Question

Acetylene reacts with hydrogen in the presence of a catalyst to form ethane according to the following reaction:

${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{\to }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\left(g\right)$

The pressure of a mixture of acetylene and an excess of hydrogen decreases from $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{atm}$to $\mathbf{0}\mathbf{.}\mathbf{042}\mathbf{}\mathbf{atm}$in a vessel of a given volume after the catalyst is introduced, and the temperature is restored to its initial value after the reactionreaches completion. What was the mole fraction of acetylene in the original mixture?

Short answer

The mole fraction of acetylene in the original reaction mixture is$0.193$.

Step-by-step solution

Given information

Initial total pressure$\left({\mathrm{P}}_{\mathrm{i}}\right)=0.100\mathrm{atm}$

Initial total pressure$\left({\mathrm{P}}_{\mathrm{f}}\right)=0.042\mathrm{atm}$

Law of partial pressure

The law of partial pressure states that the partial pressure of the gas is equal to the product of the mole fraction and the total pressure of the gaseous mixture.

${\mathrm{P}}_{\mathrm{i}}={\mathrm{x}}_{\mathrm{i}}{\mathrm{P}}_{\mathrm{total}}$

Partial pressure of acetylene

When the reaction attains completion the total pressure is$0.042\mathrm{atm}$$0.042\mathrm{atm}$

Since only ethane molecules are present in the container we can say that the partial pressure of the ethane molecules in the reaction mixture is$0.042\mathrm{atm}$

Therefore, the pressure exerted by the hydrogen and acetylene in the reaction mixture$=0.100\mathrm{atm}-0.0042\mathrm{atm}=0.058\mathrm{atm}$

Total number of moles when only acetylene and hydrogen is present$=3$

Therefore,

$$\begin{gathered} {\mathrm{x}}_{{\mathrm{H}}_2}=\frac{2}{3} \\ {\mathrm{x}}_{{\mathrm{C}}_2{\mathrm{H}}_2}=\frac{1}{3} \end{gathered}$$

Therefore the partial pressure of acetylene is:

role="math" localid="1662101008708" $\begin{array}{rcl}{\left({\mathrm{P}}_{\mathrm{i}}\right)}_{{\mathrm{C}}_2{\mathrm{H}}_2}& =& {\mathrm{x}}_{{\mathrm{C}}_2{\mathrm{H}}_2}\mathrm{P}\\ & =& \frac{1}{3}\times 0.058\mathrm{atm}\\ & =& 0.0193\mathrm{atm}\end{array}$

Mole fraction of acetylene

From Dalton’s law of partial pressure, we can calculate the mole fraction of acetylene in the original reaction mixture.

$\begin{array}{rcl}{\left({\mathrm{P}}_{\mathrm{i}}\right)}_{{\mathrm{C}}_2{\mathrm{H}}_2}& =& {\mathrm{x}}_{{\mathrm{C}}_2{\mathrm{H}}_2}\mathrm{P}\\ {\mathrm{x}}_{{\mathrm{C}}_2{\mathrm{H}}_2}& =& \frac{{\left({\mathrm{P}}_{\mathrm{i}}\right)}_{{\mathrm{C}}_2{\mathrm{H}}_2}}{P}\\ & =& \frac{0.0193}{0.1}\\ & =& 0.193\end{array}$