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.

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.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.