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In the reaction in equilibrium

$$\ce{2HI} \leftrightarrow \ce{H2 + I2}$$

The value $K_c$ is $0.0198$ at $721$ K. Calculate $\ce{[HI]}$ at $721$ K if $\ce{H2} = 0.0120$ M and $\ce{I2} = 0.0150$ 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

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

But the options are

$0.00211$

$0.133$

$0.0334$

$0.006567$

$0.0953$

While it is pretty close to $0.0334$, 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?

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

$$\mathrm{K_c} = {[\ce{H2}][\ce{I2}]\over[\ce{HI}]^{2}}$$

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

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

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    $\begingroup$ Well I can see that this method is way better. Do you know what went wrong with my particular procedure, though? $\endgroup$ – Voldemort Aug 20 '15 at 15:43
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    $\begingroup$ The ICE method takes you from initial conditions to equilibrium (final) conditions: in your problem, above, we're already at equilibrium, so the business with subtracting $x$ from each product concentration is not called for, as you're given those concentrations (0.012 M and 0.015 M). You might think of this problem as just the final step in the ICE procedure, perhaps, where the Change has already happened. $\endgroup$ – Todd Minehardt Aug 20 '15 at 15:53

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