Question

When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay: HCOOH(g) →CO2(g) + H2 (g) The rate of reaction is monitored by measuring the total pressure in the reaction container. Time (s) . . . P (torr) 0 . . . . . . . . . 220 50 . . . . . . . . 324 100 . . . . . . . 379 150 . . . . . . . 408 200 . . . . . . . 423 250 . . . . . . . 431 300 . . . . . . . 435 At the start of the reaction (time = 0), only formic acid is present. What is the formic acid pressure (in torr) when the total pressure is 364? Hint: use Dalton's law of partial pressure and the reaction stoichiometry.

Answer 1

`p_(HCOOH)=76torr`

Step-by-step explanation:

Hello.

In this case, since the total pressure at any point of the experiment is written via the Dalton's law:

`p=p_(HCOOH)+p_(CO_2)+p_(H_2)`

We can also write the partial pressure of formic acid in terms of its initial pressure (220 torr) and the change
`x` as the time goes by:

`p_(HCOOH)=220-x`

Thus, based on the stoichiometry, since it is a first-order decay and all the stoichiometric coefficients are 1, we can infer that the decrease in the partial pressure of formic acid equals the increase in the partial pressure of both carbon dioxide and hydrogen, therefore we can write:

`p_(CO_2)=p_(H_2)=x`

In such a way, we write the Dalton's law as shown below:

`p=220torr-x+x+x\\\\p=220torr+x`

Thus, at the point in which the total pressure is 364 torr, the change is:

`x=364torr-220torr=144torr`

It means, that the partial pressure of formic acid is:

`p_(HCOOH)=220torr-144torr\\\\p_(HCOOH)=76torr`

Best regards.