That is all that is left in the equilibrium constant expression. Jim Clark Chemguide. Mole fraction If you have a mixture of gases A, B, C, etc , then the mole fraction of gas A is worked out by dividing the number of moles of A by the total number of moles of gas.
Partial pressure The partial pressure of one of the gases in a mixture is the pressure which it would exert if it alone occupied the whole container.
It is easy to see this visually: Gas A is creating a pressure its partial pressure when its molecules hit the walls of its container. The more important relationship is the second one: Learn it! A good example of a gaseous homogeneous equilibrium is the conversion of sulfur dioxide to sulfur trioxide at the heart of the Contact Process: Writing an expression for K p We are going to start by looking at a general case with the equation: If you allow this reaction to reach equilibrium and then measure or work out the equilibrium partial pressures of everything, you can combine these into the equilibrium constant, K p.
The Contact Process equilibrium You will remember that the equation for this is: K p is given by: The Haber Process equilibrium The equation for this is:. The equilibrium produced on heating carbon with steam Everything is exactly the same as before in the expression for K p , except that you leave out the solid carbon. The equilibrium produced on heating calcium carbonate This equilibrium is only established if the calcium carbonate is heated in a closed system, preventing the carbon dioxide from escaping.
Contributors and Attributions Jim Clark Chemguide. For heterogeneous equilibrium the partial pressure of pure solids and liquids are not included in the equilibrium constant expression.
As for example the heating of calcium carbonate in a closed vessel to form calcium oxide and carbon dioxide. Here only carbon dioxide is in gas phase. The equilibrium constant for this reaction by excluding any terms for solid is.
Assuming that all the gases in equilibrium reaction obey the ideal gas equation, the partial pressure p of a gas is. Thus the partial pressures of individual gases A, B, C and D are:. It covers an explanation of the terms mole fraction and partial pressure , and looks at K p for both homogeneous and heterogeneous reactions involving gases. The page assumes that you are already familiar with the concept of an equilibrium constant, and that you know about K c - an equilibrium constant expressed in terms of concentrations.
Important: If you have come directly to this page via a search engine, you should first read the page on equilibrium constants - K c before you go on - unless you are already fully confident about how to write expressions for K c.
You will find a link back to this page at the bottom of the K c page. Before we can go any further, there are two terms relating to mixtures of gases that you need to be familiar with. If you have a mixture of gases A, B, C, etc , then the mole fraction of gas A is worked out by dividing the number of moles of A by the total number of moles of gas. The mole fraction of gas A is often given the symbol x A.
The mole fraction of gas B would be x B - and so on. For example, in a mixture of 1 mole of nitrogen and 3 moles of hydrogen, there are a total of 4 moles of gas.
The partial pressure of one of the gases in a mixture is the pressure which it would exert if it alone occupied the whole container. The partial pressure of gas A is often given the symbol P A. The partial pressure of gas B would be P B - and so on. There are two important relationships involving partial pressures.
The first is again fairly obvious. Viewed 16k times. Improve this question. Gamira Gamira 1 1 gold badge 2 2 silver badges 6 6 bronze badges. Equilibrium constant can only appear dimensionless when it so happens that the units are cancelled out.
Also, KC reads as potassium carbide. Add a comment. Active Oldest Votes. Improve this answer. Excuse me? I think mol is dimensionless so I'm only considering the dimensions of litres. Actually, thermodynamic equilibrium constant is defined in terms of activities fugacities and not just activity coefficients! So, you see that it is the activity fugacity which is dimensionless.
Ask Atkins if you don't believe me. Karsten Theis Karsten Theis
0コメント