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Darya [45]
3 years ago
11

Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon

dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Chemistry
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

All of the above processes have a ΔS < 0.

Explanation:

ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.

The question requests us to identify the process that has a negative change of entropy.

carbon dioxide(g) → carbon dioxide(s)

There is  a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.

water freezes

There is  a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.

propanol (g, at 555 K) → propanol (g, at 400 K)

Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.

This reaction highlights a drop in temperature which means a negative change in entropy.

methyl alcohol condenses

Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.

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Which two changes would make this reaction reactant-favored?
Vera_Pavlovna [14]

Two changes would make this reaction reactant-favored

C. Increasing the temperature

D. Reducing the pressure

<h3>Further explanation</h3>

Given

Reaction

2H₂ + O₂ ⇒ 2H₂0 + energy

Required

Two changes would make this reaction reactant-favored

Solution

The formation of H₂O is an exothermic reaction (releases heat)

If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)  

While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient  

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As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)

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4 0
2 years ago
Your lab partner named this compound 3-methyl-4-n-propylhexane, but that is not correct.
loris [4]

<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

<u>Explanation:</u>

The IUPAC nomenclature of alkanes are given as follows:

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  • For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
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We are given:

An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms

The image of the given alkane is shown in the image below.

Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane

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