Aerodynamics Questions and Answers – Chemical Equilibrium in High Temperature Air

This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Chemical Equilibrium in High Temperature Air”.

1. Which of these reactions is not carried out for the formation of nitric oxide?
a) Dissociation reaction of oxygen molecule
b) Dissociation reaction of nitrogen molecule
c) Shuffle reaction of nitric oxide molecule
d) Dissociative – recombination reaction of nitrogen and oxygen
View Answer

Answer: d
Explanation: The reaction taking place below 9000 K for the formation of nitric oxide are dissociation reaction and bimolecular exchange reaction also known as shuffle reaction. They are as follows:

           O2 + M ⇌ 2O + M
           N2 + M ⇌ 2N + M
           NO + M ⇌ N + O + M
           O2 + N ⇌ NO + O

The first two reaction are the dissociation of oxygen and nitrogen molecule respectively. The third and the fourth equations are the shuffle reactions.

2. The dissociative – recombination reaction below yields in NO.

        N + O ⇌ NO + e- 

a) True
b) False
View Answer

Answer: b
Explanation: The reaction above is a kind of dissociate – recombination reaction in which the nitrogen and oxygen result in NO+ ion instead of NO which combines with the electron producing the dissociated product N + O.

3. What is the equation of rate of formation of oxygen atoms in the following chemical reaction? (M is the collision molecule)

            O2 + M → 2O + M
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a) \(\frac {d[O_2]}{dt}\) = k[O2][M]
b) \(\frac {d[O]}{dt}\) = 2k[O2][M]
c) \(\frac {d[O_2]}{dt}\) = 2k[O][M]
d) \(\frac {d[O]}{dt}\) = k[O][M]
View Answer

Answer: b
Explanation: For the chemical reaction where a collision particle M collides with the oxygen molecule for achieving equilibrium condition, the rate of formation of the oxygen atom is given by:
\(\frac {d[O]}{dt}\) = 2k[O2][M]
Where, [O] is the number of moles of oxygen atom per unit volume of mixture
[O2] is the number of moles of oxygen molecule per unit volume of mixture
k is the reaction rate constant.

4. For the equilibrium reaction, what is the relation between the forward and reverse reaction rate constant?

          O2 + M ⇌ 2O + M 

a) kf = kb\(\frac {[O]}{[O_2]}\)
b) kf = kb\(\frac {[O]^{*2}}{[O_2]^{*}}\)
c) kf = kb\(\frac {[M]^{*}}{[O_2]}\)
d) kf = kb\(\frac {[O_2]}{[O]} \)
View Answer

Answer: b
Explanation: For the reaction which is in chemical equilibrium, the rate of change of the oxygen atom is zero i.e. \(\frac {d[O]}{dt}\) = 0. At equilibrium, [O2] = [O2]*, [M] = [M]* and [O] = [O]* where the asterix lets us know the equilibrium condition.
The net rate of reaction is the net sum of both forward and reverse reaction resulting in:
\(\frac {d[O]}{dt}\) = 2kf [O2][M] – 2kb [O]2 [M]
After substituting the equilibrium conditions in the above equation, we get
0 = 2kf [O2]* [M]* – 2kb [O]*2 [M]*
2kf [O2]* [M]* = 2kb [O]*2 [M]*
kf = kb \(\frac {[O]^{*2}}{[O_2]^{*}}\)

5. What is the value of equilibrium constant based on concentrations if the forward and reverse rate constant of a non – equilibrium reaction are 1.85 × 10-4 and 2.34 × 10-5?
a) 6.45
b) 7.91
c) 8.43
d) 2.31
View Answer

Answer: b
Explanation: Given, kf = 1.85 × 10-4 , kb = 2.34 × 10-5
The equilibrium constant based on concentrations for non – equilibrium is same as derived for the equilibrium condition since it relates the forward and reverse rate constants. It is given by:
kc = \(\frac {k_f}{k_b}\)
Substituting the values, we get,
kc = \(\frac {1.85 × 10^{- 4}}{2.34 × 10^{- 5}}\) = 7.91

6. 2H2 + O2 → 2H2O is an elementary chemical reaction.
a) True
b) False
View Answer

Answer: b
Explanation: The reaction between hydrogen and oxygen molecule to form water molecule involves a series of chemical reactions instead of one single step. Thus, it is not an elementary chemical reaction.

Sanfoundry Global Education & Learning Series – Aerodynamics.

To practice all areas of Aerodynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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