Two Chem Questions...

[Gurgle]

1. A 1.000 Liter Vessel is filled with 1.000 moles of N2, 2.000 moles of H2, and 3.000 moles of NH3, when the reaction N2(g) + 3H2(g) ==> 2NH3 comes to equilibrium it is observed that the concentration of NH3 is 1.86 mol/liters what is the numerical value of kc. ANSWER:. 043

I set up an ICE table but still can't solve the problem. Then I tried flipping the entire equation and got the answer by mistake.

2. A gas occupies 1800 milliliters at 142.2 C at 740 torr. What volume will it occupy at 80.2 C at the same pressure.
This question seems really simple and straightforward so I tried V1/T1 =V2/T2 1800/415.2=X/353.2 and came out with 1533 or something, however the answer key lists the answer as 2100.

Can you guys show me how to get the answers to the problems? Thanks in advance.

[Kenon]

#2's answer key is flat out wrong. The only way I could think it would make ANY sense at all would be is if the temperature increased by 80.2 degrees C, in which case, it would be 2148 mL anyways.

#1. Change in mols of NH3 is -1.14 mol. Thus, N2 is +.57 and H2 is +1.71

Kc in this direction is:

[unparseable or potentially dangerous latex formula]

Remember that Kc is based upon how you write the equation as well. Had you have wrote it 2NH3 <=> N2 + 3H2 it would have netted something like 23 just because of the way it works out in the problem. In the way I rewrote it, the products are favored over the reactants, while in the original it is the other way. Just keep this in mind.

[Gurgle]

Thank you very much!

[DCook]

#2

Actually, the answer is right because you MUST change the temperature in Centigrade to Kelvin

So you just use the gas laws to get:

V1/T1=V2/T2 and solve for T2

[Kenon]

Sorry, but the answer still is wrong. C to K doesn't change the fact that the proportion is still [unparseable or potentially dangerous latex formula] where [unparseable or potentially dangerous latex formula] and all of the variables are greater than 0.
[unparseable or potentially dangerous latex formula]. Thus, if [unparseable or potentially dangerous latex formula], then [unparseable or potentially dangerous latex formula] where C is some constant greater than 0. Thus, we can assume that V2 is less than V1. By the answer key, this is not true.

With numbers it is fairly easy to prove this to the way the OP did initially, but I felt like showing it with numbers.