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qaws [65]
3 years ago
7

Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reactio

n of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:
2NO(g)+O2(g)--->2NO2(g)
2NO2(g)+2SO2--->2NO(g)+2SO3(g)

Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.
Chemistry
2 answers:
vekshin13 years ago
4 0

Answer 1) In the given reaction of sulfuric acid


2NO_{(g)} + O_{2}_{(g)} ---->  2NO_{2}_{(g)}


2NO_{2}_{(g)} + 2SO_{2}_{(g)} ---->  2NO}_{(g)} + 2SO_{3}_{(g)}


On addition of nitrogen monoxide gas the reaction rate increases and more amount of product is formed.


So, it is clear that NO is the catalyst in this reaction.


Answer 2) This can be proven that NO is catalyst because it increases the rate of the reaction, but it is not consumed during the reaction, and it also gets regenerated at the end of reaction.


Hence, nitrogen mono oxide is considered as the catalyst in the given reaction.


Answer 3) It increases the rate of reaction by decreasing the activation energy of the reaction. Also it can be clearly seen in this reaction the NO is reacting with oxygen to lower the energy of activation. So, it is providing an alternative pathway for proceeding the reaction. This all confirms the assumptions of NO being the catalyst.


2NO_{(g)} + O_{2}_{(g)} ---->  2NO_{2}_{(g)}

ludmilkaskok [199]3 years ago
3 0

NO is the catalyst.

NO is the catalyst because it increases the rate of the reaction but is not consumed during the reaction.

NO increases the rate of the reaction by lowering the activation energy.

The reaction of NO with O2 provides an alternative reaction pathway with a lower activation energy.

are the answers for edg

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Answer:

1.5

Explanation:

Given that :

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Comparing the ratios :

Compound A:

S : O = 6.00 : 5.99

S/0 = 6.0g S / 5.99g O

Compound B :

S : O = 8.60 : 12.88

S / O = 8.60g S / 12.88g O

Mass Ratio of A / mass Ratio of B

(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)

(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)

(6 × 12.88) / (5.99 × 8.60)

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