............N2O4 ==> 2NO2
E............72......29
Kp = pN2O4/(pNO2)^2
Substitute and solve for Kp.
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
3 answers
........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.
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.
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 answer