A mixture of $1\ \mathrm{g}$ of $\ce{H2(g)}$ and $1.06\ \mathrm{g}$ $\ce{H2S(g)}$ in a $0.5\ \mathrm{L}$ flask come to equilibrium at $1670\ \mathrm{K}$. At equilibrium, there are $8\times10^{-6}\ \mathrm{mol}$ of $\ce{S2(g)}$ present. Determine $K_\mathrm{p}$?

I set up an ICE table: $$\ce{2H2 +S2<=>2H2S}$$ $$\begin{array}{|l|l|l|} \ce{H2 & S2 & H2S} \\ \hline 1\ \mathrm{M}& 0\ \mathrm{M}& 0.062\ \mathrm{M} \\ 2x & x & 2x \\ ? & 1.6\times10^{-5}\ \mathrm{M} & ?\end{array}$$


What's wrong with my solution?

  • $\begingroup$ Just where did that $\ce{S2}$ come from? At the beginning, there was none; also, it is spent in the reaction... unless the reaction goes in the reverse direction! $\endgroup$ – Ivan Neretin May 3 '16 at 19:31
  • $\begingroup$ I made the reaction reversible. Doesn't make sense otherwise. I also added a parenthesis to balance in denominator. $\endgroup$ – MaxW May 3 '16 at 19:41
  • $\begingroup$ The S2 molarity does not appear to be correct. it should be 0.5 microMolar not 16 microMolar $\endgroup$ – Jeanno May 3 '16 at 19:45
  • $\begingroup$ @IvanNeretin It seems you are right. I'll try the other way. $\endgroup$ – Young May 3 '16 at 19:46
  • $\begingroup$ @MaxW thank you so much. I'm not a code person. Not Chm major LOL $\endgroup$ – Young May 3 '16 at 19:47

The problem lies in that our information of the system is in terms of concentration, so you actually calculated $K_\mathrm{c}$.

$$\ce{2H2 +S2<=>2H2S}$$ $$\begin{array}{|l|l|l|} \ce{H2 & S2 & H2S} \\ \hline 0.992\ \mathrm{M}& 0\ \mathrm{M}& 0.062\ \mathrm{M} \\ 2(1.6\times 10^{-5}) & (1.6\times 10^{-5}) & -2(1.6\times 10^{-5}) \\ 0.992+2(1.6\times 10^{-5}) & 1.6\times10^{-5}\ \mathrm{M} & 0.062-2(1.6\times 10^{-5})\end{array}$$

$$K_\mathrm{c}=\frac{(0.062\ \mathrm{M}-3.2\times 10^{-5}\ \mathrm{M})^2}{(1.6\times10^{-5}\ \mathrm{M})\cdot(0.992\ \mathrm{M}+3.2\times 10^{-5}\ \mathrm{M})^2}=245\ \mathrm{M^{-1}}$$

Using the relation $$K_\mathrm{p}=K_\mathrm{c} (RT)^{\Delta n},$$ where $$\Delta n=(\sum \text{coefficient of gaseous products})-(\sum\text{coefficient of gaseous reactants}),$$ we can calculate $K_\mathrm{p}$: $$K_\mathrm{p}=(245\ \mathrm{M^{-1}})\cdot(8.3144598\ \mathrm{J\ K^{-1}\ mol^{-1}}\cdot 1670\ \mathrm{K})^{-1}=0.0176\ \mathrm{kPa^{-1}}$$

  • $\begingroup$ I only calculated Kc lol.. $\endgroup$ – Young May 3 '16 at 23:05

You messed up the Change part in your ICE table.

The 'I' values are correct. However, if something is being formed , then the thing that is forming it disappears by the stoichiometrically corresponding amount. (Conservation of mass). In this case, $\ce{S_2}$ is being formed. The partial pressure of it increases by $x$. The partial pressure of $\ce{H_2}$ increases with it by $2x$. However, the partial pressure of $\ce{H_2S}$ decreases by $2x$. Thus, the numerator of your $K_p$ expression is actually $(0.062-2*1.6*10^{-5})^2$.

  • $\begingroup$ But I got the same result. And 240 is still not correct. $\endgroup$ – Young May 3 '16 at 21:59
  • $\begingroup$ Oh yeah, you just calculated $K_c$ $\endgroup$ – Yunfei Ma May 3 '16 at 22:34

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