At 750°C, the Kp for the reaction 2 CO + O2➞ 2 CO2 is 2.72 x 10^11. ( 2.72 times ten to the eleventh) At that temperature, CO and O2 are added to a constant volume vessel. The initial pressure of the CO is 2.00 atm, and that of the O2 is 1.50 atm. Calculate the pressures of all three components in the equilibrium mixture.
Kp = p(CO2)^2/p(CO)^2*p(O2)
Make an ICE chart.
initial:
CO2 = 2
P2 = 1.5
CO2 = 0
change:
CO2 = 2x
O2 = -x
CO = -2x
equilibrium:
CO2 = +x
CO = 2-2x
O2 = 1.5-x
Substitute into the Kp exression and solve for x.
2x will be CO2
2-2x will be CO
1.5-x will be O2
do you have to use limiting reagent for this problem
To calculate the pressures of all three components (CO, O2, CO2) in the equilibrium mixture, we will make use of the ideal gas law and the expression for the equilibrium constant (Kp) at a given temperature. Here's how we can go about it:
Step 1: Write the balanced equation for the reaction:
2 CO + O2 ➞ 2 CO2
Step 2: Calculate the number of moles of CO and O2 initially present:
According to the initial pressures given, we can use the ideal gas law (PV = nRT) to calculate the number of moles. The ideal gas law equation can be rearranged as follows:
n = PV / RT
For CO:
n_CO = (P_CO * V) / (R * T)
= (2.00 atm * V) / (R * T) ---(1)
For O2:
n_O2 = (P_O2 * V) / (R * T)
= (1.50 atm * V) / (R * T) ---(2)
Note: V is the volume of the vessel, R is the ideal gas constant, and T is the temperature in Kelvin.
Step 3: Calculate the reaction quotient (Qp):
The reaction quotient (Qp) is calculated by substituting the partial pressures of the components into the expression of Kp. For the given reaction, Qp can be written as:
Qp = (P_CO2)^2 / (P_CO)^2 * (P_O2)
Step 4: Calculate the partial pressures of the equilibrium mixture:
Knowing the expression for Kp, we can relate it to the reaction quotient (Qp) as follows:
Kp = Qp
Substituting the partial pressures into the equation for Qp, we get:
(2.72 x 10^11) = (P_CO2)^2 / (P_CO)^2 * (P_O2)
We now have two unknowns, P_CO2 and P_CO. However, since the volume of the vessel is constant, the number of moles of CO and CO2 does not change. Therefore, we can write the following relationship between the partial pressures:
P_CO2 = 2 * P_CO ---(3)
Using equations (1), (2), and (3), we can solve for the unknowns.
Step 5: Substitute and solve for the unknowns:
Considering equation (1), we can rewrite it as:
n_CO = (2.00 atm * V) / (R * T)
Since the coefficients of the balanced equation are 2, we can say:
2 * n_CO = (2.00 atm * V) / (R * T)
Thus, n_CO = (1.00 atm * V) / (R * T)
Similarly, for equation (3), we can write:
n_CO2 = 2 * n_CO
= 2 * (1.00 atm * V) / (R * T)
= (2.00 atm * V) / (R * T)
Now, we can substitute n_CO and n_CO2 into the expression for Qp:
(2.72 x 10^11) = (P_CO2)^2 / (P_CO)^2 * (P_O2)
= [(2.00 atm * V) / (R * T)]^2 / [(1.00 atm * V) / (R * T)]^2 * (1.50 atm)
= (4.00 / 1.00) * (1.50 atm)
= 6.00 * 1.50 atm
= 9.00 atm
Simplifying the equation, we get:
(2.72 x 10^11) = 9.00 atm
Step 6: Solve for P_CO:
P_CO = sqrt((2.72 x 10^11) * 9.00 atm)
Step 7: Solve for P_O2:
P_O2 = (P_CO)^2 * (P_CO2) / (sqrt((2.72 x 10^11) * 9.00 atm))
Step 8: Calculate P_CO2:
P_CO2 = 2 * P_CO
Now that we have obtained the values for P_CO, P_O2, and P_CO2, we have calculated the pressures of all three components in the equilibrium mixture.