Question

C: CaCO3 (s) ----> CaO (s) + CO2 (g) delta H= +178 kJ/mol
D: PCl3 (g) + Cl2 (g) -----> PCl5 (g) delta H= - 88 kJ/mol

Classify these changes as either LEFTWARD SHIFT, RIGTHWARD SHIFT, or NO SHIFT

System C increase temperature
System C decrease temperature
System D increase temperature
System D decrease temperature

Answer 1

since ?H for CaCO3 (s)< ----> CaO (s) + CO2 (g) is positive(endothermic) so this reaction can be written in this way...
CaCO3 (s) + Heat <-----> CaO (s) + CO2 (g) ..(Treat Heat as a reagent)
if now you increase the temperature i.e. Heat is supplied to the system then according to Le Chateller's principle equilibrium will shift to that side that absorbs heat i.e. in forward direction or rightward shift ...
similalrly if temperature is decreased equilibrium shifts to backward direction or leftward shift ...

for system D ?H is negative (exothermic) ...so this reaction can be written in this way :
PCl3 (g) + Cl2 (g) <----> PCl5 (g) + Heat
if temperature is increased then equilibrium will shift in that direction which absorbs heat i.e. in backward direction or leftward shift...
and decreasing temperature will shift the equilibrium to right..

...................

And

Assume that these are meant to be equilibria.
in which case, C is intended to be a shift to the left. However, you might point out to teacher that as entropy is negative (inc in the no moles of gas) then increasing temp would tend to oppose the positive value of the enthalpy change, so that the free energy (delta G) would become increasingly negative. Since no information is given on the temperature of this rxn, then it is not possible to say.

In D, there are 2 moles on the left an d one on the right, entropy is positive, the term -(TdeltaS) becomes more negative as temp increases, in this case, teh rxn is favorable at all temperatures. If it is at equilibrium, an inc in temp would certainly shift the balance in the endothermic direction, to the left.