# Chemical Reaction Engineering Questions and Answers – Rate Laws and Stoichiometry Definitions

This set of Chemical Reaction Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Rate Laws and Stoichiometry Definitions”.

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

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

2. Arrhenius first suggested the temperature dependence of the rate constant.
a) True
b) False

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

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

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.

4. 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⁡$$(\frac{k1}{k2}) = \frac{E}{R} (\frac{T1}{T2})$$

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⁡$$(\frac{k1}{k2}) = \frac{E}{R} (\frac{1}{T2}-\frac{1}{T1}).$$

5. 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$$e^{-\frac{1}{RT}}$$

Explanation: According to transition state theory the relationship between temperature and rate constant is given by = koT$$e^{-\frac{E}{RT}}.$$
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6. Value of m to obtain the collision theory relation for temperature dependency of rate constant is ________
a) 1
b) 0
c) -1
d) 0.5

Explanation: The general equation for temperature dependency is k=koTm $$e^{-\frac{E}{RT}}$$
If m = 0 → Arrhenius equation
m = 1 → transition state theory
m = 0.5 → Collision theory.

7. Calculate the activation energy for the following data using transition state theory.

 T in deg C k 0 0.002 80 0.08

a) 30.07 KJ/mol
b) 50.43 KJ/mol
c) 34.37 KJ/mol
d) 82.31 KJ/mol

Explanation: When we have two values of k and T
k1=koT1$$e^{-\frac{E}{RT1}}$$ and k2=koT2$$e^{-\frac{E}{RT2}}$$
Modifying it gives
ln⁡(k1)=ln⁡(ko)+ln⁡(T1) – $$(\frac{E}{R})\frac{1}{T1}$$ and ln⁡(k2)=ln⁡(ko)+ln⁡(T2) – $$(\frac{E}{R})\frac{1}{T2}$$
On further simplification we get ln⁡$$(\frac{k1}{k2})$$=ln⁡$$(\frac{T1}{T2}) – \frac{E}{R}(\frac{1}{T1}-\frac{1}{T2})$$
ln⁡$$(\frac{0.002}{0.08})$$=ln⁡⁡$$(\frac{273}{353})-\frac{E}{8.314}(\frac{1}{273} – \frac{1}{353})$$
E = 34.377 KJ/mol.

8. The reaction rate of a bimolecular reaction at 300K is 10 times the reaction rate at 150K. Calculate the activation energy using collision theory.
a) 4928 J/mol
b) 5164 J/mol
c) 3281 J/mol
d) 1296 J/mol

Explanation: T1 = 150K, T2 = 300K and k2 = 10k1
When we have two values of k and T
k1=ko$$\sqrt{T1}e^{-\frac{E}{RT1}}$$ and k2=ko$$\sqrt{T2}e^{-\frac{E}{RT2}}$$
Modifying it gives
ln⁡(k1)=ln⁡(ko)+0.5ln⁡(T1) – $$(\frac{E}{R})\frac{1}{T1}$$ and ln⁡(k2)=ln⁡(ko) – $$(\frac{E}{R})\frac{1}{T2}$$
On further simplification we get ln⁡⁡$$(\frac{k1}{k2}) = 0.5ln⁡(\frac{T1}{T2}) \frac{E}{R} (\frac{1}{T1}-\frac{1}{T2})$$
ln⁡$$(\frac{k1}{10k1})$$=0.5ln⁡$$(\frac{150}{300}) – \frac{E}{8.314}(\frac{1}{300} – \frac{1}{150})$$
E = 4928 J/mol.

9. For the reaction -rA = $$\frac{1670[A][Po]}{6+CA} \frac{kmol}{ml.hr}$$ write the units for constants 6 and 1670.
a) kmol/ml and 1/hr
b) 1/hr and 1/hr
c) kmol and kmol
d) data insufficient

Explanation: To get the final units
-rA = $$\frac{1670[\frac{kmol}{ml}][\frac{kmol}{ml}]}{6+\frac{kmol}{ml}}$$
The units of constants have to be 1/hr for 1670 and kmol/ml for 6.

10. Which theory is the basis for Arrhenius equation?
a) Collision theory
b) Kinetic theory of gases
c) Ideal gas law
d) Charles and Boyles law

Explanation: Arrhenius equation is based on Kinetic theory of gases.

Sanfoundry Global Education & Learning Series – Chemical Reaction Engineering.

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