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Gaseous xenon tetrafluoride at partial pressure of $\pu{2 kPa}$ and hydrogen at partial pressure of $\pu{10 kPa}$, are exploded in an enclosed container producing xenon and hydrogen fluoride: $$\ce{XeF4 (g) + 2 H2 (g) -> Xe (g) + 4 HF (g)}$$ Find out the total pressure of mixture of gases in the enclosed container.

Assuming $T$ and $V$ are constant, we can assume that the pressure is directly related to the amount of substance, hence $\pu{1 kPa}$ of $\ce{XeF4}$ reacts with $\pu{2 kPa}$ of $\ce{H2}$.

This means that only $\pu{4 kPa}$ of $\ce{H2}$ is needed to react with $\pu{2 kPa}$ of $\ce{XeF4}$.

Now, to find total pressure of mixture of gases I did

$$p = \pu{2 kPa}~\text{(from Xe)} + \pu{8 kPa}~\text{(from HF)} = \pu{10 kPa}$$

Why is this wrong? Why must I add $\pu{10 kPa}$ by $\pu{6 kPa}$ which is the unreacted hydrogen?

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    $\begingroup$ Pressure is an intensive property. An expression "$\pu{1 kPa}$ of $\ce{XeF4}$ reacts with $\pu{2 kPa}$ of $\ce{H2}$" makes no sense. $\endgroup$
    – andselisk ♦
    Dec 5, 2017 at 1:29
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    $\begingroup$ Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". $\endgroup$
    – Martin - マーチン ♦
    Dec 5, 2017 at 5:14

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You are correct up to the last part. The total pressure would be the sum of all the partial pressures in the reaction vessel.

We know that $$P=\left(\frac{RT}{V}\right)n$$ and therefore $n$ varies directly with $P$ as $T$, $R$, and $V$ are constant. For the sake of the problem (it won't matter) we can just say that $x=\left(\frac{RT}{V}\right)$. Therefore, $P=xn$ and the starting amounts of substance are equal to $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ and $\frac{10}{x}\pu{ mol}$ $\ce{H2}$.

From the balanced equation and the starting materials and stoichiometry, we know that $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ would react with $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ to form $\frac{2}{x}\pu{ mol}$ $\ce{Xe}$ and $\frac{8}{x}\pu{ mol}$ $\ce{HF}$ to make $\frac{10}{x}\pu{ mol}$ product.

However, you only reacted $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ of the $\frac{10}{x}\pu{ mol}$ total, so there is $\frac{6}{x}\pu{ mol}$ unreacted $\ce{H2}$ left in the reaction vessel.

$$\begin{multline} \frac{10}{x}\text{ amount of substance of product } + \frac{6}{x}\text{ amount of substance of unreacted }\\ = \frac{16}{x}\text{ total amount of substance} \end{multline}$$

From $P=xn$, the fact that this occurred instantly, and $T$, $R$, and $V$ are constant, we then know that the final pressure is $\ce{16 kPa}$.

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  • $\begingroup$ I would suggest to have a look at How can I format math/chemistry expressions here? regarding formatting guidelines. And again, pressure is an intensive property, one cannot say that much kPa reacts, it's like counting green and red apples by color. $\endgroup$
    – andselisk ♦
    Dec 5, 2017 at 1:30
  • $\begingroup$ Thank you for the link, I'm new here and didn't know where to look. In this case, because the reaction is happening in an explosion, and T and V are constant, the number of moles relates directly to the pressure by some constant $$P=(RT/V)n$$ For the sake of the problem, we can carry $RT/V$ through in our stoichiometry, then re-convert it out at the end. In the answer, I merely used this shortcut (which I should have made known) and did not say this. $\endgroup$
    – Chemgoat
    Dec 5, 2017 at 3:56
  • $\begingroup$ I get that, but the terminology you and OP are using is just wrong, even though the answer seems to be correct. Comparing quantities using intensive parameters in this context makes little sense. That's all. $\endgroup$
    – andselisk ♦
    Dec 5, 2017 at 4:00
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    $\begingroup$ Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". $\endgroup$
    – Martin - マーチン ♦
    Dec 5, 2017 at 5:18

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