Question

The decomposition of Hydrogen Iodide on finely divided gold atis zero order with respect to HI. The rate defined below is constant at $\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{}}\frac{\mathbf{mol}}{\mathbf{Ls}}$.

$\begin{array}{l}\mathbf{2}\mathbf{HI}\left(g\right){\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{I}}_{\mathbf{2}}\left(g\right)\\ \mathbf{Rate}\mathbf{=}\frac{\mathbf{-}\mathbf{d}\left[H\right]}{\mathbf{dt}}\mathbf{=}\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\frac{\mathbf{mol}}{\mathbf{Ls}}\end{array}$

(a) If the initial HIconcentration was$\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\frac{\mathbf{mol}}{\mathbf{L}}$, calculate the concentration ofHI at 25minutes after the start of the reaction.

(b) How long will it take for all of the 0.250 MHIto decompose?

Short answer

(a) The concentration ofHIat25minutes is0.007 M.

(b) It will take 2083.33 seconds for all of the 0.250 MHI to decompose.

Step-by-step solution

Calculating the concentration of HI at 25 minutes.

Since we have a zero-order reaction, we can determine the reaction constant through rate law:
$\begin{array}{l}\mathit{r}\mathit{a}\mathit{t}\mathit{e}\mathbf{=}\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\frac{\mathbf{mol}}{\mathbf{Ls}}\\ \left[HI\right]=-Kt+{\left[HI\right]}_0=-1.20\times {10}^{-4}{\mathrm{molLs}}^{-1}\times 1500\mathrm{s}+0.250\mathrm{M}\\ \left[HI\right]=0.07\mathrm{M}\end{array}$

Calculating the time required for 0.250 MHI to decompose.

Using the same equation,

Full decomposition will then take:

$\begin{array}{l}t=\frac{\left[HI\right]}{K}\\ t=\frac{0.250\mathrm{M}}{1.20\times {10}^{-4}{\mathrm{molLs}}^{-1}}\\ t=2083.338\mathrm{seconds}\end{array}$