1
$\begingroup$

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

$$K_\mathrm{p}=\frac{(0.062+2*1.6\times10^{-5})^2}{(1.6\times10^{-5})*(1-2*1.6\times10^{-5})^2}=240$$

What's wrong with my solution?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $K_\mathrm{p}$: $$K_\mathrm{p}=(245\ \mathrm{M^{-1}})\cdot(8.3144598\ \mathrm{J\ K^{-1}\ mol^{-1}}\cdot 1670\ \mathrm{K})^{-1}=0.0176\ \mathrm{kPa^{-1}}$$

$\endgroup$
  • $\begingroup$ I only calculated Kc lol.. $\endgroup$ – Young May 3 '16 at 23:05
1
$\begingroup$

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 $(0.062-2*1.6*10^{-5})^2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.