Shown below is the chemical equation for the thermal decomposition of calcium carbonate.
CaCO3(s) <-> CaO(s) + CO2(g)
H°rxn = 175 kJ
How will the amount (not concentration) of CaCO3(s) change with the following stresses?
CO2(g) is removed?
Decrease, Increase or No Change
CaO(s) is added?
Decrease, Increase or No Change
The temperature is increased.
Decrease, Increase or No Change
The volumes of the container is decreased.
Decrease, Increase or No Change
All of these questions are Le Chatelier's Principle. This really is a simple concept. Tell me what you don't understand about it and perhaps I can set you up to answer all of them.
CO2(g) is removed:
The removal of CO2(g) will shift the equilibrium to the right, according to Le Chatelier's principle. This means that more CaCO3(s) will decompose to produce CaO(s) and CO2(g) to restore equilibrium. Therefore, the amount of CaCO3(s) will decrease.
CaO(s) is added:
Adding CaO(s) will increase the concentration of CaO(s) in the system. According to Le Chatelier's principle, this will shift the equilibrium to the left, reducing the amount of CaO(s) produced. Since CaO(s) is produced from the decomposition of CaCO3(s), a decrease in CaO(s) will result in a decrease in the decomposition of CaCO3(s). Therefore, the amount of CaCO3(s) will decrease.
The temperature is increased:
An increase in temperature favors the endothermic direction of the reaction, as indicated by the positive ΔH°rxn value. Therefore, increasing the temperature will shift the equilibrium to the left. Since CaCO3(s) decomposes to produce CaO(s) and CO2(g) in the forward reaction, the amount of CaCO3(s) will decrease.
The volumes of the container is decreased:
Decreasing the volume of the container will increase the pressure in the system. According to Le Chatelier's principle, an increase in pressure will favor the direction with fewer moles of gas. In the given equation, the forward reaction produces one mole of gas (CO2(g)), while the reverse reaction has no gas molecules. Therefore, increasing pressure by decreasing the volume will favor the reverse reaction, leading to a decrease in the amount of CaCO3(s). Consequently, the amount of CaCO3(s) will decrease.
To determine how the amount of CaCO3(s) will change with the given stresses, we need to consider Le Chatelier's principle. This principle states that if a system at equilibrium is subjected to a stress, it will respond in a way that minimizes the effect of that stress.
1. If CO2(g) is removed:
The equilibrium will shift to the right to compensate for the loss of CO2. This means that more CaCO3(s) will decompose to form more CO2(g) and restore the balance. Thus, the amount of CaCO3(s) will decrease.
2. If CaO(s) is added:
The addition of CaO(s) does not directly influence the decomposition of CaCO3. It only provides more CaO to react with CO2 if it is produced. Therefore, the amount of CaCO3(s) will remain unchanged.
3. If the temperature is increased:
The thermal decomposition of CaCO3 is endothermic, meaning it requires heat input. According to Le Chatelier's principle, increasing the temperature will favor the endothermic reaction. Therefore, the system will tend to shift towards the left to absorb more heat, resulting in a decrease in the amount of CaCO3(s).
4. If the volumes of the container are decreased:
Changes in pressure or volume affect the equilibrium position if gases are involved. However, in this reaction, only one gaseous component (CO2) is present. Decreasing the volume will increase the pressure, which will lead to a shift in the equilibrium to the side with fewer moles of gas molecules. In this case, the right side of the equation has one mole of CO2(g), while the left side has one mole of CaCO3(s). So, reducing the volume will have no effect on the amount of CaCO3(s).
In summary:
- If CO2(g) is removed, the amount of CaCO3(s) will decrease.
- If CaO(s) is added, the amount of CaCO3(s) will remain unchanged.
- If the temperature is increased, the amount of CaCO3(s) will decrease.
- If the volumes of the container are decreased, the amount of CaCO3(s) will remain unchanged.