Calculating [HI] given Kc, [H2], and [I2] for the reaction 2HI <-> H2+I2


> 
> In the reaction in equilibrium
> 
> 
> $$\ce{2HI <=> H2 + I2}$$
> 
> 
> The value $K\_c$ is $0.0198$ at $\pu{721 K}$. Calculate $\ce{[HI]}$ at $\pu{721 K}$ if $\ce{[H2]} = \pu{0.0120 M}$ and $\ce{[I2]} = \pu{0.0150 M}$.
> 
> 
> 


We use an ICE box. Since $K < 1$, the equilibrium is to the left. We are spending products to create reagents, so...


$$\begin{bmatrix}&\ce{[HI]} & \ce{[H2]} & \ce{[I2]}\\
\text{Initial} & 0 & 0.0120 & 0.0150\\
\text{Change} & +2x & -x & -x\\
\text{Equilibrium} & 2x & 0.0120 - x & 0.0150 - x
\end{bmatrix}$$


The equilibrium is given by


$$0.0198 = \frac{(0.0120-x)(0.0150-x)}{(2x)^2}$$


Solving yields:


$$x = 0.01907$$


Therefore


$$\ce{[HI]} = \pu{0.0381 M}$$


But the options are



> 
> $\pu{0.00211M}$
> 
> 
> $\pu{0.133M}$
> 
> 
> $\pu{0.0334M}$
> 
> 
> $\pu{0.006567M}$
> 
> 
> $\pu{0.0953M}$
> 
> 
> 


While it is pretty close to $\pu{0.0334M}$, I can't see how could I have made my calculation any more accurate (or if they were the ones who were inaccurate).


Was my procedure correct?




You're close. Using the expression for $K\_c$, you just need to solve for $[\ce{HI}]$:


$$K\_c = {[\ce{H2}][\ce{I2}]\over[\ce{HI}]^{2}}$$


$$0.0198 = {(\pu{0.012 M})(\pu{0.015 M})\over[\ce{HI}]^{2}}$$


$$[\ce{HI}] = \sqrt{{(\pu{0.012 M})(\pu{0.015 M})\over 0.0198}} = \pu{0.0953 M}$$


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