# 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?

• Consider the change in free energy $\Delta G =\Delta H -T\Delta S$. As A is a solid as long as it is in excess, over its stoichiometric amount, its saturated vapour pressure is constant at constant T. – porphyrin Aug 1 '16 at 19:34