chemistry help [SOLVED]

[thubby]

"H2+I2<-->2HI

if 1 mole of H2 and 2 moles of I2 are introduced to a 3L flask, how many moles of HI will be present at equilibrium?"

this has been driving me nuts. i set up the ICE table, but i end up with negative HI (which is impossible). help:smallsigh:

edit: forgot to mention that Kc=49
(this is all from experiments, that number happened to be buried in another equation)

[Sucrose]

Solution Process:

First, set up the equation, as you have done. Note that all samples are gases at room temperature, meaning that unless your the equation for the equilibrium constant is
K(eq)=[HI(eq)]^2/([H2(eq)][I2(eq)])
Now, you've already set up the ICE diagram, so you should know that
HI(eq)=2x
H2(eq)=(1/3)-x
I2(eq)=(2/3)-x
So, now you need to find K(eq). Since you've not provided it, I shall assume that you are expected to find it from essentially base principles.

For this, you will need the equation K(eq)=e^(-Delta G/(RT)), and the equation DG=DH-TDS.

DH, the per-mole change in enthalpy over the reaction, is twice the molar change in enthalpy of formation of hydrogen iodide.

Similarly, DS, the per-mole change in entropy over the reaction, is twice the change in entropy of one mole hydrogen iodide being formed from one half mole of hydrogen gas and one half mole of iodine gas.

Both DH and DS should be calculable from tables in your chemistry textbook. Enthalpy of formation table, and Molar Entropy table.

After finding DH and DS, calculate DG. Once you have DG, calculate K(eq) using the formula K(eq)=e^(-Delta G/(RT)).

R is the universal gas constant, and T is the temperature, in Kelvin, that you are running the reaction at.

Once you have K(eq), set that numerical value against
(2x)^2/((1/3-x)*(2/3-x)).

Solve for x, and then substitute it back into the expressions for HI, I2, and H2 at equilibrium, and you have your equilibrium values. Since K(eq) literally cannot be negative, you should have no issues with a negative value for HI.

A trick that may simplify the last step for you, if K(eq) is quite small, is assuming that x will be essentially negligible when compared against 1/3 or 2/3. That allows you to simplify the expression to K(eq)=Value=(2x)^2/(2/9)= (4x^2)*9/2=18x^2. However, after you solve for x using this assumption, you then need to actually look carefully at the value you found for x, and verify that it is indeed much smaller than 1/3.

[rogueboy]

Solution Process:

First, set up the equation, as you have done. Note that all samples are gases at room temperature, meaning that unless your the equation for the equilibrium constant is
K(eq)=[HI(eq)]^2/([H2(eq)][I2(eq)])
Now, you've already set up the ICE diagram, so you should know that
HI(eq)=2x
H2(eq)=(1/3)-x
I2(eq)=(2/3)-x
So, now you need to find K(eq). Since you've not provided it, I shall assume that you are expected to find it from essentially base principles.

[HI(eq)] should actually be 1-x, and [I2(eq)] should be 2-x. We're looking at the number of moles present. 1 mole - x moles = 1-x moles. 1 mole / 3 moles - x moles = (1/3) - (x moles)... which doesn't actually make any sense. Other than that, advice looks good (at least to my half-asleep, organic chemist's brain).

Final equation ends up being
K(eq) = (2x)^2/[(1-x)*(2-x)] = 4 x^2 / (2-3x+x^2), if my ability to do basic algebra is currently intact.

[Sucrose]

[HI(eq)] should actually be 1-x, and [I2(eq)] should be 2-x. We're looking at the number of moles present. 1 mole - x moles = 1-x moles. 1 mole / 3 moles - x moles = (1/3) - (x moles)... which doesn't actually make any sense. Other than that, advice looks good (at least to my half-asleep, organic chemist's brain).

Final equation ends up being
K(eq) = (2x)^2/[(1-x)*(2-x)] = 4 x^2 / (2-3x+x^2), if my ability to do basic algebra is currently intact.

Actually, I was going with the fact that it's in a 3L flask. Been a while since I've worked with gases in these sorts of equations, so I was thinking that perhaps the number of moles per volume of space would be relevant... in fact, I kinda still do. Should work basically the same way as a liquid solution, right? I mean, if it were partial pressures of gases, then that would already be accounting for how spread out they are, so dividing by volume again would be redundant. But for moles? Can't say I see it...

When you wake, could you explain why the volume that the gas is spread out in is not relevant?:smallconfused:

...Speaking of, I did forget a step. The equilibrium values you would calculate with my method, if I did not err in assuming you should and did convert to concentration, would be concentrations. The number of moles of hydrogen iodide you would produce would be 6x, rather than 2x.

[thubby]

i don't know if ICE works with anything other than concentrations, so i did what you did and got the same equation.

i think i screwed up the algebra between there and the quadratic equation.
just to make sure i've got this right this time
[2x]^2/([1/3-x][2/3-x])=49 works out to 11.25X^2-12.25x+2.7+2/9, right?

if thats right then x=~.311, which would result in almost no reactants, but lots of product at equilibrium, which fits with Kc being 49

edit: yup. the math works out to about 49 (rounding and truncating account for the inaccuracy)

[rogueboy]

Actually, I was going with the fact that it's in a 3L flask. Been a while since I've worked with gases in these sorts of equations, so I was thinking that perhaps the number of moles per volume of space would be relevant... in fact, I kinda still do. Should work basically the same way as a liquid solution, right? I mean, if it were partial pressures of gases, then that would already be accounting for how spread out they are, so dividing by volume again would be redundant. But for moles? Can't say I see it...

When you wake, could you explain why the volume that the gas is spread out in is not relevant?:smallconfused:

We're both correct, as it turns out! While concentration is the norm (especially for liquids), since we're looking at a fixed volume, we can look at moles instead in this case. Essentially, we're multiplying everything by the volume (without being concerned about what that volume is), so it all works out in the end. Plugging 0.933 (the molar change) into my equation as x, I get K(eq) = 48.7; close enough to be a rounding error rather than a math error.

[Sucrose]

We're both correct, as it turns out! While concentration is the norm (especially for liquids), since we're looking at a fixed volume, we can look at moles instead in this case. Essentially, we're multiplying everything by the volume (without being concerned about what that volume is), so it all works out in the end. Plugging 0.933 (the molar change) into my equation as x, I get K(eq) = 48.7; close enough to be a rounding error rather than a math error.

Ah, right. Suppose that could simplify things. Think it might be a bit different for reactions that result in a net change in number of gases/aqueous species, but so long as that's held constant, you don't need to worry about it. Still a bit muzzy-headed, and haven't bothered to do the math myself, so maybe it could work for a constant volume even then.

[rogueboy]

Ah, right. Suppose that could simplify things. Think it might be a bit different for reactions that result in a net change in number of gases/aqueous species, but so long as that's held constant, you don't need to worry about it. Still a bit muzzy-headed, and haven't bothered to do the math myself, so maybe it could work for a constant volume even then.

As long as the volume is held constant - by ideal gas assumptions and constant moles of gas or not creating solvent or constant volume containers - you should be able to ignore the volume. If the volume isn't constant, things get far more complicated, since you now have to account for the changing volume as well as the number of moles. Not something I think I've ever seen, actually...