So I have this question that I was assigned, and I am having trouble with step 2:

Hydrogen sulfide can be removed from natural gas via the following process: $$\ce{2H2S(g) + SO2(g) <=> 3S(s) + 2H2O(g)}\ \ \Delta H = \pu{-145 kJ mol^-1}$$

  1. Write the equilibrium constant expression for this reaction.
  2. Calculate the equilibrium constant, when $\pu{1.00 mol}$ of $\ce{H2S}$ and $\pu{1.00 mol}$ of $\ce{SO2}$ in a $\pu{1.00 L}$ vessel at $\pu{373 K}$ to give $\pu{0.50 mol}$ of water vapour under equilibrium conditions.
  3. Identify FOUR factors that would maximise the removal of $\ce{H2S(g)}$ in this reaction.

I have the answers for this, and this is the step I got to:


So, I know in the top part of the fraction, it is given that at equilibrium $\pu{0.5 mol}$ of water is present. I get that.

But it also says that $\pu{1 mol}$ of $\ce{H2S}$ and $\pu{1 mol}$ of $\ce{SO2}$ reacted. I'm not sure how these lead to values of $0.5$ and $0.75$?

It says there are $\pu{0.5 mol}$ of water at equilibrium. Does that mean that the $\pu{1 mol}$ of $\ce{H2S}$ and $\ce{SO2}$ are not all used? Thus since the ratio of water to $\ce{H2S}$ is $1:1$ it is $0.5$ in the fraction, but can nothing explain the $0.75$?


Firstly, let's write the equilibrium expression for this reaction: $$\ce{ K =\frac{[H2O]^2}{[H2S]^2[SO2]}}$$ So we know that initially $\ce{[H2O]}$ = $\ce{[SO2]}$ = $\mathrm{1~M}$ and $\ce{[H2O]}$ = $\mathrm{0~M}$. We also know that at the end, $\ce{[H2O]}$ = $\mathrm{0.5~M}$.

Now the stoichiometric ratios of $\ce{~H2O:SO2:H2O} = 2:1:2$. Therefore the decrease in the concentration of $\ce{H2S}$ should equal the increase in the concentration of $\ce{H2O}$. Also the decrease in the concentration of $~\ce{SO2}$ should be half of the decrease in the concentration of $\ce{H2S}$. Therefore $\ce{[H2S]}$ = $\mathrm{0.5~M}$ and $\ce{[SO2]}$ = $\mathrm{0.75~M}$

Therefore, plugging in these values into the equilibrium constant, you should get the required answer.


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