Answer for PART A and PART B Entropy and the Second Law of Thermodynamics Accord

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Answer for PART A and PART B

Entropy and the Second Law of Thermodynamics

According to the second law of thermodynamics, all reactions proceed spontaneously in the direction that increases total entropy. The total entropy is defined as

%u0394Stotal=%u0394Ssystem+%u0394Ssurroundings

The direction of a spontaneous change is always determined by the sign of the total entropy change:

-The reaction is spontaneous if %u0394Stotal>0.

-The reaction is nonspontaneous if %u0394Stotal<0.

-The reaction is at equilibrium if %u0394Stotal=0.

The value of %u0394Stotal depends on %u0394Ssystem and %u0394Ssurroundings, which are defined by %u0394Ssystem=S(products)%u2212S(reactants) and %u0394Ssurroundings= %u2212%u0394Hsystem/T

Spontaneity is therefore affected by the enthalpy of the system and the Kelvin temperature.

CaO(s)+H2O(l)%u21CCCa(OH)2(s)

Assuming the commonly used standard-state temperature of 25%u2218C, calculate %u0394Stotal for this reaction using table from the table below.

Substance S%u2218
[J/(K%u22C5mol)] %u0394H%u2218f
(kJ/mol) CaO(s) 39.9 %u2212635.1 H2O(l) 69.9 %u2212285.8 Ca(OH)2(s) 83.4 %u2212986.1

Explanation / Answer

A.)deltaHsys = Sum of npH0f (products) -  Sum of nrH0f(reactants)             = [1mol(-986.1kJ/mol)] - [1mol(-635.1kJ/mol) - 1mol(-285.8kJ/mol)]             = - 986.1kJ + 635.14kJ + 285.8kJ             = - 65.16kJ deltaS surr = 65200 J/ 298 K = 219 J/K deltaSsys =Sum of npS0(products) - Sum of nrS0(reactants)            = [1mol(83.4J.mol-1.K-1) ] - [ 1mol(39.9J.mol-1.K-1 ) + 1mol (69.9J.mol-1.K-1)]            = -26.4J.K-1   deltaS total=219 -26.4=192.8>0 So The reaction is spontaneous if deltaStotal>0. So ANSWER is 193 J/K
B.)deltaSvap = deltaHvap / Tb         112.9J/K.mol = (38.0 x 10^3 J /mol ) / Tb So boiling point of the solvent Tb = 336.5 K                                                             = 63.5 oC          So the solvent boils under the reaction conditions(75 C) . So the solvent is not suitable . Anyprocess with negative entropy change is not feasible . So the solvent is not suitable because deltaStotal < 0 So ANSWER is D

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