# Calculating the value of equilibrium constant in terms of pressure

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

I set up an ICE table:

$$\begin{array}{ccc} \ce{& &2 H2 &+ &S2 &<=> &2 H2S} \\ \text{Initial} & &\pu{1 M} && 0 && \pu{0.062 M} \\ \text{Change} & &2x && x && 2x \\ \text{Equilibrium / Final} & &? && \pu{1.6E-5 M} && ? \end{array}$$

From the $$\ce{S2}$$ column we see that $$x = \pu{1.6E-5 M}$$, so I calculated the equilibrium constant as follows:

$$K_p = \frac{(0.062 + 2\times\pu{1.6E-5})^2}{(\pu{1.6E-5})(1-2\times\pu{1.6E-5})^2} = 240$$

What is wrong with my solution?

The problem lies in that our information of the system is in terms of concentration, so you actually calculated $$K_c$$.
$$\ce{2H2 +S2<=>2H2S}$$ $$\begin{array}{|l|l|l|} \ce{H2 & S2 & H2S} \\ \hline 0.992\ \mathrm{M}& 0\ \mathrm{M}& 0.062\ \mathrm{M} \\ 2(1.6\times 10^{-5}) & (1.6\times 10^{-5}) & -2(1.6\times 10^{-5}) \\ 0.992+2(1.6\times 10^{-5}) & 1.6\times10^{-5}\ \mathrm{M} & 0.062-2(1.6\times 10^{-5})\end{array}$$
$$K_c=\frac{(0.062\ \mathrm{M}-3.2\times 10^{-5}\ \mathrm{M})^2}{(1.6\times10^{-5}\ \mathrm{M})\cdot(0.992\ \mathrm{M}+3.2\times 10^{-5}\ \mathrm{M})^2}=245\ \mathrm{M^{-1}}$$
which is numerically the same as the value you obtained, except that I used a more precise molecular mass for $$\ce{H2}$$.
Using the relation $$K_p=K_c(RT)^{\Delta n},$$ where $$\Delta n=(\sum \text{coefficient of gaseous products})-(\sum\text{coefficient of gaseous reactants}),$$ we can calculate $$K_p$$: $$K_p=(245\ \mathrm{M^{-1}})\cdot(8.3144598\ \mathrm{J\ K^{-1}\ mol^{-1}}\cdot 1670\ \mathrm{K})^{-1}=0.0176\ \mathrm{kPa^{-1}}$$