Question

Can you explain how to find the change in entropy from my data for these processes? (Process 5 change in entropy might be wrong)

Calla(s) HaO > Caa+ (aq) + 2C1- (aq) Temp. on thermometer rose Process S Equation : Sign of AHO Sign of Aso Sign of AGO I s Tempe happen Sº increases as the temperature increases (Temp. on Thermometer rose) Something happened to cause the temp to rise YES NO NONE Cindependent of T Product favored? Temperature dependent? Product favored at Process ba Equation : 2Al(s)+ 3 cucla(aq) → 2 A1 Clz (aq) + 3 Cu (s) Sign of AHO Temp. on thermometer rose Sign of Aso Sign of AGO Something happened → Red / Brown particles of Cu formed'in beaker. Product favored? Temperature dependent? YES E D Product favored at Process 6b Equation : a AlCl3(aq) + 3 Cu (s) –> 2Al(s) + 3 Cucla laq) Sign of AHO + This is the reverse lopposite process of ba, so heat must be absorbed. Sign of ASO Sign of AGO + Nothing happened per the reaction as written. There were no Alor Cucla (blue-green sitin) formede Product favored? DI Temperature dependent? Product favored at TTI

Answer 1

As we know entropy increases in following order

Solid<liquid<gas

In process 5, there are three moles of aqueous sol in product side that means entropy increases i.e delta s is positive

For delta g to be negative, temperature must be low.

Process 6a; delta s is negative and deltaH is also negative hence delta g to be negative there is no role of temperature,it could be high or low

Process6b; delta s is positive because in product 3moles of liq while in reactant side 2 miles of liq. delta H is Positive, for this reaction delta g can not be negative hence reaction can't be favoured in forward direction as it is the reverse of 6a.