Lime (CaO) is used to prevent SO2 from escaping smokestacks of coal burning power plants by the formation of solid CaSO4•2H2O also known as the mineral gypsum. One of the important reactions in the overall process is the dissociation of CaCO3(s).
b. If a 12.0 g sample of solid CaCO3 is placed in an evacuated vessel at 725oC, what will the pressure of CO2 be when the system reaches equilibrium?
c. Which of the following actions cause an increase in the pressure of CO2 in the vessel described in question 2 above? Make your choice then justify your answer.
i. Addition of He gas
ii. Addition of SO3 gas
iii. Addition of more CaCO3 solid
iv. Increasing the volume of the vessel
d. If a 12.0 g sample of solid CaCO3 is placed in a vessel at 750oC in which the pressure of CO2 is 2.5 atm what mass of lime will form?
For a, do you have a Kc or Kp? Or does the question assume ALL of the CaCO3 decomposes?
I have part b if that helps
100 g of CacO3 give 44 g of CO2
so 12 g will give 5.28 g of CO2
so moles of CO2 = 0.12 moles
Volume = 0.12 X 22.4 L = 2.688 L
PV = nRT
Pressure = 0.12 X 0.0821 X 998 / 2.688 = 3.65 atm
To find the answers to these questions, we need to use the principles of chemical equilibrium and stoichiometry.
b. To determine the pressure of CO2 at equilibrium, we need to calculate the number of moles of CO2 produced from the 12.0 g sample of CaCO3. The balanced equation for the dissociation of CaCO3 is:
CaCO3(s) ⇌ CaO(s) + CO2(g)
The molar mass of CaCO3 is 100.087 g/mol. So, we can calculate the number of moles of CaCO3:
moles of CaCO3 = mass / molar mass
= 12.0 g / 100.087 g/mol
= 0.1199 mol
According to the balanced equation, the stoichiometric ratio of CaCO3 to CO2 is 1:1. Therefore, the number of moles of CO2 produced will be the same as the number of moles of CaCO3, which is 0.1199 mol.
Now, we can use the ideal gas law to calculate the pressure of CO2 at equilibrium. The ideal gas law equation is:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
Given that the vessel is evacuated, the initial pressure of CO2 is zero. The temperature is given as 725 °C, which is 998 K.
Using the ideal gas law, we can rearrange the equation to solve for P:
P = (nRT) / V
Since the vessel is evacuated, the volume will be determined by the size of the vessel. However, the volume does not affect the pressure at equilibrium. Therefore, we can use any value for the volume.
Let's assume that the volume is 1 L. Plugging in the values, we get:
P = (0.1199 mol * 0.0821 L·atm/(mol·K) * 998 K) / 1 L
= 9.7847 atm
So, the pressure of CO2 at equilibrium will be approximately 9.78 atm.
c. To determine which actions cause an increase in the pressure of CO2 in the vessel, we need to consider Le Chatelier's principle. According to this principle, when a system at equilibrium is subjected to a change, it will shift to counteract the change and establish a new equilibrium.
i. Addition of He gas:
The addition of an inert gas (such as He) at a constant volume does not affect the equilibrium position because it does not change the concentration of any of the reactants or products. Therefore, the pressure of CO2 will remain unchanged.
ii. Addition of SO3 gas:
The addition of SO3 gas will have no direct effect on the reaction between CaCO3 and CO2. It will not change the concentration of CO2, so the pressure of CO2 will remain the same.
iii. Addition of more CaCO3 solid:
According to Le Chatelier's principle, adding more CaCO3 solid will increase the concentration of CaCO3 and disturb the equilibrium. To counteract this increase, the reaction will shift to the right to produce more CaO and CO2. As a result, the pressure of CO2 will increase.
iv. Increasing the volume of the vessel:
Increasing the volume at a constant temperature will cause the system to shift in the direction that produces more moles of gas. In this case, the reaction produces one mole of CO2 gas. Therefore, increasing the volume will favor the formation of more CO2, leading to an increase in the pressure of CO2.
So, the actions that cause an increase in the pressure of CO2 in the vessel are the addition of more CaCO3 solid and increasing the volume of the vessel.
d. To determine the mass of lime (CaO) that forms from the given sample of CaCO3, we need to use stoichiometry and the balanced equation:
CaCO3(s) ⇌ CaO(s) + CO2(g)
From question b, we found that the number of moles of CO2 produced from the 12.0 g sample of CaCO3 is 0.1199 mol. Since the stoichiometric ratio of CaCO3 to CaO is 1:1, the number of moles of CaO will also be 0.1199 mol.
The molar mass of CaO is 56.077 g/mol. So, we can calculate the mass of CaO:
mass of CaO = moles of CaO * molar mass
= 0.1199 mol * 56.077 g/mol
= 6.744 g
Therefore, the mass of lime (CaO) that will form is approximately 6.744 g.