Flipping the reaction

[Russell Chuang 1J]

What do we do with the equilibrium constant if we flip the reaction?

[Sophia Zhao]

When you flip the reaction you do 1/Kc.

[Kayla Arellano 1K]

What do we do with the equilibrium constant if we flip the reaction?

When using the reverse reaction, we find find the reciprocal of the equilibrium constant.

[Shannon Lau 14B - 1H]

Like my peers said, when we flip the reaction we find the reciprocal of the equilibrium constant = 1/Kc.

This is very similar to flipping the reaction when given enthalpy but in that case we just change the sign a good example of this occurring during enthalpy would be textbook problem 4D.19.

[Myra Goraya Dis 2E]

If you flip the reaction, you would use a new kc value which is equal to 1/kc

[Sarah Wang 1I]

You use the reciprocal of K, which means you set the equation to 1/K.

[Caitlin Gee 2K]

Use the reciprocal of the equilibrium constant

[Ruben Adamov 1E]

Flipping the reaction means your new equilibrium constant is the reciprocal of the old one. 1/Kc

[Alexander Moroz 1B]

If Ka is [H3O+]/[OH-] then the K for the reverse reaction would be the former reactants/products, so [OH-]/[H3O+]. Therefore the flipped K is just the reciprocal of the original K.

[Natalie Coughlin 1I]

When you flip a reaction, you will want to flip the equilibrium constant as well, by taking it's reciprocal. So you would do 1/k.

[Julia Zahra]

If you flip or reverse the reaction the equilibrium constant is divided by 1. so 1/K

[Lindsey Walter 3E]

When we flip the reaction, we find the reciprocal of the equilibrium constant! Therefore, if the original equilibrium constant was Kc, then the equilibrium constant for the flipped reaction would be 1/Kc.

[loganchun]

The equilibrium constant will be its reciprocal (1/Kc) when the reaction is flipped. You essentially divide the equilibrium constant by 1 to get the new value.