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.

Options

A. 1.787 atm
B. 0.878 atm
C. 0.787 atm
D. 1.878 atm

Answer

C

Explanation

`{:(,2H_(2(g)),+,O_(2(g)),rarr,2H_(2)O_((g))),("Initial mole"," 2a",," a",," 0"),("Final mole",(2a-2x),,(a-x),," 2x"):}`
Given `2x=2axx(80)/(100)=1.6 a`
`therefore x = 0.8a`
Thus, after the reaction `H_(2)` left `=2a-1.6a=0.4a` mole
`O_(2)` left = 0.2a mole
`H_(2)O` formed = 1.6 a mole
`therefore` Total mole at `120 .^(@)C` in gaseous phase
`=0.4a+0.2a+1.6a=2.2a` ltbRgt Now, given at initial conditions P = 0.8 atm, T = 293 K
`PxxV=nRT`
`0.8V=3a xx R xx293`
`therefore V=(3a xx R xx 293)/(0.8)`
The volume of container remains constant.
Using, `PxxV=nRT`
`therefore P xx (3a xx R xx 293)/(0.8)=2.2a xx R xx 393 " " [T=393 K]`
`P=(393xx0.8xx2.2)/(3xx292)=0.787` atm