Equilibrium pressures question?

If $5.00 \cdot {10}^{-} 1$ atm of $H I$ gas, $0.00100$ atm of ${H}_{2}$ gas, and $0.00500$ atm of ${I}_{2}$ gas are present in a $5.00$ L flask, calculate the equilibrium pressures of all 3 gases once equilibrium is established ?

Jun 26, 2018

${P}_{{H}_{2}} = \text{0.0402 atm}$
${P}_{{I}_{2} , e q} = \text{0.0442 atm}$
${P}_{H I , e q} = \text{0.4216 atm}$

Well, from your notes, you seem to have ${K}_{p} = 100$ (in implied units of $\text{atm}$). So, let's set up the reaction and ICE table based solely on the numbers you have given here:

$\text{H"_2(g) " "" "+" "" " "I"_2(g)" " rightleftharpoons " " 2"HI} \left(g\right)$

$\text{I"" "0.00100" "" "" "0.00500" "" "" } 0.500$
$\text{C"" "+x" "" "" "" "+x" "" "" "" } - 2 x$
$\text{E"" "0.00100+x" "0.00500+x" } 0.500 - 2 x$

Here, we have written that the reaction goes in reverse. Now that you have ${K}_{p}$,

${K}_{p} = 100 = {P}_{H I}^{2} / \left({P}_{{H}_{2}} {P}_{{I}_{2}}\right)$

$= {\left(0.500 - 2 x\right)}^{2} / \left(\left(0.00100 + x\right) \left(0.00500 + x\right)\right)$

Not much we can do here other than solve it in full; ${K}_{p}$ is not small.

$100 = {\left(0.500 - 2 x\right)}^{2} / \left(5 \times {10}^{- 6} + 0.00600 x + {x}^{2}\right)$

Multiply through by the denominator.

$5 \times {10}^{- 4} + 0.600 x + 100 {x}^{2} = {\left(0.500 - 2 x\right)}^{2}$

Expand the right-hand side.

$5 \times {10}^{- 4} + 0.600 x + 100 {x}^{2} = 0.250 - 2 \cdot 0.500 \cdot 2 x + 4 {x}^{2}$

Get this into standard quadratic form:

$- 0.2495 + 0.600 x + 100 {x}^{2} = - 2 x + 4 {x}^{2}$

$96 {x}^{2} + 2.600 x - 0.2495 = 0$

Now,

$a = 96$
$b = 2.600$
$c = - 0.2495$

so that

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- \left(2.600\right) \pm \sqrt{{2.600}^{2} - 4 \left(96\right) \left(- 0.2495\right)}}{2 \left(96\right)}$

This, as-given, has:

$x = 0.0392 , - 0.0662$

When we plug these in, however, only $x = 0.0392$ seems to work.

$\textcolor{b l u e}{{P}_{{H}_{2}} = 0.00100 + \left(0.0392\right) = \text{0.0402 atm}}$
$\textcolor{b l u e}{{P}_{{I}_{2} , e q} = 0.00500 + \left(0.0392\right) = \text{0.0442 atm}}$
$\textcolor{b l u e}{{P}_{H I , e q} = 0.500 - 2 \left(0.0392\right) = \text{0.4216 atm}}$

Jun 26, 2018

use an ICE table, then find pressures with ${K}_{p} = 100$?