A mixture in which the mole ratio of `H_(2)` and `O_(2)` is `2:1` is used to prepare water by the reaction.
`2H_(2(g))+O_(2(g))rarr2H_(2)O_((g))`
The total pressure in the container is `0.8 atm` at `20^(@)C` before the reaction. Determine the final pressure at `120^(@)C` after reaction assuming `80%` yield of water.

Explanation

`2H_(2)(g)+O_(2)(g)to2H_(2)O(g)`
`{:(i nitial mol,2a,a,0),(Fi nal mol,(2a-2x),(a-x),2x):}`
Since `80%` of water is obtained, so.
`2x=2axx(80)/(100)=1.6a`
`thereforex=0.8a`
Thus, after the reaction,
" mol of "`H_(2)` left`=(2a-2xx0.8a)=0.4a`
" mol of "`O_(2)` left `=(a-0.8a)=0.2a`
" mol of "`H_(2)O` formed`=2xx0.8a=1.6a`
Total moles at `373K` in gaseous phase
`=0.49+0.2a=1.6a`
`=0.49+0.2a+1.6a=2.2a`
`thereforePV=nRT`
`V=(nRT)/(P)=(3axxRxx293)/(0.8)`
Since the volume of container remains contant, therefore, the pressure after the reaction is,
`P=(nRT)/(P)=(2.2cancel(a)xxcancel(R)xx393)/(3cancel(a)xxcancel(R)xx((293)/(0.8))`
`=(2.2xx393xx0.8)/(3xx293)`
`=0.787 atm`