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The equilibrium constant `K_(P)` for the reaction `N_(2)O_(4)(g) hArr 2NO_(2)(g)` is `4.5 atm`. What would be the average molar mass (in `g//mol`) of an equilibrium mixture at a total pressure of `2 atm` of `N_(2)O_(4)` and `NO_(2)` formed by the dissociation of pure `N_(2)O_(4)` ?A. 69B. 57.5C. 80.5D. 85.5 |
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Answer» Correct Answer - B `underset(P-Palpha)underset(P)(N_(2))O_(4)(g)hArrunderset(2Palpha)(2NO_(2)(g))` is 4.5 `P-Palpha+2Palpha=2` `P+Palpha=2` `4.5=(4P^(2)alpha^(2))/(P(1-alpha))` `4.5=(4Palpha^(2))/(1-alpha)` `4.5(1-alpha)=4Palpha^(2)` `P=(4.5(1-alpha))/(4alpha^(2))` `P(1+alpha)=2` `(4.5(1-alpha)(1+alpha))/(4alpha^(2))=2` `4.5(1-alpha^(2))=8alpha^(2)` `4.5-4.5alpha^(2)=8alpha^(2)` `4.5=12.5alpha^(2)` `alpha=sqrt((4.5)/(12.5))` `alpha=0.6` `alpha=(M-EMM)/(EMM(n-1))` `0.6EMM=92-EMM` `1.6EMM=92` `EMM=(92)/(1.6)` `=57.5` |
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