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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?

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  • $\begingroup$ For the record, at the temperature presented, the single equilibrium reaction cited is but one of many reactions. Missing are those involving the hydrogen atom and also the iodine atom, see aip.scitation.org/doi/10.1063/1.1730174 . $\endgroup$ – AJKOER Mar 1 at 19:25
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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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    $\begingroup$ 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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