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brilliants [131]
1 year ago
5

in the next three problems, we will attempt to solve the problem: what is the standard entropy change for the combustion of 1 mo

le of diamond at room temperature?
Chemistry
1 answer:
hichkok12 [17]1 year ago
7 0

The standard entropy change for the combustion of 1 mole of diamond at room temperature is 6.22  J / K mol.

The reaction is given as :

C (diamond)  + O₂   ----->  CO₂

At room temperature the value of S°  is given as follows :

C = 2.38 J / K mol

O₂ = 205.2 J/Kmol

CO₂ = 213.8 J/Kmol

The  standard entropy change is given as :

ΔS° = ∑S° product  - ∑ S° reactant

ΔS° = ( 213.8 ) - ( 2.38 + 205.2)

ΔS° = 213.8 - 207.58

ΔS° = 6.22  J / K mol

Thus, The standard entropy change for the combustion of 1 mole of diamond at room temperature is 6.22  J / K mol.

To learn more about entropy change here

brainly.com/question/28380336

#SPJ4

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Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

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