# What is the concentration of H2 for the equation H2+I2=2HI if the equilibrium constant is 64 and the initial concentrations are both 1.9

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

$$8=\frac{x}{(1.9-x)}$$

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

$$1.9-1.688=.212$$

But it’s telling me this answer is wrong

• $x$ moles of reactant consumed will yield $2x$ moles of product, so the first equation needs to be modified. – William R. Ebenezer Mar 26 at 7:00
• So you get 0.212 sausages per square meter, or what is your unit? – 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}
• Why is it 2X^2, instead of just X^2 – Hunter Mar 26 at 21:00
• When $\pu{x~mol}$ of $\ce{H2}$ reacted ,$\pu{ 2x ~mol}$ of $\ce{HI}$ formed ,so $[\ce{HI}]_\mathrm{e}=\pu{2x~M}$ and $[\ce{HI}]^2_\mathrm{e}=(2x)^2$ . – Adnan AL-Amleh Mar 26 at 21:18