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InterviewSolution
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At `450^(@)C` the equilibrium constant `K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` was found to be `1.6xx10^(-5)` at a pressure of `200` atm. If `N_(2)` and `H_(2)` are taken in `1:3` ratio. What is `%` of `NH_(3)` formed at this temperature? |
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Answer» Correct Answer - A::B::C `{:(,N_(2),+,3H_(2),hArr,2NH_(3)),("moles before eq",1,,3,,0),("moles after eq",(1-x),,3(1-x),,2x):}` `:. K_(p)=((2x^(2))(4-2x)^(2))/((1-x)[3(1-x)]^(3)xxP^(2))` `:. K_(p)=((2x^(2))(4-2x)^(2))/((1-x)[3(1-x)]^(3)xxP^(2))` `K_(p)=(16xx x^(2) xx(2-x)^(2))/(27(1-x)^(4)xxP^(2))` or, `1.6xx10^(-5)=(16x^(2)xx(2-x)^(2))/(27(1-x)^(4)xx(200)^(2))` `(x^(2)(2-x)^(2))/((1-x)^(4))=(1.6xx10^(-5)xx27xx(200)^(2))/(16)` `=(16xx10^(-6)xx27xx(200)^(2))/16` `:. (x(2-x))/((1-x)^(2))=200xx10^(-3)xxsqrt(27)=1.039` `:. x=0.301` `:.` mole of `NH_(3)` formed `=2xx0.301=0.602` Total "moles" at equilibrium `=4-2xx0.301=3.398` `:. %` of `NH_(3)` at equilibrium `=0.602/3.398xx100=17.76%` |
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