Chemical Reaction Engineering Questions and Answers Part-12

1. Space time is defined as ____
a) Time required to process unit volume of feed
b) Time required for the reaction to occur
c) Time required to produce unit volume of product
d) Time to process the entire volume of feed entering

Answer: a
Explanation: Space time is the time spent by a unit volume of reactants inside the reactor. It is the performance measure for a flow reactor.

2. The unit of space velocity is ____
a) sec
b) sec-1
c) m3
d) m-3

Answer: b
Explanation: Space velocity has the dimensions of time-1. A space velocity of 2 hr-1 implies that 2 reactor volumes are fed per hour.

3. Space time as a function of molar feed rate and initial concentration is given as ____
a) \(\frac{VC_{A0}}{F_{A0}} \)
b) \(\frac{C_{A0}}{VF_{A0}} \)
c) \(\frac{VF_{A0}}{C_{A0}} \)
d) \(\frac{C_{A0}}{F_{A0}} \)

Answer: a
Explanation: Space time is the ratio of the product of reactor volume and initial concentration to the feed rate. \(\frac{VC_{A0}}{F_{A0}} \) has the same unit as that of space time.

4. If V represents volume and Q represents volumetric flow rate, then which of the following is a valid expression for space velocity?
a) \(\frac{V}{Q} \)
b) \(\frac{Q}{V} \)
c) Q × V
d) \(\frac{V}{C_{A0}} \)

Answer: b
Explanation: Space velocity is the ratio of volumetric flow rate to volume. It has the unit hr-1.

5. State true or false.
Space velocity is the number of unit volumes of feed that can be treated per unit time.
a) true
b) false

Answer: a
Explanation: Space velocity is the inverse of space time. It gives a measure of what volume of feed can be processed in a unit time.

6. Hydrogenation of oil is an example of ___________ reactor
a) Homogeneous
b) Heterogeneous
c) Autocatalytic
d) Insufficient data

Answer: b
Explanation: The reactants are of different phases i.e. Hydrogen (Gas phase) and oil (Liquid phase). Ni catalyst (Solid) is also used.

7. Arrhenius first suggested the temperature dependence of the rate constant.
a) true
b) false

Answer: a
Explanation:
k = \(koe^{\frac{-E}{RT}}\)
This is known as Arrhenius equation and it gives the relation between temperature and rate constant.

8. The slope of the line in the graph gives the ___________
a) Activation energy
b) Rate constant
c) Frequency factor
d) Insufficient data

Answer: a
Explanation: The Arrhenius equation k=\(koe^{\frac{-E}{RT}}\) can be modified as
ln⁡(k)=ln⁡(ko) – \((\frac{E}{R})\frac{1}{T} \)
The slope of the line gives E/R from which Activation energy can be determined.

9. Arrhenius equation to determine the activation energy.
a) ln⁡\((\frac{k1}{k2}) = \frac{E}{R} (\frac{1}{T2}-\frac{1}{T1}) \)
b) ln⁡\((\frac{k1}{k2}) = \frac{E}{R} (\frac{1}{T2}+\frac{1}{T1}) \)
c) ln⁡\((\frac{k1}{k2}) = -\frac{E}{R} (\frac{1}{T2}-\frac{1}{T1}) \)
d) ln⁡\((\frac{k1}{k2}) = \frac{E}{R} (\frac{T1}{T2}) \)

Answer: a
Explanation: When we have two values of k and T
k1 = ko\(e^{-\frac{E}{RT1}}\) and k2 = ko\(e^{-\frac{E}{RT2}} \)
Modifying it gives
ln⁡(k1) = ln⁡(ko) – \((\frac{E}{R})\frac{1}{T1}\) and ln⁡(k2) = ln⁡(ko) – \((\frac{E}{R})\frac{1}{T2} \)
On further simplification we get ln⁡\((\frac{k1}{k2}) = \frac{E}{R} (\frac{1}{T2}-\frac{1}{T1}). \)

10. Temperature dependence of rate constant according to transition state theory is ______
a) k=ko\(e^{-\frac{E}{RT}} \)
b) k=\(e^{-\frac{E}{RT}} \)
c) k=koT\(e^{-\frac{E}{RT}} \)
d) k=ko\(e^{-\frac{1}{RT}} \)

Answer: c
Explanation: According to transition state theory the relationship between temperature and rate constant is given by = koT\(e^{-\frac{E}{RT}}. \)