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Old 02-24-2007, 07:30 PM   #28 (permalink)
pig
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jess,

because i'm a nerd, let's go ahead and hit that stoich junk. Tell me if this helps:

write the equation as :

aH2 + bI2 => cHI, where a, b, and c are constants.

you're given that dI2/dt = -.0037 mol/L/s

you want d(HI)/dt.

so, you write d(HI)/dt = - d(I2)/dt * c/b

The idea is that for every b moles of I2, there will always be c moles of HI.

a=1
b=1
c=2.

Many times in reaction chemistry, we take the (-) sign out of the conversion equation I put up, and write

a=-1
b=-1
c=2

to reflect that H2 and I2 are being consumed. It also makes the math more standardized. But as long as you keep track that some stuff is being generated, and some other stuff is being consumed, then it all comes down to the same thing. Production of HI shouldn't be a negative number.
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Last edited by pig; 02-24-2007 at 08:19 PM.. Reason: Automerged Doublepost
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