second-order reaction

n = 0, zeroth-order reaction

n can be fractional

Half-life − 𝑙𝑛 1 −

ln 2 𝑘𝐴 Note: Half-life of a 1st-order reaction is independent of the reactant’s initial concentration First-Order Reaction ChE 2115: Chemical In terms of conversion, 𝑋𝐴 = 1 −

Integrating, we get − ln 1 − 𝑋𝐴 = 𝑘𝐴𝑡 + 𝐶 At t = 0, X A

, thus − ln 1 = 𝐶 − 𝒍𝒏 𝟏 − 𝑿𝑨 = 𝒌𝑨𝒕

Integrating, we get ln 𝐶𝐴 = −𝑘𝐴𝑡 + 𝐶 At t = 0, CA = CA0, thus ln 𝐶𝐴0 = 𝐶 ln 𝐶𝐴 = −𝑘𝐴𝑡 + ln 𝐶𝐴 𝒍𝒏

At 500 0 C, cyclopropane rearranges to propene. The reaction is first-order and the rate constant is 6.7 X 10

• 4 s - 1 . (a) Calculate its half-life at 500 0 C. (b) Calculate the time it will take for 90% of cyclopropane to react. First-Order Reaction ChE 2115: Chemical EXAMPLE

Half-life 1 𝐶𝐴

Second-Order Reaction ChE 2115: Chemical In terms of conversion, 𝐶𝐴 = 𝐶𝐴0 1 − 𝑋𝐴 , 𝑑𝐶𝐴 = −𝐶𝐴0𝑑𝑋𝐴 𝐶𝐴0𝑑𝑋𝐴 𝐶 𝐴 2 1 − 𝑋𝐴 2

Integrating knowing that XA = 0 at t =0, we get 𝟏 𝑪𝑨𝟎

2 𝑑𝐶𝐴 𝐶𝐴 2

Integrating, we get 𝟏 𝑪𝑨

Half-life 𝐶𝐴

Zeroth-Order Reaction ChE 2115: Chemical In terms of conversion, 𝐶𝐴 = 𝐶𝐴0 1 − 𝑋𝐴 , 𝑑𝐶𝐴 = −𝐶𝐴0𝑑𝑋𝐴 𝐶𝐴0𝑑𝑋𝐴 = 𝑘𝐴𝑑𝑡 Integrating knowing that XA = 0 at t =0, we get 𝑪𝑨𝟎𝑿𝑨 = 𝒌𝑨𝒕

Note: Rate of a 0th-order reaction is independent of the reactant’s concentration.

Integrating, we get

Half-life 1 1 2

𝑛− 1

𝑛− 1

1 −𝑛 − 1 𝑘𝐴 𝑛 − 1 𝐶𝐴 𝑛− 1 n th

• Order Reaction ChE 2115: Chemical

𝑛 𝑑𝐶𝐴 𝐶𝐴

𝑛 =^ −𝑘𝐴𝑑𝑡

Integrating, we get 𝟏 𝑪𝑨 𝒏−𝟏

𝒏−𝟏

Try solving these problems: For what values of n (order of the reaction) will the reaction reach completion at finite amount of time? For what values of n (order of the reaction) will the half-life of the reaction increase with increasing initial reactant concentration?

The reaction of butadiene gas (C

4

H

6

) to yield C

8

H

12

gas is described by the equation:

2C

4

H

6

(g) ⟶ C

8

H

12

(g)

This “dimerization” reaction is second order with a rate constant, k

but

, equal to

5.76 X 10

−

L mol

−

min

−

under certain conditions. If the initial concentration of

butadiene is 0.200 M, what is the concentration after 10.0 min?

n th

• Order Reaction ChE 2115: Chemical EXAMPLE

Consider the reaction 𝑎𝐴 + 𝑏𝐵 → 𝑐𝐶 Conversion ChE 2115: Chemical 𝑁𝐴0 − 𝑁𝐴 𝑎 = 𝑁𝐵0 − 𝑁𝐵 𝑏 𝑁𝐵 = 𝑁𝐵0 − 𝑏 𝑎 𝑁𝐴0 − 𝑁𝐴 𝑁𝐵 = 𝑁𝐵0 − 𝑏 𝑎 𝑁𝐴0𝑋𝐴 𝑁𝐴0 − 𝑁𝐴 𝑎 = 𝑁𝐶 − 𝑁𝐶 𝑐 𝑁𝐶 = 𝑁𝐶0 + 𝑐 𝑎 𝑁𝐴0 − 𝑁𝐴 𝑁𝐶 = 𝑁𝐶0 + 𝑐 𝑎 𝑁𝐴0𝑋𝐴 For a constant volume reaction 𝐶𝐵 = 𝐶𝐵0 − 𝑏 𝑎 𝐶𝐴0𝑋𝐴 𝑴𝑩𝑨 = 𝑪𝑩𝟎 𝑪𝑨𝟎 → 𝐶𝐵0 = 𝑀𝐵𝐴𝐶𝐴 𝑪𝑩 = 𝑪𝑨𝟎 𝑴𝑩𝑨 − 𝒃 𝒂 𝑿𝑨 𝐶𝐶 = 𝐶𝐶0 + 𝑐 𝑎 𝐶𝐴0𝑋𝐴

