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So I have a simple reaction:

$\ce{A(g) <=> B(g)}$

To start with, I had $0.037~\mathrm{mole}$ of $\ce{A}$, and 100 minutes I have $\approx0.03616~\mathrm{mol}$ of $\ce{A}$ and $\approx0.00088~\mathrm{mol}$ of $\ce{B}$.

I reached these values by myself, but now I'm told to calculate the Yield at this point. I'm aware of the equation Fractional Yield = actual yield / theoretical yield. But is it important if the reaction is reversible? In my case it is, would that change the theoretical yield? Or is it always 100%? Is even the fractional yield what I'm asked here? It just says: "calculate the yield of this reaction after 100 minutes have passed".

It's my first problem that includes yield and i would love an explanation. I've searched on google and wikipedia so I know a little bit about it.

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Well, all chemical reactions are reversible at least to some extent. In the case here with your imaginary compounds and your one-way arrow, I suspect that whoever wrote this problem didn't care too much about whether B would turn back into A.

How did you calculate how many moles of A and B you had after 100 minutes?

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  • $\begingroup$ I know it's a first grade reaction because I was given the velocity constant in 1/seconds, so I put the data in the equation and was left with [A] = 0.395moles/liter * e^-0.2377 (-constant*time). [A] = 0.03847 moles/liter so the concentration of [B] must be 0.00088. Volume according to ideal gas law was 0.94 liters so I should have 0.00088*0.94 = 0.00083 moles of the gas B. Maybe the yield equation is moles of product / moles of reactant? I really don't know $\endgroup$ – Gaspa79 May 16 '14 at 1:10
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    $\begingroup$ Yield should be actual yield over theoretical yield. Moles of product over moles of reactant is not yield. $\endgroup$ – Dissenter May 16 '14 at 1:13
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    $\begingroup$ I think the question is just asking for total yield, not a percent yield. $\endgroup$ – Dissenter May 16 '14 at 1:17
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    $\begingroup$ Total yield or yield is just how much you get. Percent or fractional yield is how much you get over how much you expected to get. $\endgroup$ – Dissenter May 16 '14 at 1:19
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    $\begingroup$ Yes! It would be useful though if I could see the original problem in its entirety. $\endgroup$ – Dissenter May 16 '14 at 3:32

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