I've been assigned the following problem at school:

Given the following reaction:

$$\ce{C (s) + CO2 (g) <=> 2 CO (g)}$$

Find the mole fraction of $\ce{CO2}$ provided that, in equilibrium, $K_\mathrm{p} = 14.1$.

Our teacher's solution is $x(\ce{CO2}) = 0.324$ (mole fraction).

I suspect that the problem as is provides insufficient information to arrive at the solution above. Assuming that the total pressure of the system is $\pu{10 atm}$, the resulting mole fraction is indeed $0.324$.

Is the problem truly incomplete or can it be solved without knowing the total pressure?

  • 3
    $\begingroup$ You seem to be comfortable enough working out the answer when you're given a pressure. Why not try working it out without explicitly choosing a pressure? Use $p$ as a symbol instead of putting in something like 10 atm, and if you find that all the $p$'s cancel out in your answer, then it follows that you don't actually need to know the pressure. $\endgroup$ – orthocresol Jan 11 at 20:42
  • 1
    $\begingroup$ I have attempted this. The final expression is: 14.1x = (1-x)^2*PT. The total pressure does not seem to cancel out. Here, x is the mole fraction. X seems to be a function of PT. The value of X changes with other values of PT. $\endgroup$ – Grego Jan 11 at 20:44
  • $\begingroup$ Took me a while to understand what that equation meant, but yeah, you are correct! So, the question is indeed incomplete. [More common notation would be something like $p_\text{tot}$; PT suggests some pressure multiplied by temperature...] $\endgroup$ – orthocresol Jan 11 at 20:48
  • $\begingroup$ You can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details. $\endgroup$ – orthocresol Jan 11 at 20:50
  • $\begingroup$ Ok! Thanks for the quick reply! My bad. I'll keep this in mind for the next question. $\endgroup$ – Grego Jan 11 at 20:51

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.