So for this, I set up the equation =

$$64= \frac{(x)^2}{(1.9-x)^2}$$

I took the sqrt of both sides and got


So then I did $$8(1.9-x)=x$$

And simplified I got $$15.2=9x$$

So $$x=1.688$$

So to find concentration it’s Initial- Change


But it’s telling me this answer is wrong

  • 2
    $\begingroup$ $x$ moles of reactant consumed will yield $2x$ moles of product, so the first equation needs to be modified. $\endgroup$ – William R. Ebenezer Mar 26 at 7:00
  • 1
    $\begingroup$ So you get 0.212 sausages per square meter, or what is your unit? $\endgroup$ – mcocdawc Mar 26 at 7:32
  • let the amount of $\ce{H2}$ reacted = $\pu{x~mol}$
  • Use the molar ratio to detemine the amount of $\ce{I2}$ reacted and the amount of $\ce{HI}$ formed: $$\begin{align} &\text{The amount of }\ce{H2}~\text{reacts to reach equilibrium }&=\pu{x~mol}\\ &\text{The amount of }\ce{I2}~\text{reacts to reach equilibrium}&=\pu{x~mol}\\ &\text{The amount of }\ce{HI}~\text{formed at equilibrium }&=\pu{2x~mol}\\ \end{align}$$
  • Use the following table to determine the amount and the concentration of each species at equilibrium : $$\begin{align} \ce{&H2 &&+ &&O2 &<=> &&2HI\\ I~~~~ &1.9 &&&&1.9 &&&0 \\ C ~~~ &-x &&&&-x &&&2x \\ E ~~~ &(1.9-x) &&&&(1.9-x) &&&2x} \end{align}$$
  • Th concentrations at equilibrium : $$[\ce{H2}]_\mathrm{e}=(1.9-x) , [\ce{I2}]_\mathrm{e}=(1.9-x) , [\ce{HI}]_\mathrm{e}= \pu{2x~M} $$
  • Sustitute the concentrations in the following formula: $$\begin{align} K_\mathrm{C} = \frac{ [\ce{HI}]_\mathrm{e} } {[\ce{H2}]_\mathrm{e}[\ce{I2}]_\mathrm{e} }\\ 64= \frac{(2x)^2}{(1.9-x)(1.9-x)} \end{align}$$

  • Take the square root of both sides : $$\sqrt{64}=\sqrt{\frac{(2x)^2}{(1.9-x)^2}}$$

$$8=\frac{2x}{(1.9-x)}$$ - Solve for $x : x=1.52$

  • Calculate the concentrations at equilibrium : $$\begin{align} [\ce{H2}]_\mathrm{e}&=(1.9-x)&=(1.9-1.52)&=\pu{0.38~M} \\ [\ce{I2}]_\mathrm{e}&=(1.9-x)&= (1.9-1.52)&=\pu{0.38~M} \\ [\ce{HI}]_\mathrm{e}&= \pu{2\times{x}}&= 2\times{1.52}&=\pu{3.04~M} \end{align} $$
  • 1
    $\begingroup$ Why is it 2X^2, instead of just X^2 $\endgroup$ – Hunter Mar 26 at 21:00
  • $\begingroup$ When $\pu{x~mol}$ of $\ce{H2}$ reacted ,$\pu{ 2x ~mol}$ of $\ce{HI}$ formed ,so $[\ce{HI}]_\mathrm{e}=\pu{2x~M}$ and $[\ce{HI}]^2_\mathrm{e}=(2x)^2$ . $\endgroup$ – Adnan AL-Amleh Mar 26 at 21:18

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