Engineering Questions with Answers - Multiple Choice Questions

# Aerodynamics – Chemical Equilibrium in High Temperature Air

1 - Question

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
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

2 - Question

The dissociative – recombination reaction below yields in NO.
N + O ⇌ NO + e-
a) True
b) False
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 - Question

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
a) d[O2]dt = k[O2][M]
b) d[O]dt = 2k[O2][M]
c) d[O2]dt = 2k[O][M]
d) d[O]dt = k[O][M]
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: 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 - Question

For the equilibrium reaction, what is the relation between the forward and reverse reaction rate constant?
O2 + M ⇌ 2O + M
a) kf = kb[O][O2]
b) kf = kb[O]∗2[O2]∗
c) kf = kb[M]∗[O2]
d) kf = kb[O2][O]
Explanation: For the reaction which is in chemical equilibrium, the rate of change of the oxygen atom is zero i.e. 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: 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 [O]∗2[O2]∗

5 - Question

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