InterviewSolution
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InterviewSolution
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At `700 K`, hydrogen and bromine react to form hydrogen bromine. The value of equilibrium constant for this reaction is `5xx10^(8)`. Calculate the amount of the `H_(2), Br_(2)` and `HBr` at equilibrium if a mixture of `0.6 mol` of `H_(2)` and `0.2 mol` of `Br_(2)` is heated to `700K`. |
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Answer» Correct Answer - A::B::D `H_(2)+Br_(2) hArr 2HBr` `{:("Initial",0.6,0.2,0),("At equilibrium",(0.6-x)/V,(0.2-x)/V,(2x)/V "mole"),("mol. conc.",,,):}` `K=(((2x)/v)^(2))/(((0.6-x)/v)((0.2-x)/v))` `(4x^(2))/((0.6-x)(0.2-x))=5xx10^(8)` `(x^(2)-0.8x-0.12)xx5xx10^(8)=4x^(2)` Neglecting `4x^(2)` in comparison to `5xx10^(8)x^(2)`, we get `x^(2)-0.8-0.12=0` `x=(0.8 +- sqrt((0.8)^(2)-4xx0.12))/2` `=(0.8 +- 0.693)/(2)` `=0.7465` and `0.0535` `x=0.7465` is impossible hence, `x=0.0535` `[H_(2)]_(eq)=0.6-0.0535=0.5465` mol `[Br_(2)]_(eq)=0.2-0.0535=0.1465` mol `[HBr]_(eq)=2xx0.0535=0.1070` mol |
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