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Question:For the following equilibrium system, which of the changes will form more CaCO3?

CO2(g)+Ca(OH)2(s)→CaCO3(s)+H2O(l)∆Ho=-113kJ

(a) Decrease temperature at constant pressure (no phase change)

(b) Increase volume at constant temperature

(c) Increase partial pressure of CO2

(d) Remove one-half of the initial CaCO3

Short Answer

Expert verified
  1. Decrease in temperature gives more CaCO3 in product side.
  2. Increase in volume doesn’t form more CaCO3
  3. Increase in partial pressure doesn’t give change in product side. Both forward and backward species kept constant at its concentration.
  4. Equilibrium shifts towards left side, to reactant side. So, no chance for more CaCO3 synthesis.

Step by step solution

01

Decrease in temperature

Given reaction is exothermic because∆Ho is negative. If the enthalpy change is negative, then the reaction must be negative. There occurs a temperature decrease in heat releasing reactions at constant pressure. Exothermic reaction are forward reactions. So, decrease in temperature favors more yield of CaCO3.

02

Increase in volume

Decrease in pressure means there is an increase in volume. So, that reaction proceed left side to yield more gaseous molecules. Thus, increase in volume doesn’t favor more CaCo3 in product side.

03

Increase in partial pressure

Partial pressures are mainly appeared in reaction quotient. They didn’t have any effect in reaction. So, addition of inert gas doesn’t affect the changes in partial pressure. So, the product side and reactant side remain same.

04

Step 4: Removing one-half of the initial CaCO3

If a product concentration is decreased or it removed, then the reaction will shift towards backward direction. That means equilibrium shifts left side to reduce the disturbance occurred there. So, there is no chance of more product yield.

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