InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
One mole of `H_(2)O` and one mole of `CO` are taken in a `10 litre` vessel and heated to `725 K`. At equilibrium, `40 per cent` of water (by mass) reacts with carbon monoxide according to the equation, `H_(2)O_((g))+CO_((g))hArrH_(2(g))+CO_(2(g))` Calculate the equilibrium constant for the reaction. |
|
Answer» Correct Answer - `0.44` The given reaction is : `{:(,H_(2)O_((g)),+,CO_((g)),hArr,H_(2(g)),+,CO_(2(g))),("Initial conc.",1/10M,,1/10M,,0,,0),("At equilibrium",(1-0.4)/(10)M,,(1-0.4)/(10)M,,(0.4)/(10)M,,(0.4)/(10)M),(,=0.06M,,=0.06M,,=0.04M,,=0.04M):}` Therefore, the equilibrium constant for the reaction, `K_(c) = ([H_(2)][CO_(2)])/([H_(2)O][CO])` `= (0.4 xx 0.04)/(0.06 xx 0.06)` `= 0.444` (approximately) |
|