[sodium.dioxid]

Say I have a solid in a container decomposing into gas. I double the amount of solid present. Why doesn't the concentration of the gas go up. I think I know why: doubling the amount of solid will reduce the volume available for gas (pressure increases). The system responds by reducing the pressure by favoring solid formation. Thus, the concentration/volume ratio is maintained. Can someone confirm this for me? This would mean that rate=k for the solid. Is this true?

[Dan]

Can you write an equilibrium constant for the reaction:

A(s) B(g)

[sodium.dioxid]

Can you write an equilibrium constant for the reaction:

A(s) B(g)

This is not the problem. I know that the constant is K_eq=B. Yes, the equation says it is independent of the solid. I was trying to understand the "why".

[Dan]

Oh, I see. Let's see:

doubling the amount of solid will reduce the volume available for gas (pressure increases). The system responds by reducing the pressure by favoring solid formation. Thus, the concentration/volume ratio is maintained. Can someone confirm this for me?

I think you're basically there. I would say that when considering this problem, you can ignore the volume of the solid. Reason being, the density of the solid >> gas, so changes in solid volume are negligibly small in comparison to the gas volume.

From there the argument can follow what you have said. The concentration of a gas (ideal) is directly proportional to its pressure. As more gas is formed, pressure increases. As this happens, you can imagine high pressure pushing the gaseous molecules back together again, forming solid (Le Chatelier). Increasing pressure favours solid formation, and eventually you get to a point where the rate of solid formation is the same as rate of gas formation: equilibrium.

The solid does not appear in the rate equation because we approximate its volume to constant.

[sodium.dioxid]

Thank you!

[juanrga]

Can you write an equilibrium constant for the reaction:

A(s) B(g)

This is not the problem. I know that the constant is K_eq=B. Yes, the equation says it is independent of the solid. I was trying to understand the "why".

I do not think that the concentration of gas in independent of the amount of solid. Put zero amount of solid and you will obtain zero concentration of gas.

The equation K_eq=B = B/A (because A=1) says that I will find B for each unit of A.

[sodium.dioxid]

Can you write an equilibrium constant for the reaction:

A(s) B(g)

This is not the problem. I know that the constant is K_eq=B. Yes, the equation says it is independent of the solid. I was trying to understand the "why".

I do not think that the concentration of gas in independent of the amount of solid. Put zero amount of solid and you will obtain zero concentration of gas.

The equation K_eq=B = B/A (because A=1) says that I will find B for each unit of A.

Please refer to the discussion titled "Does rate of dissolution depend on the size of the solid?".