For the reaction below, Keq = 0.0064 (at 825C). If you place 12.011 g Carbon and 5 atm pressure of water vapour in a 2.0 L reaction flask, calculate the pressure of carbon dioxide gas that will be present once the reaction comes to equilibrium.

C(s) + 2 H2O(g) ↔ CO2(g) + 2 H2(g)

Here is what I have done so far:

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\begin{array}{l}K_{eq}=\frac{P\left(CO_2\right)\cdot P\left(H_2\right)}{P\left(H_2O\right)}=0.0064\ \ \left(at\ 825\ \text{°C}\right)\\\frac{\left(x\right)\left(2x\right)}{\left(5-2x\right)}=0.0064\\\frac{2x^2}{5-2x}=0.0064\\2x^2=0.0064\left(5-2x\right)\\2x^2=0.032-0.0128x\\2x^2+0.0128x-0.032=0\\x\ =0.1233\\∴\ \text{The pressure of carbon dioxide is 0.1233 atm}\end{array}

I have gotten the answer to be 0.1233 atm but the answer is supposed to 0.3 atm. So, did I do something wrong here? Could someone help me out here if possible?

  • $\begingroup$ Using photos/screenshots of ( even handwritten) text instead of typed text itself is strongly discouraged. It is impossible to index/search/reuse it in answers referring to it. In a case of handwritten text, it puts extra burden on responders to properly decipher it. That all may lead to the question being ignored or even closed. Consider copy/paste or retyping and using eventually MathJax for expressions and formulas. $\endgroup$
    – Poutnik
    Nov 12, 2020 at 19:26
  • $\begingroup$ Similarly as in your prior question, you ignore power coefficients of pressures in the equilibrium expression. $\endgroup$
    – Poutnik
    Nov 12, 2020 at 19:30
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    $\begingroup$ If you find your mistake yourself, it gives you more, than if you are just told. BTW, the site prefers teaching fishing to giving fish. $\endgroup$
    – Poutnik
    Nov 12, 2020 at 20:27
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    $\begingroup$ @JabanBaroose the term $P_{\ce{H2O}}$ in the denominator of the equation should have a power of 2, since the coefficient of $\ce{H2O}$ in balanced chemical equation is 2. $\endgroup$
    – Eyy boss
    Nov 19, 2020 at 6:21
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    $\begingroup$ @Eyyboss Thank you for your reply. I have already got the answer that I was looking for as it was a very simple error that I made. Since Poutnik here couldn't help me out, I asked an online friend of mine who really helped me clarify the mistake that I made. But I greatly appreciate you for giving me an answer, even if it is too late. $\endgroup$ Nov 21, 2020 at 21:31

1 Answer 1


I'm not sure if I'm late to answering your question but I had the same problem with my calculations and I figured out what I did wrong. You forgot in your Keq equation the exponents: (x)(2x)^2/(5-2x)^2 = 0.0064 and then you square root both sides and go from there until you get a quadratic equation and like you did before just use the quadratic formula to solve for x. Hope this helps


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