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:

$$x = 0.01907$$


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

But the options are






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?


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

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

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

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

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.