Question

What is the sign of △S for each of the following processes? I. The separation of gaseous molecules of UF6, into 238UF6 and 235UF6 at constant temperature and pressure. II. The dissolving of I2(s) in CCl4(l).

Answer 1

Step-by-step explanation:

Entropy is the degree of randomness or disorderliness of as system. The more random a system tends, the more positive the entropy. Gases have the highest entropy whereas solids have little to no entropy.

When a reaction goes from one initally with a lower entropy to one with a higher entropy, the change in entropy, ΔS is postive. ΔS is negative when a reaction goes from one with a higher entropy to one with lower entropy.

When there is no entropy change, both the reactants and products are in the same state.

In the first reaction, UF₆ is in a gaseous phase and it was seperated into two gaseous products, 238-UF₆ and 235-UF₆. Therefore, ΔS is zero i.e no change in entropy of the system.

In the second reaction, solid I₂ is dissolved in CCl₄. The solid substance goes into solution to form a liquid medium. This is an increase in the randomness or entropy of the system. Therefore ΔS is +ve.