Calculating Partial Pressure
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Calculating Partial Pressure
When calculating partial pressure from given equilibrium constants, how come the stoichiometric coefficients of the reaction don't play a role?
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Re: Calculating Partial Pressure
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.
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Re: Calculating Partial Pressure
In the equilibrium constant equation for both Kc and Kp, the stoichiometric coefficients become the exponent.
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Re: Calculating Partial Pressure
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.
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.
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Re: Calculating Partial Pressure
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.
So if your coefficient is 2 -- then Kc^2.
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Re: Calculating Partial Pressure
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.
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Re: Calculating Partial Pressure
Stoichiometric coefficients of the reaction do play a role in calculating partial pressure because they become the exponents required to balance a chemical equation.
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