I have the following chemical equation:

$$\ce{Sb2S3 + 3H2 <=> 2 Sb + 3H2S}$$

I have $\pu{1000 grams}$ of $\ce{Sb2S3}$ that reacts with $\pu{10 grams}$ of $\ce{H2}$ in a reactor at $\pu{713 K}$.

I would like to calculate $K_p$ (equilibrium constant using partial pressures).

I know that $K_p$ is equal to the quotient of the partial pressure of $\ce{H2S}$ cubed divided by partial pressure of $\ce{H2}$ cubed, and that I need to use the law of perfect gases.

$$K_p = \frac{P_{\ce{H2S}}^3}{P_{\ce{H2}}^3}$$

I have calculated the amount of moles of $\ce{H2S}$ to be $\pu{2.264 mol}$ and the amount for $\ce{H2}$ is $\pu{2.02 mol}$.

However there is just one problem I don't have the volume in this exercise, since it wasn't given. Thus, I am a bit stumped.

  • $\begingroup$ Use the principle of partial pressures, Volume cancels out.. $\endgroup$ Jun 15 '21 at 4:17
  • $\begingroup$ It does not seem this problem can be solved with the information given, please add information so that one can arrive at the number of moles stated here. Or, incorporate clarifications which you stated in comments into your post. $\endgroup$
    – Buck Thorn
    Jun 16 '21 at 0:42

So you need $$K_p = \frac{p_{\ce{H2S}}^3}{p_{\ce{H2}}^3}$$ where $p_{\ce{H2S}}$ and $p_{\ce{H2}}$ are partial pressures of $\ce{H2S}$ and $\ce{H2}$ respectively.

Partial pressure is nothing but the mole fraction of the substance times the total pressure. You already have mole fractions, so that part is done.

Now notice that in the expression of $K_p$ , you have cube divided by cube, so the term of total pressure will get cancelled out.

So, you will end up having $$K_p = \frac{p_{\ce{H2S}}^3}{p_{\ce{H2}}^3}=\frac{(2.264 )^3}{(2.02 )^3}=1.407$$

  • $\begingroup$ Thank you very much for your answer. However, in my book the answer given is 0,426. Sadly I came to the same answer that you have given me but it appears to be wrong. Edit: the amount of mols I gave you for H2 is actually wrong, my mistake, it's 2.685 mol. the answer I get is 0.6, which still isn't right though :/ $\endgroup$ Jun 15 '21 at 15:59
  • $\begingroup$ @StormCaster Initially I just put the values you had given without thinking much. Now when I think about it, how are you finding these values? Using the atomic mass of Sb as 122 g and sulphur as 32 g, we get the moles of $\ce{Sb_2S_3}$ as $\dfrac{1000}{122\times 2+ 32 \times 3}=2.941$ and moles of $\ce{H_2}$ as 10/2 =5. Now how will you find the equilibrium concentrations of these without knowing anything else? Are you sure you are not missing any other detail? May I know which book are you using? $\endgroup$ Jun 15 '21 at 18:00
  • $\begingroup$ Well, This is embarassing... In the end it turns out there are 2,1302 moles of H2S and 2,8198 moles of H2. so Kp is equal to (2,1302^3)/(2,8198)^3 which is equal to 0,4311 (the answer given in my book is 0,426 which is close enough I guess) Thank you for your help, sorry for wasting your time! $\endgroup$ Jun 15 '21 at 19:10

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