Calculating Partial Pressure

[DArcy Perlman 1C]

When calculating partial pressure from given equilibrium constants, how come the stoichiometric coefficients of the reaction don't play a role?

[Tawny Vu 1A]

They should play a role. The stoichiometric coefficients become the power of the exponent in the equilibrium constant expression, which can then be rearranged to find the partial pressures desired.

[Tara Cumiskey 3K]

In the equilibrium constant equation for both Kc and Kp, the stoichiometric coefficients become the exponent.

[Warren Jolicoeur 1B]

The reason why stoichiometric coefficients play a role in equilibrium as exponents is because mathmatically the chemical equation multiplies everything together, including the coefficients. For example
OG equation 2A + B = C

it is not C/ ((A+A) * B)

it is instead C/((A*A) *B)

Hope that makes sense from a math perspective. We multiply things that interact together, that includes the same molecule.

[Kayla Tran 3C]

The stoichiometric coefficients are used as the exponents.

[Nandini_Parmar_1I]

They do. If there is a coefficient in front of it, it is usually squared to that coefficient.

So if your coefficient is 2 -- then Kc^2.

[Jessica Servoss 1H]

When you are given a reaction, a Kp and partial pressures for the reactants, the stoichiometric coeffs. are very important when forming your equation. Kp = partial pressures of products / partial pressures of reactants, but if there are stoichiometric coeffs. in your original chemical equation, you need to use those and raise the partial pressures to the number they are.

[Guadalupe_3B]

Stoichiometric coefficients of the reaction do play a role in calculating partial pressure because they become the exponents required to balance a chemical equation.