For the reaction below, Kp = 1.16 at 800.°C.
CaCO3(s) CaO(s) + CO2(g)
If a 25.0-g sample of CaCO3 is put into a 10.2-L container and heated to 800.°C, what percent of the CaCO3 will react to reach
equilibrium?
CaCO3(s) ==> CaO(s) + CO2
I...25.0.......0.........0
C.......................x
E........................x
Kp = pCO2 = 1.16
Therefore, p CO2 = 1.16 atm. Plug that into PV = nRT and solve for n.
Convert 25.0 g CaCO3 to mols, then calculate the percent decomposed.
Well, let's calculate it in a fun way!
First, we need to convert the mass of CaCO3 to moles. Since its molar mass is 100.09 g/mol, we have:
25.0 g / 100.09 g/mol = 0.2499 mol
Now, let's assume x moles of CaCO3 react. As a result, x moles of CaO and x moles of CO2 will be formed. So, the total moles of gas produced is 2x.
Since we have a 10.2 L container, we can use the ideal gas law to determine the number of moles of CO2 at equilibrium:
PV = nRT
Where P is the pressure (we'll take it as 1 atm), V is the volume (10.2 L), n is the number of moles of CO2 (2x), R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin (800°C is 1073 K).
1 atm * 10.2 L = (2x) * 0.0821 L·atm/(mol·K) * 1073 K
10.2 L = 172.9542x
Now, let's solve for x:
x = 10.2 L / 172.9542
x ≈ 0.059 mol
The percent of CaCO3 that reacted to reach equilibrium is:
(0.059 mol / 0.2499 mol) * 100%
Tell me, did you hear about the mathematician who's afraid of negative numbers? He will stop at nothing to avoid them! So, we have:
(0.059 mol / 0.2499 mol) * 100% ≈ 23.61%
Approximately 23.61% of the CaCO3 will react to reach equilibrium.
To determine the percentage of CaCO3 that will react to reach equilibrium, we need to compare the initial pressure of CO2 to the equilibrium pressure of CO2.
First, we need to calculate the number of moles of CaCO3 in the 25.0 g sample. The molar mass of CaCO3 is 100.09 g/mol.
Number of moles of CaCO3 = mass / molar mass = 25.0 g / 100.09 g/mol = 0.2498 mol ≈ 0.250 mol (to 3 significant figures)
Since the stoichiometry of the reaction is 1:1 between CaCO3 and CO2, the number of moles of CO2 produced at equilibrium will also be 0.250 mol.
To calculate the equilibrium partial pressure of CO2, we can use the ideal gas law:
P = nRT/V
Where:
P is the pressure of CO2
n is the number of moles of CO2
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
V is the volume of the container
The temperature needs to be converted to Kelvin by adding 273.15:
T = 800°C + 273.15 = 1073.15 K
Now we can plug in the values and solve for the equilibrium partial pressure of CO2:
P = (0.250 mol) * (0.0821 L·atm/(mol·K)) * (1073.15 K) / (10.2 L)
P ≈ 2.13 atm (to 3 significant figures)
Since Kp is the ratio of the equilibrium partial pressure of products to the equilibrium partial pressure of reactants, we can set up the following equation:
Kp = P(CO2) / P(CaCO3)
Given that Kp = 1.16 and P(CO2) = 2.13 atm, we can solve for P(CaCO3):
1.16 = 2.13 / P(CaCO3)
Rearranging the equation to solve for P(CaCO3):
P(CaCO3) = 2.13 / 1.16
P(CaCO3) ≈ 1.84 atm (to 3 significant figures)
Now, we can calculate the percentage of CaCO3 that will react:
Percentage of CaCO3 that will react = (Initial pressure - Equilibrium pressure) / Initial pressure * 100%
Percentage of CaCO3 that will react = (1.84 atm - 2.13 atm) / 1.84 atm * 100%
Percentage of CaCO3 that will react ≈ -0.1587 * 100%
Percentage of CaCO3 that will react ≈ -15.87%
Since the percentage is negative, it implies that no CaCO3 will react to reach equilibrium at this temperature and pressure.
To calculate the percent of CaCO3 that will react to reach equilibrium, we need to determine the amount of CaCO3 that will react at equilibrium and compare it to the initial amount of CaCO3.
First, we need to convert the mass of CaCO3 to moles since the equilibrium constant (Kp) is expressed in terms of partial pressures, which is directly related to the number of moles.
Step 1: Calculate the number of moles of CaCO3
To convert mass to moles, we need to use the molar mass of CaCO3. The molar mass of CaCO3 is:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 oxygen atoms)
Total molar mass = 40.08 + 12.01 + (16.00 * 3) = 100.09 g/mol
Number of moles of CaCO3 = mass / molar mass
Number of moles of CaCO3 = 25.0 g / 100.09 g/mol
Step 2: Calculate the pressure of CO2 at equilibrium using the ideal gas law.
PV = nRT
In this case, we are given the volume (V) of the container (10.2 L), the temperature (T) in Kelvin (800°C = 1073 K), and the equilibrium constant (Kp = 1.16).
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K).
Rearranging the equation, we can solve for the number of moles of CO2 (n):
n = (Kp * PV) / RT
Note: The pressure (P) is in atm units, so we need to convert the volume from liters to atm.
1 atm = 101325 Pa = 1013.25 hPa = 1013.25 J/cm^3 = 1.1013 J/cm^3 = 1.1013 × 10^6 dyn/cm^2 = 760 mmHg = 760 torr.
Step 3: Calculate the percent of CaCO3 that will react
To calculate the percent of CaCO3 that will react at equilibrium, we need to compare the number of moles of CaCO3 initially to the number of moles of CaCO3 that will react at equilibrium.
Percent of CaCO3 reacted = (moles reacted / initial moles) * 100
By following these steps, you should be able to determine the percent of CaCO3 that will react to reach equilibrium.