Question

A vessel contains dimethyl ether at a pressure of $0.4\text{ }atm.$ Dimethyl ether decomposes as $C{{H}_{3}}OC{{H}_{3\left( g \right)}}\to C{{H}_{4\left( g \right)}}+C{{O}_{\left( g \right)}}+{{H}_{2\left( g \right)}}.$ The rate constant of decomposition is $4.78\times {{10}^{-3}}mi{{n}^{-1}}.$ Calculate the ratio of initial rate of diffusion to rate of diffusion after $4.5$ hours of initiation of decomposition. Assume the composition of gas present and gas diffused to be the same.

Answer 1

We know that gas particles have a tendency to diffusion because they have kinetic energy. The rate of diffusion is faster at high temperature as particles of gases have greater kinetic energy. Gases move from high concentration regions to low concentration regions. We can calculate the rate of diffusion by the formula, rate of diffusion $=$ amount of gas passing through an area unit of time.

Complete answer:
Rate of diffusion depends on the difference in the difference in the concentration of gradient as diffusion takes place from high concentration to low concentration. As per Graham’s law that states “at constant pressure and temperature, the rate of diffusion takes place from high concentration to low concentration. Complete answer: As per Graham’s law that states “at constant pressure and temperature,
The rate of diffusion $\alpha \dfrac{1}{\sqrt{m}}$ and we have the chemical reaction $C{{H}_{3}}OC{{H}_{3\left( g \right)}}\to C{{H}_{4\left( g \right)}}+C{{O}_{\left( g \right)}}+{{H}_{2\left( g \right)}}.$
The pressure of dimethyl ether after 4.5 hours will be $A={{A}_{0}}{{e}^{-kt}}=0.4{{e}^{-4.7\times {{10}^{-3}}\times 4.5\times 60}}=0.11$
Total pressure after $4.5$ hours will be $0.4+2\left( 0.4-0.11 \right)=0.98mm.$
Thus, average molecular weight after $4.5$ hours will be $\dfrac{\left( 0.4\times 46 \right)+\left( 0.29\times 16 \right)+\left( 0.29\times 28 \right)+\left( 0.29\times 2 \right)}{0.98}=18.86$
Now that we have find the ratio of initial rate of diffusion to rate of diffusion after $4.5$ hours of initiation of decomposition will be
$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}\times \dfrac{{{P}_{1}}}{{{P}_{2}}}=\sqrt{\dfrac{18.86}{46}}\times \dfrac{0.4}{0.98}=0.261$
Therefore, the ratio of initial rate of diffusion to rate of diffusion after $4.5$ hours of initiation of decomposition $0.261$

Additional Information:
Rate of diffusion can be defined as;
- Volume or number of moles of gas diffused.
- Distance travelled by gas per unit time through a tube of uniform cross section.

Note:
We know about the term diffusion which is a t movement of atoms or molecules from high concentration to low concentration .we have approached this problem easily as there is given in the problem that the ratio of rates of diffusion of gases.