# Starting with \tt{SO_3(g)} at \sf{"1.00 atm"}, what will be the total pressure when equilibrium is reached with the following reaction at \sf{"700 K"}?

## $\setminus \texttt{2 S O 3 \left(g\right) \setminus \leftrightarrow 2 S {O}_{2} \left(g\right) + {O}_{2} \left(g\right)}$ $\setminus \textsf{{K}_{p} = 1.6 \times {10}^{-} 5 \text{ @ 700 K}}$ (Has this been posted before? I apologize if yes)

Aug 6, 2018

About $\text{1.02 atm}$ to two decimal places.

Well, let's think about what we want at the end. As in Dalton's law of partial pressures for ideal gases,

${P}_{\text{tot}} = {P}_{S {O}_{3}} + {P}_{S {O}_{2}} + {P}_{{O}_{2}}$

All of these quantities are equilibrium quantities. Once we know all three (actually, we will find that we only need one quantity), we know how to find the total pressure.

Now it is practical to write the ICE table, since we have an idea of what path to follow. In terms of $\text{atm}$, we have:

$2 {\text{SO"_3(g) " "rightleftharpoons" " 2"SO"_2(g) + "O}}_{2} \left(g\right)$

$\text{I"" "1.00" "" "" "" "" "0" "" "" "" } 0$
$\text{C"" "-2x" "" "" "" "+2x" "" } + x$
$\text{E"" "1.00-2x" "" "" "2x" "" "" } x$

Now what we find is that the total pressure is given by:

${P}_{\text{tot}} = \left(1.00 - 2 x\right) + \left(2 x\right) + \left(x\right) = 1.00 + x$

And so, if we find $x$, we are basically done.

$1.6 \times {10}^{- 5} = \frac{{P}_{S {O}_{2}}^{2} {P}_{{O}_{2}}}{{P}_{S {O}_{3}}^{2}}$

$= \frac{{\left(2 x\right)}^{2} x}{1.00 - 2 x} ^ 2$

We can tell that ${K}_{P}$ is small (on the order of ${10}^{- 5}$ or less), so $x$ is small compared to $\text{1.00 atm}$. Therefore,

$1.6 \times {10}^{- 5} \approx \frac{4 {x}^{3}}{1.00} ^ 2$

Thus,

$x \approx {\left(\frac{1.6 \times {10}^{- 5} \cdot {\left(1.00\right)}^{2}}{4}\right)}^{1 / 3}$

$=$ ${0.015}_{9}$ $\text{atm}$ to two sig figs.

and so,

color(blue)(P_"tot") = "1.00 atm" + 0.015_9  $\text{atm}$

$=$ $\textcolor{b l u e}{\text{1.02 atm}}$ to two decimal places.