CHEMICAL ENGINEERING | Chemical Reaction Engineering

Carbon particles accummulated on the catalyst used in the gas oil cracking lies in the category of __________ poison.

Byadmin August 11, 2022

Question: Carbon particles accummulated on the catalyst used in the gas oil cracking lies in the category of __________ poison.
[A].

deposited

[B].

chemisorbed

[C].

selectivity

[D].

stability

Answer: Option A

Explanation:

No answer description available for this question.

CHEMICAL ENGINEERING | Chemical Engineering Basics

Elimination of brittleness resulting from welding of saw blades is done by __________ of the welded portion.

Question: Elimination of brittleness resulting from welding of saw blades is done by __________ of the welded portion.
[A].

annealing
[B].
toughening
[C].
work hardening
[D].
tempering

Answer: Option A

Explanation:

No answer description available for this question.

Read More Elimination of brittleness resulting from welding of saw blades is done by __________ of the welded portion.
CHEMICAL ENGINEERING | Chemical Process

Hydrolysis of sugar is called

Question: Hydrolysis of sugar is called
[A].

hydration
[B].
inversion
[C].
esterification
[D].
none of these

Answer: Option B

Explanation:

No answer description available for this question.

Read More Hydrolysis of sugar is called
CHEMICAL ENGINEERING | Fuels and Combustion

The calorific value of a gas of composition CO2 = 20%, H2 = 50%, CH4 = 30%, is y. If the composition of gas is changed to H2 = 50%, CH4 = 30%, CO2 = 10%, N2 = 10%, then the calorifice value will be

Question: The calorific value of a gas of composition CO2 = 20%, H2 = 50%, CH4 = 30%, is y. If the composition of gas is changed to H2 = 50%, CH4 = 30%, CO2 = 10%, N2 = 10%, then the calorifice value will be
[A].

y
[B].
0.95 y
[C].
1.05 y
[D].
1.8 y

Answer: Option A

Explanation:

No answer description available for this question.

Read More The calorific value of a gas of composition CO2 = 20%, H2 = 50%, CH4 = 30%, is y. If the composition of gas is changed to H2 = 50%, CH4 = 30%, CO2 = 10%, N2 = 10%, then the calorifice value will be
CHEMICAL ENGINEERING | Fluid Mechanics

In case of laminar flow of fluid through a circular pipe, the

Question: In case of laminar flow of fluid through a circular pipe, the
[A].

shear stress over the cross-section is proportional to the distance from the surface of the pipe.
[B].
surface of velocity distribution is a paraboloid of revolution, whose volume equals half the volume of circumscribing cylinder.
[C].
velocity profile varies hyperbolically and the shear stress remains constant over the cross-section.
[D].
average flow occurs at a radial distance of 0.5 r from the centre of the pipe (r = pipe radius).

Answer: Option B

Explanation:

No answer description available for this question.

Read More In case of laminar flow of fluid through a circular pipe, the
CHEMICAL ENGINEERING | Chemical Reaction Engineering

For all positive reaction orders for a particular duty,

Question: For all positive reaction orders for a particular duty,
[A].

mixed reactor is always larger than the plug-flow reactor.
[B].
ratio of the volume of the mixed reactor to that of the plug-flow reactor decreases with order.
[C].
reactor size is independent of the type of flow.
[D].
density variation during reaction affects design.

Answer: Option A

Explanation:

No answer description available for this question.

Read More For all positive reaction orders for a particular duty,
CHEMICAL ENGINEERING | Chemical Reaction Engineering

At a given temperature, K1, K2 and K3 are equilibrium constants for the following reactions 1, 2, 3 respectively.CH4(g) + H2O(g) CO(g) + 3H2(g), CO(g) + H2O(g) CO2(g) + H2(g)CH4(g) + 2H2O(g) CO2(g) + 4H2(g)Then K1, K2 and K3 are related as:

Question: At a given temperature, K1, K2 and K3 are equilibrium constants for the following reactions 1, 2, 3 respectively.CH4(g) + H2O(g) CO(g) + 3H2(g), CO(g) + H2O(g) CO2(g) + H2(g)CH4(g) + 2H2O(g) CO2(g) + 4H2(g)Then K1, K2 and K3 are related as:
[A].

K3 = K1.K2
[B].
K3 = (K1.K2)0.5
[C].
K3(K1+K2)/2
[D].
K3 = (K1.K2)2

Answer: Option A

Explanation:

No answer description available for this question.

Read More At a given temperature, K1, K2 and K3 are equilibrium constants for the following reactions 1, 2, 3 respectively.CH4(g) + H2O(g) CO(g) + 3H2(g), CO(g) + H2O(g) CO2(g) + H2(g)CH4(g) + 2H2O(g) CO2(g) + 4H2(g)Then K1, K2 and K3 are related as:

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