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Consider the following changes:

(a) COCl₂​(g)→CO(g) + Cl₂​(g)
(b) N₂​(g)→N₂​(l)
(c) CO(g) + H2₂O(g) →H₂​(g)+CO₂(g)
(d) Ca₃​P₂(s)+6H₂​O(l) →3Ca(OH)₂​(s)+2PH₃​(g)
(e) 2CH₃​OH(l) +3O₂​(g) →2CO₂​(g) +4H₂​O(l)
(f) I₂(s)→I₂​(g)

At constant temperature and pressure, in which of these changes is work done by the system on the surroundings?

User Lochemage
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1 Answer

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Final answer:

In the given changes, the reaction (f) 2CH₃OH(l) +3O₂​(g) →2CO₂​(g) +4H₂​O(l) does work on the surroundings. The combustion of methanol (CH₃OH) is exothermic and results in an increase in volume, leading to work done on the surroundings.

option e is the correct

Step-by-step explanation:

In the given changes, the reaction that does work on the surroundings is (f) 2CH₃OH(l) +3O₂​(g) →2CO₂​(g) +4H₂​O(l). In this reaction, the system releases energy in the form of heat and work.

The combustion of methanol (CH₃OH) to form carbon dioxide (CO₂) and water (H₂O) is exothermic, meaning heat is released. Additionally, during this reaction, the volume of the system increases as the gaseous products are formed, resulting in work being done on the surroundings.

User Kaloyan Stamatov
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