Question

(a) Calculate the change in entropy when 5.00 moles of ozone are compressed isothermally to one quarter of the origional volume. Treat ozone as an ideal gas.

(b) Calculate the residual entropy at T close to 0 K for 1.00 mol of bromobenzene, (C6H5Br).

(c) What is the sign on ∆S for the following reaction: NaCl (s) --> Na+ (aq) + Cl- (aq)? Explain why.

Answer 1

1.

If the compression is conducted isothermally, then T is constant. If T is constant, then the change in the internal energy of the gas (∆U) is zero, since the energy of an ideal gas depends only on the temperature.

∆U = 0 = q + w. Since the process can be carried out isothermally, dq = –dw = PdV

∆S = ∫dq/T = –∫dw/T = +(1/T)∫PdV = +(1/T)∫(nRT/V)dV = +(nR)∫dV/V where Vf = ¼Vi

∆S = nR•ln(Vf/Vi) = nR•ln(¼) = –(5 mol)(8.314 J/mol•K)•ln4
∆S = –23.61 J/K

2.

This molecule is a benzene ring with one of the 6 equivalent hydrogens replaced by a bromine atom. There are six choices for which carbon atom to bind the bromine to, and each of these possibilities has the same energy, so there are 6 possible configurations for. Using the Boltzmann entropy formula, the residual entropy for one mole of bromobenzene is then:

S_resid = (1*mol)*R*ln(6)

where R is the universal gas constant.

= (1 mol) * 8.314* 1.7917

= 14.89 J/K

3.

Delta S is the measure of entropy change within your system. Entropy is just a thermodynamic word for chaos. So if you think about things scattered all about, which side of the reaction has more "chaos". The easy answer is, whichever one has more particles. If you reaction goes from 3 separate elements like...

NaCl is (ideally) a perfectly ordered crystal, and in dissolving the NaCl one is forming a disordered mixture. Thus, most would immediately say that yes, delta S > 0 makes sense. So its +ve sign on the S

Hope this leads you in the right direction!