InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
For the reaction : `CO(g) +H_(2)O(g) hArr CO_(2)(g) +H_(2)(g)` the value of `K_(c) = 4.24 at 600 K.` Calculate the equilibrium concentration of `CO_(2), H_(2), CO` and `H_(2)O` at 800 K, if only CO and `H_(2)` are present initially at a concentration of 0.10 M each. |
|
Answer» `{:(,CO(g),+,H_(2)O(g),hArr,CO_(2)(g),+,H_(2)(g)),("Initial molar conc": ,0.1M,,0.1M,,0,,0),("Eqm. molar conc" : ,(0.1-x),, (0.1-x),,xM,,xM):}` `K_(c) =[[CO_(2)][H_(2)]]/[[CO][H_(2)O]]=(x xx x)/((0.1-x)(0.1-x)) = (x^(2))/((0.1-x)^(2))` `(x^(2))/((0.1-x)^(2))=4.24 or (x)/((0.1-x))=(4.24)^(1//2)=2.06` , `x= 2.06(0.1-x) or 3.06x = 0.206 or x = (0.206)/(3.06) = 0.067` Eqm. conc of CO =0.1 -0.067 = 0.033 M Eqm. conc. of `H_(2)O` = 0.1 - 0.067 = 0.033 M Eqm. conc. of `CO_(2)` = 0.067 M Eqm. conc. of `H_(2)O` = 0.067 M` |
|