Variation $1$:

Given initial concentrations of reactants and concentration of product formed at equilibrium.

$1\mathrm{.}0$ mole of both $S{O}_2$ and $O_2$ are loaded into a $2\mathrm{.}0L$ flask. At equilibrium, the concentration of $S{O}_3$ is $0\mathrm{.}150M.$ Find the value of $K_c,$ and the percent completion.

getsusal upareuts.

$2S{O}_{2(g)}+O_{2(g)}2S{O}_{3(g)}$

limiting reoant $(2$ xosfust os $0_2)$

Step #$1$: chonge quicntities to if heeded. $c=\frac{h}{v}$

$[S{O}_2]\frac{1\mathrm{.}0(mole)}{2\mathrm{.}0ba}=0\mathrm{.}5mo\frac{l}{L}$

$[O_2]=\frac{1\mathrm{.}0(mole)}{2\mathrm{.}0(L)}=0\mathrm{.}5mo\frac{l}{L}$

Step #$2$: set up an ice chart.

$0$ products alwoys increoses, hos a positive chage$-$

reactents ore usel up till we get to equilibrium

to find change in $[]$ use molor ratias.

$S{O}_{2}S{O}_3$

$\frac{2}{x}=\frac{2}{0\mathrm{.}150}$

$\frac{1}{x}=\frac{O_2}{2}$

$\frac{1}{x}=\frac{\frac{S{O}_3}{2}}{0\mathrm{.}150}$

Step #$3$: Find $KC*$use eqvolves.

$\%$ completition $=\frac{\text{a}\text{c}\text{t}\text{u}\text{a}\text{l}\text{y}\text{i}\text{e}\text{l}\text{d}}{\text{t}\text{h}\text{e}\text{o}\text{r}\text{e}\text{t}\text{i}\text{c}\text{a}\text{l}\text{y}\text{i}\text{e}\text{l}\text{d}}100$

$0\mathrm{.}50M$ Theor

$=\frac{0\mathrm{.}150100}{0\mathrm{.}50}$

$=30\%$

Actual yield $-$ the omount of product produced at equilibsium.

$-S{O}_3=$ equilbrium $=0\mathrm{.}150M$

$\%$ completion $=\frac{A}{T}100$

$=\frac{0\mathrm{.}150100}{0\mathrm{.}50}$

$=30\%$

Theoretical vield$-$ how much we could have produced under ideul contitans if all the reaction's weregsed ap

we need to look at our limiting reosent $longrightarrowS{O}_2.$