How does one create a "product-favored reaction equilibrium"?


> 
> For the following reaction equilibrium, which of the following actions
>  would not only temporarily increase the yield of product, but also
>  result in a more product-favored reaction equilibrium overall (i.e.,
>  increase the Keq)? 
> 
> 
> $\ce{2A(s) + 3B(g) ↔ A2B3(g)}$ 
> 
> 
> ΔH = −50 kJ/mol
> 
> 
> (1) compressing (decreasing the volume of) the reaction vessel and
>  increasing the temperature
> 
> 
> *Correct answer: (2) compressing (decreasing the volume of) the
>  reaction vessel and decreasing the temperature*
> 
> 
> (3) expanding (increasing the volume of) the reaction vessel and
>  increasing the temperature
> 
> 
> (4) expanding (increasing the volume of) the reaction vessel and
>  decreasing the temperature
> 
> 
> **Wrong answer I'm curious about: (5) increasing the amount of A(s)
>  and decreasing the temperature**
> 
> 
> 


I know that increasing the amount of A(s) will not affect the equilibrium because it is a solid and solids exist at a constant concentration in general, so we ignore them in equilibrium constants. 


I think that there are actually two parts to this question: 



> 
> For the following reaction equilibrium, which of the following actions
>  would not only **temporarily increase the yield of product**, but also
>  result in a more product-favored reaction equilibrium overall (i.e.,
>  **increase the Keq**)? 
> 
> 
> 


To temporarily increase the yield, we must increase the pressure/decrease the volume of the container. But to actually change the K so that it is increased, we must change the temperature. Since this reaction is exothermic, decreasing the temperature will do the trick. 


Is my reasoning valid? 


