K

[Luyan Zhang - 2D]

For Kc and Kp, would the constant be the same no matter whether you plug in the concentration or the partial pressure?

[nehashetty_2G]

Yes, it should be! Because, the only reason we use Kp versus Kc is based on the phases of the reactants/products (and whether the concentration or partial pressure is give to us for gases).

[Vincent Leong 2B]

The only thing that affects the equilibrium constant or the equilibrium conditions of the reaction is temperature. Understanding the E in the ICE table also helps when explaining why this is or how initial concentration/partial pressures will impact the concentrations/partial pressures of products and reactants at equilibrium.

[Natalie Benitez 1E]

I'm not sure if the constant Kc will be the same as Kp but I do know that they are used on whether you were given concentrations or if you were given partial pressures.

[Sydney Myers 4I]

Since the concentration and the pressure can be related by a factor of R*T (using the ideal gas law), the K should be the same whether you use partial pressures of temperature.

[Abigail_Hagen2G]

It will be the same, because even if you use partial pressure instead of concentration, you would still be able to convert between the two.

[Manav Govil 1B]

K is the equilibrium constant, which can be measured by either comparing the concentrations of the reactants and the products in the reaction, or the partial pressures of the reactants and products in the said reaction. Since K is a constant, whenever it is used (as Kc or Kp), it should equate. With this knowledge, you can potentially find missing partial pressures or concentrations if proved information of the latter.

[Astrid Lunde 1I]

Yes it will still be the same, but Kc is in terms of Moles and Kp is in terms of Bars.

[205150314]

If Kc and Kp are suppose to be the same, then why does the chart on 5G.2 have different values for some K and Kc?

[Ruth Glauber 1C]

Yes I think so!

[gabbymaraziti]

Kc and Kp are the same, they are just used for reactions in different states.