Kp
Lajoy Tucker
Teacher
Contents
Gaseous Equilibria And Kp
Many reactions occur where both the products and reactants are in the gaseous phase. It is easier practically to determine the pressure of the gaseous mixture rather than the concentration of the gases involved ∴ a different equilibrium constant, is used.
The composition of a gaseous mixture, in an equilibrium can be determined using the equilibrium constant, .
is the partial pressure of that gas. Square brackets, [ ], must not be used in the equation.For the reaction:
For the reaction:
Rule for Exponents: The partial pressure of each substance is raised to a power which is the same as the numbers in the stoichiometric ratio.
Partial pressure:
is the contribution a particular gas makes towards the total pressure (PT) in an equilibrium mixture. It is the same pressure that the pure gas would exert in the same volume at the same temperature.
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The value of the equilibrium constant Kp is only dependent on the temperature at which the equilibrium is established. It is independent of the total pressure of the substance in the equilibrium.
No answer provided.
Kp Calculation Explained
Dalton’s Law Of Partial Pressures
Dalton’s Law states - that in a gaseous mixture that does not react chemically, the total pressure () is equal to the sum of all the partial pressures of the constituent gases.
Partial pressure of gas A in a mixture of gas A and B
Dalton’s Law applies to gaseous equilibria because although the gases are constantly reacting, the overall amount of each gas at equilibrium does not change ∴ the partial pressures do not change.
Partial pressure - is the contribution a particular gas makes towards the total pressure (PT) in an equilibrium mixture. It is the same pressure that the pure gas would exert in the same volume at the same temperature.
Partial pressure of a gas in a mixture is equal to the product of the mole fraction () of that gas and the total pressure ()
Total number of moles:
No answer provided.
Using Dalton’s Law Of Partial Pressures
For the reaction
In the Haber Process, a mixture 20% , 55% and 25% by volume has a total pressure in the equilibrium mixture of 9.80 x 104 N m-2. What is the partial pressure of each gas? Calculate Kp. (N.B. this is a very low pressure for this reaction - below atm).
The value for is extremely small, illustrating that the position of equilibrium lies heavily over to the left. Very little ammonia is formed. Therefore the operating conditions will need to be changed.
Calculating Kp
When 1.0 mol of hydrogen gas and 1.0 mol of iodine gas are allowed to reach equilibrium in a 1.0 flask at 723 at a pressure () of 1.01 x 105 (atmospheric pressure i.e. 1 atm), the amount of hydrogen iodide at equilibrium is 1.56 mol. Calculate .
2HI | |||
Initial amount | 1.00 | 1.00 | 0.00 |
Change | -a | -a | +2a |
Eq. amount | 0.22 | 0.22 | 1.56 |
Eq. amount | 0.22/2 = 0.11 | 0.22/2=0.11 | 1.56/2=0.78 |
Mole fraction | 0.11 x PT | 0.11 x PT | 0.78 x PT |
Partial pressure | 11110 | 11110 | 78780 |
If a = the amount of reactant changed into product then 2a is the amount of Hl formed from the equation.
The value for is relatively large, illustrating that the position of equilibrium lies heavily over to the right. This is not a surprise due to the high operating temperature for this equilibrium.
In the Haber Process nitrogen and hydrogen are mixed in a ration of 1:3. At equilibrium, at 873 and 10 atm, the percentage of ammonia in the mixture of gases is 15 %. Calculate .
Initial amount | 1.00 | 3.00 | 0.00 |
Change | -a | -3a | +2a |
Eq. amount | 1-a | 3-3a | 2a |
Eq. amount | 0.74 | 2.22 | 0.52 |
Mole fraction | 0.74/3.48 = 0.21 | 2.22/3.48=0.64 | 0.52/3.48=0.15 |
Partial pressure | 0.21 x PT | 0.64 x PT | 0.15 x PT |
Partial pressure | 2.1 | 6.4 | 1.5 |
Let be the fraction of that has reacted at equilibrium:
No answer provided.
Gaseous Dissociation Equilibria
Thermal Dissociation - is the reversible breakdown of a substance when heated in a closed system, on cooling the compound returns back to its original state.
During the dissociation of a gas, in a closed container, the pressure of the all the gases in the equilibrium is greater than expected. This is due to a greater number of moles are formed.
Degree of Dissociation (α) - is the fraction of the gas originally present that has dissociated. It is a number between 0 and 1.
If the dinitrogen tetroxide is 20% dissociated
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Initial amount | 1.00 | 0.00 |
Change | ||
Eq. amount | 0.8 | 0.4 |
Mole fraction | 0.8/1.2 = 0.667 | 0.4/1.2=0.333 |
Partial pressure | 0.667 x PT | 0.333 x PT |
Partial pressure | 67333 | 33667 |