Question

The pressure of a mixture of acetylene and an excess of hydrogen decreases from 0.100 to 0.042 atm in a vessel of a given volume after the catalyst is introduced, and the temperature is restored to its initial value after the reac- tion reaches completion. What was the mole fraction of acetylene in the original mixture

Answer 1

Step-by-step explanation:

The given reaction equation is as follows.

`C_(2)H_(2)(g) + 2H_(2)(g) \rightarrow C_(2)H_(6)(g)`

Let us assume that the partial pressure of
`C_(2)H_(2)` be 'x' atm. Hence, the partial pressure of
`H_(2)` = (0.1 - x) atm

And, amount of
`H_(2)` left unreacted is as follows.

`(0.1 - x) - 2 * x` = 0.1 - 3x

Now, the pressure of
`C_(2)H_(6)` = partial pressure of reacted
`C_(2)H_(2)` = x

So, the pressure of product will be as follows.

Pressure of the product = partial pressure of
`C_(2)H_(6)` + partial pressure of unreacted
`H_(2)`

0.042 = x + (0.1 - 3x)

2x = 0.058

or, x = 0.029 atm

Therefore, mole fraction of
`C_(2)H_(2)` = partial pressure of
`\frac{C_(2)H_(2)}{\text{Total pressure of the gas mixture}}`

=
`(0.029)/(0.1)`

= 0.29

Thus, we can conclude that the mole fraction of acetylene in the original mixture is 0.29.