Question

A 4.72-g sample of methanol (CH₃OH) was placed in an otherwise empty 1.00-L flask and heated to 250.⁰C to vaporize the methanol. Over time the methanol vapor decomposed by the following reaction:

$\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\left(g\right)\mathbf{\rightleftharpoons }\mathit{C}\mathit{O}\left(g\right)\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\left(g\right)$

After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much H₂(g) as CH₃OH(g). Calculate K for this reaction at 250.⁰C.

Short answer

The value of K for the reaction at 250⁰C will be 0.23 mol²/L²

Step-by-step solution

Determine the amount of CH3OH present 

Number of moles of CH₃OH present initially

$$\begin{gathered} =4.72gC{H}_{3}OH\times \frac{1mol}{32.04g} \\ =0.147molC{H}_{3}OH \end{gathered}$$

Applying graham’s law of diffusion, we get

$\begin{array}{rcl}\frac{Rat{e}_{H_2}}{Rat{e}_{C{H}_{3}OH}}& =& \sqrt{\frac{M_{C{H}_{3}OH}}{M_{H_2}}}\\ & =& \sqrt{\frac{32.04}{2.016}}\\ & =& 3.99\end{array}$

Determine the equilibrium constant 

Let the equilibrium moles of H₂ and CH₃OH are$\begin{array}{rcl}& & n_{H_2}\end{array}$ and $\begin{array}{rcl}& & n_{C{H}_{3}OH}\end{array}$respectively. It has been stated that the effused mixture has 33.0 times as much H₂ as CH₃OH.

$\begin{array}{rcl}33.0& =& 3.99\times \frac{n_{H_2}}{n_{C{H}_{3}OH}}\\ \frac{n_{H_2}}{n_{C{H}_{3}OH}}& =& 8.28\end{array}$

From the above information we can write

$\begin{array}{rcl}\frac{n_{H_2}}{n_{C{H}_{3}OH}}& =& 8.28\\ \frac{n_{H_2}}{n_{C{H}_{3}OH}}& =& \frac{2x}{0.147-x}\\ \frac{2x}{0.147-x}& =& 8.28\\ x& =& 0.118\end{array}$

Now the equilibrium constant will be

$\begin{array}{rcl}K& =& \frac{\left[CO\right]{\left[H_2\right]}^2}{C{H}_{3}OH}\\ & =& \frac{\left[0.118{\displaystyle \frac{mol}{L}}\right]{\left[2\times 0.118{\displaystyle \frac{mol}{L}}\right]}^2}{\left[\left(0.147-0.118\right){\displaystyle \frac{mol}{L}}\right]}\\ & =& 0.23\frac{mo{l}^2}{L^2}\end{array}$