Consider the reaction A + 2B → 3C whose rate law is given by r A = - k A C A C B where k A = 0.03 mM

• 1 min - 1 . A 5.0 L solution initially contains 2 mM of specie A and 1 mM of specie B. Calculate the time it would take to reduce the concentration of B to 0.1 mM. Conversion ChE 2115: Chemical EXAMPLE

Consider the reaction 2A + B → 3C whose rate law is given by r A = - k A C A C B 2 where k A = 0.02 mM

• 2 min - 1 . A 1.0 L solution initially contains 2 mM of specie A and 2 mM of specie B. Calculate the total moles of product C produced after 1 hr reaction time. Conversion ChE 2115: Chemical EXAMPLE

EXAMPLE

𝑑𝐶𝐴 𝑑𝑡 = − 0. 02 𝐶𝐴𝐶𝐵 2 𝐶𝐴 = 𝐶𝐴0 1 − 𝑋𝐴 = 2 1 − 𝑋𝐴 𝑑𝐶𝐴 𝑑𝑡 = −𝐶𝐴 𝑑𝑋𝐴 𝑑𝑡 = − 2 𝑑𝑋𝐴 𝑑𝑡 𝐶𝐵 = 𝐶𝐴0 𝑀𝐵𝐴 − 𝑏 𝑎 𝑋𝐴 𝐶𝐵 = 2 1 − 1 2 𝑋𝐴 = 2 − 𝑋𝐴 By substitution into the rate equation, we get − 2 𝑑𝑋𝐴 𝑑𝑡 = − 0. 02 2 1 − 𝑋𝐴 2 − 𝑋𝐴 2 Conversion ChE 2115: Chemical Consider the reaction 2A + B → 3C whose rate law is given by rA = - kACACB^2 where kA = 0.02 mM-^2 min-^1. A 1.0 L solution initially contains 2 mM of specie A and 2 mM of specie B. Calculate the total moles of product C produced after 1 hr reaction time. By rearranging the equation, it becomes 𝑑𝑋𝐴 1 − 𝑋𝐴 2 − 𝑋𝐴 2 = 0. 02 𝑑𝑡 න 0 𝑋𝐴𝑓 𝑑𝑋𝐴 1 − 𝑋𝐴 2 − 𝑋𝐴 2 = 0. 02 1 ℎ𝑟 60 𝑚𝑖𝑛 ℎ𝑟 Thus 𝑋𝐴𝑓 = 0. 8876 𝑁𝐶 − 𝑁𝐶0 = 𝑐 𝑎 𝑁𝐴0𝑋𝐴 𝑁𝐶 − 𝑁𝐶0 = 3 2

2. 0 0. 8876 = 2. 66 𝑚𝑚𝑜𝑙 𝑚𝑜𝑙𝑒𝑠 𝐶 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 = 2. 66 𝑚𝑚𝑜𝑙

EXAMPLE

𝐶𝐴 = 𝐶𝐴0 1 − 𝑋𝐴 = 2 1 − 𝑋𝐴 𝐶𝐴 = 2 1 − 𝑋𝐴 𝐶𝐵 = 𝐶𝐴0 𝑀𝐵𝐴 − 𝑏 𝑎 𝑋𝐴 𝐶𝐵 = 2 1 − 1 2 𝑋𝐴 = 2 − 𝑋𝐴 In terms of XB 𝐶𝐵 = 𝐶𝐵0 1 − 𝑋𝐵 𝐶𝐵 = 2 1 − 𝑋𝐵 2 − 𝑋𝐴 = 2 1 − 𝑋𝐵 𝑋𝐵 = 1 − 2 − 𝑋𝐴 2 = 𝑋𝐴 2 Conversion ChE 2115: Chemical Consider the reaction 2A + B → 3C whose rate law is given by rA = - kACACB^2 where kA = 0.02 mM-^2 min-^1. A 1.0 L solution initially contains 2 mM of specie A and 2 mM of specie B. At the moment the conversion of A is 0.50, what is the conversion of B? If XA = 0. 𝑋𝐵 = 0. 25

Consider the reaction 2A + B → 3C whose rate law is given by r A = - k A C A C B 2 where k A = 0.02 mM

• 2 min - 1 . A 1.0 L solution initially contains 2 mM of specie A and 2 mM of specie B. Calculate the total moles of product C produced after 1 hr reaction time. Conversion ChE 2115: Chemical EXAMPLE