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If `50%` of `CO_(2)` converts to `CO` at the following equilibrium : `1/2C(s) + 1/2 CO_(2)(g) hArr CO(g)` and the equilibrium pressure is `12` atm Calculate `K_(P)`.A. `4`B. `7.5`C. `1`D. `14` |
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Answer» Correct Answer - A `1/2C(s)+1/2CO_(2)(g) hArrCO(g)` `{:("Initial mole",1/2,0),("Eq"^(n),1/2-1/2x,x):}` `n_("total") = 0.25 + 0.5 = 0.75` `K_(P) = (P_(CO))/(P_(CO_(2))^(1//2)) = ((0.5)/(0.75)xx12)/(sqrt((0.25/(0.075)) xx 12)` `K_(p)= (0.5 xx 12)/(0.75 xx 2) = 2/3 xx 6 = 4` |
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