The following reaction is at equilibrium at a pressure of 1 atm, in a closed container.
NaOH(s) + CO2(g) NaHCO3(s); ΔH°rxn < 0
Which, if any, of the following actions will decrease the concentration of CO2 gas present at equilibrium?
A. lowering the temperature
B. adding more solid NaOH… solids don’t affect Keq.
C. increasing the volume of the container... technically, this makes sense, but think thermodynamics regarding the problem
D. adding N2 gas to double the pressure… not in reaction
why is the answer A

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  1. Write the reaction this way.
    NaOH(s) + CO2(g) ==>NaHCO3(s) + heat (and I REALLY don't like it that you can't put in an arrow.)
    A. So adding heat MUST drive the reaction to the left; therefore, lowering the T will do the opposite.
    B. right.
    C. No it doesn't make sense. Decreasing volume is increasing pressure which shifts reaction to the right(fewer moles) so decreasing pressure will shift to the left and the problem wants to decrease CO2.
    D. See C.

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