La chatelier principle -- help me please

Use Le Châtelier’s principle to describe and explain the shifts in the following
equilibrium reactions:
i) KI(s) - -> K+(aq) + I-(aq) …………… Pb(NO3)2(s) is added
ii) 4HCl(g) + O2(g) - -> 2H2O(g) + 2Cl2(g) + 113 kJ ……. Temperature is
decreased
iii) H2(g) + I2(g) - -> 2HI(g) ……..Catalyst is added


Please help me with my question. I appreciate it. I don't have any teacher to ask. Its online course.

i) KI(s) <==> K+(aq) + I-(aq) …………… Pb(NO3)2(s) is added
There is only an equilibrium if the solution is saturated in KI. Adding lead(II) ions will cause some of the I- to precipitate as PbI2. This reduces the concentration of I- and causes a shift to the right, causing additional KI solid to dissolve.

ii) 4HCl(g) + O2(g) <==> 2H2O(g) + 2Cl2(g) + 113 kJ ……. Temperature is decreased
Think of heat as a product. Reducing the temperature, reduces the heat on the right hand side and shifts the equilibrium to the right, making more products and less of the reactants.

iii) H2(g) + I2(g) <==> 2HI(g) ……..Catalyst is added
Nothing. A catalyst doesn't affect the position of the equilbrium. It means that the system will reach equilibrium quicker.