Using Le Chatelier\'s principle, indicate how each change will affect the follow

Question

Using Le Chatelier's principle, indicate how each change will affect the following reactions. Justify your answer. 2 NaOH (aq) + CO2 (g) = Na2CO3 (aq) + H2O (l) ?H(degree) = -180 kj/mol a)- Removing Na2CO3 from the reaction vessel. b)- Adding more CO2 (g) thereby increasing the pressure in the reaction vessel. Consider both the addition of reagent and the change in pressure in your answer. c)- Adding an inert gas to the reaction mixture to increase the pressure in the reaction vessel. d)- Lowering the temperature of the reaction vessel.

Explanation / Answer

You wrote: NaHCO3 => NaOH (aq) + CO2(g) ???? Are you starting with NaHCO3(s) or NaHCO3(aq)? NaHCO3(s) --H2O--> Na+ + HCO3- ..... solid sodium bicarbonate dissociates into ions Sodium ions do nothing, and are therefore spectator ions. The only reaction is the equilibrium between bicarbonate ions, hydroxide ions and dissolved CO2. This is why a solution of sodium bicarboate is slightly basic. HCO3- OH- + CO2(aq) .... this is the net ionic equation, if you're starting with a solution of sodium bicarbonate If you're starting with solid NaHCO3, then there are no spectator ions. Everything changes. NaHCO3(s) --> Na+ + OH- + CO2(g) .... this is the net ionic equation, if you're starting with solid sodium bicarbonate. =========== Follow up ============= Karine echoed my first thought, as well, but based on Sdjflk's use of the (aq) state symbol, then I supposed the reaction to be in aqueous solution, rather than heating solid sodium bicarbonate. Heating solid sodium bicarbonate will produced Na2CO3 and CO2 gas and water vapor, but I don't think that is what Sdjflk intended. Sdjflk's reaction is simply a hydrolysis reaction where HCO3- reacts in water to make a solution that is slightly basic.

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