When do you plug in partial pressure into equilibrium expression for gases?

[Meredith92]

I'm a little confused when you can just plug in the pressure or moles into the equilibrium or when you have to plug in partial pressure. Is this only when the moles of gas on either side of the reaction are the same?

For instance,
in TBR chem's chapter on equilibrium one example describes a research mixing 1 atm of H2(g) with excess iodide solid to give 2HI(g)
In the equilibrium expression they plug in 1 atm, which i guess in this case is both the partial pressure (since HI is zero) and the actual pressure....
but lets say that they mix 1 atm of H2 with .5 atm of 2HI. are these the values you plug in for the initial pressures in the ICE table? Or do you need to somehow convert them into partial pressure values, given that there are unequal moles of gas on each side? Sorry if my question doesnt make sense.. i guess im confused when the values they give us are the partial pressure, or if you need to evaluate the moles on either side of the equation in order to get the partial pressures....
thanks for your help!



EDIT: maybe the only time you have to calculate things is if they just give you the total pressure and you have to multiply the total pressure by the mole fraction to get the partial pressure and then plug that into the equil expression? if so, would you just figure out the mole fraction by the coefficients of the gases??

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[shoehornlettuce]

You are always plugging in the partial pressure of each molecule when solving these equilibrium problems for gases. When a problem tells you the initial values for all the molecules, these are partial pressures they are giving you. Sometimes they might try to complicate a problem by not giving you all the values up front. For example, say you are given the following reaction (all in the gaseous phase):

X + Y <---> XY

A problem might say "at equilibrium there are .5atm of X and .25atm of Y. What is Kp at STP?" You'd have to recognize that at STP the pressure is 1atm so the partial pressure of XY must be .25atm. You could then use X=.5, Y=.25 and XY=.25 to solve for Kp.

Also, remember that solids and liquids are not included in the equilibrium expression because the concentration does not change.

 

[Meredith92]

Thanks for your help. I feel like I've seen examples where they do something at ATP but they describe putting 1 atm of one gas and 2 of another... how would this make sense given partial pressures are fractions of the total pressure?

 

[shoehornlettuce]

Thanks for your help. I feel like I've seen examples where they do something at ATP but they describe putting 1 atm of one gas and 2 of another... how would this make sense given partial pressures are fractions of the total pressure?

At STP the pressure is always 1 atm or 760 torr. There could have been a problem where starting conditions were 1 atm and then something was added or removed from the system to increase or decrease the pressure. You might want to consider reading the TBR section on gases and then jump back to equilibrium. I think that might help clear some things up for you.