CaCO3(s) <--> CaO(s) + CO2 (g)
A 5.9-gram sample of CaCO3 (molar mass = 100.) is placed in an evacuated 2.00L container and heated to 200^oC. Equilibrium is reached when 5.15% of the CaCO3 has decomposed.
1)Write the Kp expression for this reaction and calculate its value at 200K.
2)How many carbon dioxide molecules are there in the container at equilibrium?
3)When the reaction reaches equilibrium what % of the solid in the container is CaO by mass?
4)If the volume of the container is decreased after equilibrium is reached, but kept at 200^oC, will the mass of CaO increase or decrease from its value at equilibrium? Explain.
5.9 g CaCO3 = 5.9/100 moles and
0.059 moles x 0.515 = 0.00304
...........CaCO3(s) ==> CaO(s) + CO2(g)
begin.......0.059.......0.........0
change.....-0.00304....+0.00304..+0.00304
final.....0.05586.......0.00304...0.00304
(a) Kp = pCO2 Use PV = nRT to calculate PCO2.
(b) Remember there are 6.022E23 molecules in 1 mole.
(c) %CaO = (mass CaO/mass CaCO3)*100 = ??
(d) Use Le Chatelier's Principle.
1) To write the Kp expression for the given reaction, we need to determine the balanced equation first:
CaCO3(s) --> CaO(s) + CO2(g)
The balanced equation shows that 1 mole of CaCO3 produces 1 mole of CaO and 1 mole of CO2. Therefore, the Kp expression for this reaction can be written as:
Kp = (PCO2) / (PCaO * PCaCO3)
Since the reactants and products are all gases, we can express their partial pressures in the Kp expression.
To calculate the value of Kp at 200°C, we need additional information such as the equilibrium constant (Kp) value at that temperature.
2) To determine the number of carbon dioxide molecules at equilibrium, we first need to find out how much CaCO3 decomposed. It is given that 5.15% of the initial 5.9 grams of CaCO3 has decomposed.
Amount of CaCO3 decomposed = 5.15% of 5.9 grams = (0.0515 * 5.9) grams
Now, we can calculate the moles of CO2 produced using the molar mass of CaCO3 and the balanced equation:
Moles of CO2 = (0.0515 * 5.9) grams / (100 g/mol)
Lastly, we can convert moles of CO2 to the number of molecules by multiplying with Avogadro's number (6.02 x 10^23):
Number of CO2 molecules = Moles of CO2 * Avogadro's number
3) To determine the percentage of the solid in the container that is CaO by mass at equilibrium, we need to calculate the amount of CaO formed.
The amount of CaCO3 that decomposed is already determined in question 2, but we need to find the amount of CaO produced. Since the balanced equation shows a 1:1 stoichiometry between CaCO3 and CaO, the amount of CaO formed will be the same as the amount of CaCO3 decomposed.
Percentage of CaO by mass = Amount of CaO formed / Initial mass of CaCO3 * 100%
4) If the volume of the container is decreased after equilibrium is reached, but the temperature is kept at 200°C, the system will have a change in pressure. According to Le Chatelier's principle, when the volume is decreased, the pressure increases.
Since the reaction involves more moles of gas on the reactant side than on the product side (1 mole of CO2 vs. no gas on the product side), increasing the pressure will favor the formation of more CaCO3. Therefore, the mass of CaO will decrease from its value at equilibrium due to the shift in equilibrium towards the reactant CaCO3.
1) The Kp expression for this reaction can be written as:
Kp = (P CO2) / (P CaO)
To calculate its value at 200K, we need to use the ideal gas law equation, PV = nRT. Since the container is evacuated, the initial pressure of CO2 is 0.
Let's assume x moles of CaCO3 decompose, then (0.0515-x) moles of CaCO3 and x moles of CaO are present at equilibrium.
The molar mass of CaCO3 is 100 g/mol, so the initial moles of CaCO3 can be calculated as:
moles of CaCO3 = (mass of CaCO3) / (molar mass of CaCO3) = 5.9 g / 100 g/mol = 0.059 mol
Since 5.15% of CaCO3 decomposes, we have:
0.0515 mol = (0.0515-x) mol + x mol
Simplifying the equation:
0.0515 mol = 0.0515 mol - x mol + x mol
0.0515 mol = 0.0515 mol
So, x = 0.0515 mol
Substituting the values into the Kp expression:
Kp = (P CO2) / (P CaO)
= (n CO2)(RT/V) / (n CaO)(RT/V)
= (0.0515 mol)(R)(200K) / (x mol)(R)(200K)
= (0.0515 mol) / (x mol)
Plugging in the value of x:
Kp = (0.0515 mol) / (0.0515 mol)
= 1
Therefore, the value of Kp at 200K is equal to 1.
2) From the equation, we know that 1 mole of CaCO3 decomposes to form 1 mole of CO2. Since x = 0.0515 mol of CaCO3 decomposes, we have 0.0515 mol of CO2 at equilibrium.
To calculate the number of carbon dioxide molecules, we can use Avogadro's number, which is approximately 6.022 x 10^23 mol⁻^1.
Number of CO2 molecules = (amount of CO2 in moles) * (Avogadro's number)
= (0.0515 mol) * (6.022 x 10^23 mol⁻^1)
= 3.1033 x 10^22 CO2 molecules
Therefore, there are approximately 3.1033 x 10^22 carbon dioxide molecules in the container at equilibrium.
3) To calculate the % of the solid in the container that is CaO by mass at equilibrium, we need to determine the moles of CaO formed and compare it to the total moles of solid present.
We know that x mol of CaO is formed at equilibrium. The molar mass of CaO is 56 g/mol.
Mass of CaO formed = (moles of CaO formed) * (molar mass of CaO)
= (x mol) * (56 g/mol)
The total moles of solid present at equilibrium is the sum of the remaining CaCO3 and the formed CaO:
Total moles of solid = (moles of CaCO3 remaining) + (moles of CaO formed)
= (0.0515 mol - x mol) + x mol
= 0.0515 mol
Therefore, the % of the solid in the container that is CaO by mass at equilibrium can be calculated as:
% CaO = (mass of CaO formed) / (mass of total solid) * 100%
= [(x mol) * (56 g/mol)] / [(0.0515 mol) * (100 g/mol)] * 100%
Substituting the value of x:
% CaO = [(0.0515 mol) * (56 g/mol)] / [(0.0515 mol) * (100 g/mol)] * 100%
= 56 / 100 * 100%
= 56%
Therefore, when the reaction reaches equilibrium, 56% of the solid in the container is CaO by mass.
4) If the volume of the container is decreased after equilibrium is reached, but kept at 200°C, the mass of CaO will decrease from its value at equilibrium.
According to Le Chatelier's principle, when pressure is increased (by decreasing volume), the equilibrium will shift in the direction that reduces pressure. In this case, the equilibrium will shift to the left, causing more CaO to decompose into CaCO3 and CO2. This will result in a decrease in the amount of CaO present in the container, leading to a decrease in its mass.