[Big-Daddy]

So on this thread http://www.chemicalforums.com/index.php?topic=69295.0 the problem asked (I thought to raise my point in a new thread) was:

In the reaction C2H4 (g) + H2O (g) <---> C2H5OH (g) with K=2 , 2 moles of C2H4 and 2 moles of H2O are at equilibrium in a specific temperature and in a container with volume of 2 liters. What is the yield of the reaction?

The answer generally came to y = 2/3. I can see where that idea comes from because equilibrium moles of the product is apparently 4. Here it was assumed K=K_c=[C2H5OH]/([C2H4]*[H2O]).

If instead K=K_p then I got equilibrium moles of 99.1 mol at 298 K (assumed). equilibrium moles of product = 8RT /(V·p°), 8 because K*n(C2H4)*n(H2O)=8, V in m³, p° = 10⁵ Pa.
obviously this doesn't correspond with any of the given answers, it gives yield = 98%.

So did you just decide that because T was not given, they must be referring to K_c when they quote value of K?

[Borek]

The question was incomplete and there were no correct answer regardless of whether one decided to use Kc or Kp, so in general it is a moot.

K is given for the reaction which uses gaseous water, and 298 K doesn't make much sense to me. If anything, I would assume 373K at least.

No idea what was behind curiouscat thinking.

[Big-Daddy]

The question was incomplete and there were no correct answer regardless of whether one decided to use Kc or Kp, so in general it is a moot.

K is given for the reaction which uses gaseous water, and 298 K doesn't make much sense to me. If anything, I would assume 373K at least.

No idea what was behind curiouscat thinking.

So we'll assume it was just a bad question, that's why they didn't consider the need for given temperature.

But, why is it incomplete with K_c, yield = equilibrium moles of product / initial moles of reactant = 4 / (4+2)?

[curiouscat]

No idea what was behind curiouscat thinking.

Not having to assume a Temperature.