posted by Rachel on .
Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide.
H2(g) + CO2(g) --> H2O(g) + CO(g)
(a) Laboratory measurements at 986°C show that there are 0.11 mol each of CO and H2O vapor and 0.087 mol each of H2 and CO2 at equilibrium in a 1.0 L container. Calculate the equilibrium constant for the reaction at 986°C.
I got 2.1e2. I did (.11)(.11)/(.087)(.087)
What are i doing wrong?
(b) Suppose 0.052 mol each of H2 and CO2 are placed in a 1.5 L container. When equilibrium is achieved at 986°C, what amounts of CO(g) and H2O(g), in moles, would be present? [Use the value of K calculated in part (a).]
The numbers you have for Keq are correct; I assume you are punching the wrong buttons on your calculator. For (0.11)^2/(0.087)^2 I get 1.6
For part b.
0.052moles/1.5L = 0.0347M
..............H2 + CO2 ==> H2O + CO
Substitute into Ka expression and solve for x. The question asks for moles; therefore, x is molarity and M x L = moles.
Post your work if you get stuck.
Would you get the same answer for both parts of b?
Since you started with the same molarity for each and you used x of each, then the moles of each will be the same at equilibrium.
ok so I did (x)(x)/(.03466)(.03466)= 1.6. I solved for x and got .0438. Do I have to do anything with this number to get the correct answer?
Can you check my work?
I would have rounded the 0.034666 to 0.03467 (and that's too many significant figures). However, it's (x)(x)/(0.03467-x)^2. What did you do with the -x in the 0.03467 - x term? Throw it away. You can't do that (at least not in this problem). You must solve the quadratic. I don't get anything like 0.0438M for an answer. After you find the M for x, you must subtract it from 0.03467 to find (CO2) and (H2), then multiply (CO2), (H2), (H2O), and (CO) by 1.5 to obtain moles (since the problem asks for each of these in moles.)
Sorry I am soo confused right now O.O. is there anyway u can explain this in simplier form?