Which Action Decreases CO2 at Equilibrium?

[ada0713]

Homework Statement

The following reaction is at equilibrium at one atmosphere, in a closed container
NaOH(s) + CO2(g)-> NaHCO3 (s)
which, if any, of the following actions will decrease the total amount of CO2 gas present at equilibrium?

1)decreasing the volume of the container
2) none of the above
3) removing half of the solid NaHCO3
4)adding mlore solid NaOH
5)adding N2 gas to double the pressure

The Attempt at a Solution

I chose (1) and got it wrong, so I chose (5) next but it was wrong again,
I thought that solid is not included in equilibrium constant which means that
it has little effect on equilibrium change, that's why I eliminated (3) and (4).
I guess for, (1), decreasing the volume wouldn't do much
to shift the equilibrium constant to the right.. then

I don't understand why (5) is wrong because if you add more gas,
it will double the pressure thus makeing the reactants to shift to the
right to re-establish the equilibrium constant,,

Is the answer none of the above?

Please help!

 

[chemisttree]

5 is wrong because the equilibrium concentration of CO2 in the gas phase is constant regardless of how much pressure there is from N2. The concentration of CO2 is actually the partial pressure of CO2 in that case. Your thinking on #3 is logical but not #4. Remember that the reaction occurs on the surface of the sodium hydroxide solid. What would adding more NaOH surface do?

 

[ada0713]

When you add more solid NaOH,, more CO2 molecules will react with NaOH; thus, decreasing the moles of CO2.. Am I right,,?
But still, does solid really have anything to do with this?

I don't think I understand the reason why we don't include solid in equilibrium constant