1-Dinitrogen tetroxide (N2O4) dissociates according to the equation:
N2O4 ⇌ 2NO2
An equilibrium reaction mixture at 25 ºC was found to have the partial
pressure of N2O4 as 72 kPa whilst the partial pressure of nitrogen
dioxide (NO2) was 29 kPa. Calculate Kp for the reaction.
2-Carbon monoxide and steam react to form hydrogen at 500 ºC with the
stoichiometry:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
When 10 mol of carbon monoxide and 10 mol of steam were allowed
to react to equilibrium the total pressure was found to be 1.50
atmospheres and 7.4 mol of carbon dioxide was formed. Calculate Kp
for the reaction.
3-Sulphur trioxide (SO3), a precursor to the manufacture of sulphuric
acid, is formed by oxidising sulphur dioxide with oxygen in the
presence of a catalyst.
2SO2 + O2 ⇌ 2SO3
If 610 g of sulphur dioxide (SO2) and 280 g of oxygen were used as the
initial reactants and 750 g of sulphur trioxide were formed calculate
the value of Kp for the reaction. (RAM data: O = 16.00, S = 32.00)
4-The table below gives values of the rate constant of a particular
reaction as a function of temperature. Calculate the activation energy
for the reaction.
Temperature (K)
293
313
333
353
k (s^-1)
0.0030
0.0216
0.122
0.567
............N2O4 ==> 2NO2
E............72......29
Kp = pN2O4/(pNO2)^2
Substitute and solve for Kp.
........CO + H2O ==> CO2 + H2
I.......10....10......0.....0
C........-x...-x......x.....x
E.....10-x...10-x.....x.....x
The problem tells you that CO2 is 7.4 mols (so x = 7.4); therefore, at equilibrium
CO is 10-7.4 = 2.6 = H2O
CO2 = H2 = 7.4
Then total mols = you add them.
XCO = (mol CO/total mols)
XH2O = (mol H2O/total mols)
XCO2 = (mols CO2/total mols)
XH2 = (mols H2/total mols)
Then pCO = XCO*Ptotal
pH2O = XH2O*Ptotal
pCO2 = XCO2*Ptotal
pH2 = XH2*Ptotal
Then plug partial pressures into Kp expression and solve for Kp.
mols SO2 = 610/64 = estimated 9.3
mols O2 = 280/32 = estd 9
mols SO3 formed = 750/80 = about 9
Note: You must clean up all of the numbers. I've just estimated them here.
...........2SO2 + O2 ==> 2SO3
I..........9.3.....9.......0
C..........-2x....-x.......2x
E.........9.3-2x..9-x......+2x
The problem tells you that 2x = about 9 so x must about 4.5
This allows you to calculate the mols of each reactant and product at equilibrium, then revert to problem 2 and work it the same way to solve for Kp.
Problem 4. Use the Arrhenius equation. I don't know if you are to graph the equation and use the slop to determine activation energy but probably you can use the data to solve for activation energy multiple times and average the values you obtain.
1) To calculate Kp for the reaction N2O4 ⇌ 2NO2, we need to use the expression for Kp, which is the ratio of the partial pressures of the products to the partial pressure of the reactant at equilibrium. Kp = (p(NO2)^2) / p(N2O4).
Given that the partial pressure of N2O4 is 72 kPa and the partial pressure of NO2 is 29 kPa, we can substitute these values into the equation to find Kp.
Kp = (29 kPa)^2 / 72 kPa = 29 * 29 / 72 = 841 / 72 ≈ 11.68
Therefore, Kp for the reaction N2O4 ⇌ 2NO2 is approximately 11.68.
2) To calculate Kp for the reaction CO(g) + H2O(g) ⇌ CO2(g) + H2(g), we need to use the expression for Kp, which is the ratio of the partial pressures of the products to the partial pressure of the reactants at equilibrium. Kp = (p(CO2) * p(H2)) / (p(CO) * p(H2O)).
Given that 7.4 mol of carbon dioxide (CO2) is formed and the total pressure is 1.50 atmospheres, we can find the partial pressures of CO2, CO, H2O, and H2.
The partial pressure of CO2 is (7.4 mol / 10 mol) * 1.50 atm = 1.11 atm.
Since the stoichiometry of the reaction is 1:1, the partial pressure of CO is also 1.11 atm.
The partial pressure of H2 is (1 - 1.11) atm = -0.11 atm. However, partial pressure cannot be negative, so we assume it is zero.
The partial pressure of H2O is then 1.50 atm - 1.11 atm = 0.39 atm.
Now we can substitute these values into the equation to find Kp.
Kp = (1.11 atm * 0 atm) / (1.11 atm * 0.39 atm) = 0 / (1.11 * 0.39) = 0
Therefore, Kp for the reaction CO(g) + H2O(g) ⇌ CO2(g) + H2(g) is 0.
3) To calculate the value of Kp for the reaction 2SO2 + O2 ⇌ 2SO3, we need to use the expression for Kp, which is the ratio of the partial pressures of the products to the partial pressure of the reactants at equilibrium. Kp = (p(SO3)^2) / (p(SO2)^2 * p(O2)).
Given that 610 g of SO2 and 280 g of O2 were used as the initial reactants and 750 g of SO3 were formed, we can convert the masses into moles using the molar masses of each substance.
Molar mass of SO2 = 32.00 g/mol
Molar mass of O2 = 32.00 g/mol
Molar mass of SO3 = 80.06 g/mol
Number of moles of SO2 = 610 g / 32.00 g/mol = 19.06 mol
Number of moles of O2 = 280 g / 32.00 g/mol = 8.75 mol
Number of moles of SO3 = 750 g / 80.06 g/mol = 9.37 mol
Now we can substitute the number of moles into the equation to find Kp.
Kp = (9.37 mol^2) / ((19.06 mol)^2 * 8.75 mol) = 9.37^2 / (19.06^2 * 8.75) ≈ 0.075
Therefore, Kp for the reaction 2SO2 + O2 ⇌ 2SO3 is approximately 0.075.
4) To calculate the activation energy for the reaction, we can use the Arrhenius equation, which relates the rate constant (k) to the temperature (T) and the activation energy (Ea). The equation is given by ln(k) = ln(A) - (Ea / (R * T)), where A is the pre-exponential factor and R is the gas constant (8.314 J/mol·K).
First, we need to calculate the values of ln(k) for each temperature.
ln(k1) = ln(0.0030) = -5.808
ln(k2) = ln(0.0216) = -3.828
ln(k3) = ln(0.122) = -2.109
ln(k4) = ln(0.567) = -0.569
Next, we use these values to set up a system of equations in order to find the activation energy (Ea).
ln(A) - (Ea / (R * T1)) = ln(k1)
ln(A) - (Ea / (R * T2)) = ln(k2)
ln(A) - (Ea / (R * T3)) = ln(k3)
ln(A) - (Ea / (R * T4)) = ln(k4)
Using the given values for the temperatures and rate constants, we can solve this system of equations to find the activation energy (Ea).
Once we have the activation energy, we can use it to determine the rate constant k at any other temperature using the Arrhenius equation.