Given the baking reaction 2NaHCO_3(s) Na_2CO_3 (s) + CO_2(g) + H_2O(g) If this r

Question

Given the baking reaction 2NaHCO_3(s) Na_2CO_3 (s) + CO_2(g) + H_2O(g) If this reaction is endothermic, rewrite the equation, adding in the heat term - how could I cause it to make more CO_2, increase or decrease the temperature? Explain. exothermic, heat term is on the right, endo is opposite Would adding more baking soda cause it to rise more? Explain...... Think of this if it all went to completion, would it rise more it more soda was added? What would happen to the equilibrium if carbon dioxide is removed as it is produced ? Is it removed as the cake rises? This is removing a product as it is produced See example 14.11 for adding a product and how equilibrium is shifted Does rising only depend on temperature, what about atmospheric pressure? Explain..... Pressure is a factor if gases are involved Are gases involved? See Example 14 12 What about volume? Is the volume fixed in a cake rising situation and if so. would adding more baking soda ever do any good? How would a catalyst affect the equilibrium above?

Explanation / Answer

a) If reaction is endothermic, (ie. it need heating to go in right direction): For more CO2 production, heating is required. So increase temperature.

b) If more baking soda added the more CO2 and Water will form.

c) If CO2 is removed then reaction will proceed towards right forming more CO2 and water

d) If pressure increases then reaction goes towards left forming more baking soda (as there is no gaseous species in left side of reaction). and if pressure decreases the reaction will follow right direction.

e) if volume is fixed that means pressure will rise, hence equiliibrium shifts towards right.

) There is no net effect of catalyst on the chemical equilibrium.

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