InterviewSolution
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InterviewSolution
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One mole of `H_(2)`, two moles of `I_(2)` and three moles of HI are injected in a litre flask. What will be the concentration of `H_(2) ,I_(2)` and HI at equilibrium at `490^(@)C` ? The equiibrium constant for the reaction at `490^(@)` is 45.9 |
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Answer» `{:(,H_(2),+,I_(2)(g),hArr,2 HI(g)),("Intial conc.",1,2,,,3"moles"L^(1)), (" Concs.at eqm",(1-x),,(2-x),,(3+2x) "moe"L^(1)):}` `K = ((3 +2x)^(2))/((1-x)(2-x))=(9+ 4x^(2) + 12x)/(2+ x^(2) - 3 x ) = 45*9 ` ` :. 9 + 4 x^(2) + 12 x = 91 *8 + 45*9x^(2) - 137 * 7 x ` or ` 41*9 x^(2) - 140* 7 x + 82* 8 = 0 ` ` x = (149* 7 pm sqrt(149*7)^(2)- 4 xx 41*9 xx 82 * 8 )/(2 xx 41 * 9) = (149*7 pm sqrt( 22410*09-13877 *28))/(83*8)` ` (149*7 pm 92*4)/(83*8)= 2* 89 and 0*68 ` But `x = 2* 89` is impossible . Hence , ` x= 0* 684` ` :. " Concentrations at equilibrium will be " ` ` [H_(2)] = 1- x=1 - 0*684= 0* 316 "mol" L^(-1)` ` [I_(2) ] = 2 - x = 2 - 0* 684 = 1* 316 "mol" L^(-1)` ` [HI] = 3 +2 x = 3 + 2 xx 0*684 = 4* 368 "mol"L^(-1)` |
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