InterviewSolution
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InterviewSolution
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Calculate the precent dissociation of ` H_(2) S (g) " if " 0*1 " mole of " H_(2) S " is kept in " 0*4 " litre vessel at 1000 K . For the reaction . " 2H_(2) S (g) hArr 2 H_(2) (g) + S_(2) (g), " the value of " k_(c) " is " 1*0 xx 10^(-6)` |
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Answer» `H_(2) S = (0*1 )/(0*4) "mol" L^(-1) = 0*25 "mol" L^(-1) ` Suppose degree of dissociation of `H_(2) S = alpha ` . Then ` {: (,2 H_(2)S ,hArr,2H_(2),+,S_(2)),(" Intial conc.",0*25 M,,,,),("Conc.at eqm.",0*25 (1-alpha),,0*25 alpha = 0* 125 alpha ,,(0*25)/2alpha=0*125alpha):}` `K_(c) = ([H_(2)]^(2) [S_(2)])/([H_(2)S]^(2)):. 10^(-6) = ((0*25 alpha)^(2) (0*125 alpha))/[ 0*25 (1-alpha)]^(2)` Neglecting `alpha` in comparison to 1 , we get ` 10^(-6) = ((0*25 alpha)^(2) ( 0*125 alpha))/(0*25 )^(2) ` ` 10^(-6)= 0*125 alpha^(3) or alpha^(3) = 8 xx 10^(-6) or alpha = 2 xx 10^(2) = 0*02 ` ` :. % " age dissociation "= 0*02 xx 100 = 2 % ` |
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