Consider the reaction between iodine gas and chlorine gas to form iodine monochloride:

Question

anonymous · Answer

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aaronq · Answer

You need an ICE table. It's the same process as if you were using a solution, but now it's gases.

anonymous · Answer

@@aaronq hello! That is what i did but somehow my answer was not correct :/

aaronq · Answer

hey! hm really, did you set it up like this?


 $\large K_p=\dfrac{(2x)^2}{(P_{I_2}-x)*(P_{Cl_2}-x)}$

anonymous · Answer

yes i did! @aaronq

aaronq · Answer

so, you got {x=0.2047505656136135, x=0.3209233753363223} as your possible x values.

Now, we need to determine which one to use. Notice that the second one exceeds the partial pressure of either of the initial gases (0.25 atm), so we discard that one and use 0.20475.

$\sf \large  P_{ICl}=(2x)^2=(2*0.20475)^2=0.16769025\approx 0.17~atm$

anonymous · Answer

@aaronq Hello! I checked and my book says the answer is 0.41 atm :|

aaronq · Answer

hm.. it's like they just added the x's.

x = 0.20475 atm 

2x=0.4095 atm $\approx$ 0.41 atm


hmm i haven't done a question like this in years, so it's possible that, that's how you do it. Do another one that's similar and see what happens.

anonymous · Answer

@aaronq I figured it out :) thank you!

aaronq · Answer

awesome :)