The following reaction is carried out at `101.325 kPa` at `380 K` initial amounts of `CH_(4)` and `O_(2)` as `0.01` and `0.03` mole respectively. All reactants and products are gaseous at `380 K`. After completion of the reaction the flask is colled to `283 K` at which water is completely condensed. Calculate the total pressure and partial pressures of various species at `283 K`.
`2CH_(4) + 3O_(2) rarr 2CO + 4H_(2)O`

Explanation

The reaction is :
`{:(,2CH_(4)(g),+,3O_(2)(g),rarr,2CO(g),+,4H_(2)O(l)),("Initial no. of gas",0.01 mol,,0.03 mol,,0,,0),("No. of moles at 283 K",0.0,,0.015 mol,,0.01 mol,,"condensed"):}`
Initial noo. Of gaseous moles of `380 K = 0.01 + 0.03 = 0.04 mol`
Final no. of gaseous moles at `283 K = 0.015 + 0.010 = 0.025` mol
Final (total ) pressure `= ("Final moles")/("Initial pressure") xx (1.01.325 kPa) = 63.33 kPa`.
`= ((0.025 mol))/((0.040 mol)) xx (101 .325 kPa) = 63.33 kPa`.
Partial pressure of oxygen at `283 K = ("No. of moles of oxygen")/("Total no. of moles") xx "Total pressure"`
`= ((0.015 mol))/((0.025 mol)) xx (63.33 kPA) = 37.998 kPa`.
Partial pressure of carbon monoxide at `283 K`
`("No. of moles of carbon monoxide")/("Total no. of moles") xx "Total pressure"`
`= ((0.010 mol))/((.025 mol)) xx (63.33 kPa) = 25.33 kPa`.