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vaieri [72.5K]
3 years ago
6

What is the sign of △S for each of the following processes? I. The separation of gaseous molecules of UF6, into 238UF6 and 235UF

6 at constant temperature and pressure. II. The dissolving of I2(s) in CCl4(l).
Chemistry
1 answer:
gavmur [86]3 years ago
7 0

Answer:

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.

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Solution:- Wavelength is given and we are asked to calculate the frequency. The equation that correlates frequency with wavelength is:

\nu =\frac{c}{\lambda }

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