Consider the equilibrium system #CO(g) + Fe_3O_4(s) ⇌ CO_2(g) + 3FeO(s)# How does the equilibrium position shift as a result of each of the following disturbances?
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Left Right Not CO2 is removed by adding solid NaOH.
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Left Right Not Additional Fe3O4(s) is added to the system.
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Left Right Not Dry ice is addded at a constant T.
4 Left Right Not The volume is reduced by one-half at a constant T.
- Left Right Not CO is added.
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Left Right Not CO2 is removed by adding solid NaOH.
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Left Right Not Additional Fe3O4(s) is added to the system.
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Left Right Not Dry ice is addded at a constant T.
4 Left Right Not The volume is reduced by one-half at a constant T.
- Left Right Not CO is added.
1 Answer
- Right
- Not
- Left
- Not
- Right
Explanation:
Using the Le Chatelier's principle
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The equilibrium shifts rightwards (to the products) as the concentration of one of the (gaseous) product decreases.
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Change in the amount of
#Fe_3O_4# will not affect the equilibrium as it is in the solid state. (Only change in concentration of gaseous or aqueous reactants affects the equilibrium.) -
Dry ice (solid carbon dioxide) sublimate readily to produce gaseous
#CO_2# , increasing the concentration of products. (Also, the reaction is endothermic. http://www.chemteam.info/Thermochem/Hess-3eq-prob1-10.html) -
There's an equal number of gas molecules on both sides of the reaction. Therefore changes in volume have no influence on the equilibrium. (It does speed up reactions of both directions, anyway)
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Adding
#CO (g)# will increase the concentration of the reactants.
