InterviewSolution
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InterviewSolution
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A gaseous mixture of propane, acetylene and CO_(2) is burnt in excess of air. Total 4800 KJ heat is evolved . The volume of CO_(2) formed during combustion is 224 liters at STP. The total evolved heat is used to perform two separate process: (i) Vapourising 87.5% of water (liquid) obtained in the process of burning the original mixture. (ii) Forming 3808 litres ethylenemeasured at STP from its elements. DeltaH_(H-H) = 435 KJ//mol "" DeltaH_(C-H) = 416 KJ//mol "" DeltaH_(C-C) = 347 KJ//mol DeltaH_(C-C) = 615 KJ//mol, "" DeltaH_(C-C) = 812 KJ//mol"" DeltaH_("sublimation") "of " (C,s) = 718 KJ//mol DeltaH_(f)^(@)(C_(2)g) = -394 KJ//mol""DeltaH_(f)^(@)(H_(2)O, l) = -286 KJ//mol. "" DeltaH_(f)^(@)(H_(2)O,g) = -247 KJ//mol.Total moles of hydrocarbon gases taken in the initial mixture |
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Answer» `3` & moles of `CO_(2) =Z.` Calculation of `DeltaH_(f)^(@) "of" C_(3)H_(8) (g)` `""3C(s) + 4 H_(2) (g) rarr C_(3)H_(8)(g)` `DeltaH_(f)^(@) "of" C_(3)H_(8)(g) =[3(718)+ 4(435)] - [2(347) + 8(416)]` `= 3894- 4022 =-128KJ//mol` Calculation of `DeltaH_(f)^(@) "of" C_(2)H_(2) (g)` `""2C(s) + H_(2) (g) rarr C_(2)H_(2)(g)` ` DeltaH_(f)^(@) C_(2)H_(2)(g)=[2(718) + (435)] - [(812) + 2(416)]` `=(1436 + 435) - [1644]` `=227 KJ//mol`. Calculation of `DeltaH_(f)^(@) "of" C_(2)H_(4) (g)` `2C(s) + 2H_(2) (g) rarr C_(2)H_(4)(g)` ` DeltaH_(f)^(@)"of" C_(2)H_(4)(g) = [2(718) + 2(435)] - [615 + 4(416)]` `= 2306- 2279 = 27 KJ//mol`. Calculation of `DeltaH_("Comb")^(@) "of" C_(3)H_(8)(g)` `""C_(3)H_(8)(g) + O_(2)(g) rarr 3CO_(2)(g) + 4H_(2)O(l)` `"" DeltaH_("Comb")^(@) "of" C_(3)H_(8) = [3 DeltaH_(f^(@))(CO_(2)) + 4Delta_(F^(@)) (H_(2)O,l)] - DeltaH_(F^(@)) (C_(3)H_(8),g)` `"" = [3(-394)+ 4(-286)] - (-128) =-2198 KJ//mol`. Calculation of `DeltaH_("Comb")^(@) "of" 2C_(2)H_(2)(g)` `C_(2)H_(2)(g) + 2.50_(2)(g) rarr 2CO_(2)(g) + H_(2)O(l)` `DeltaH_("Comb")^(@) " of " C_(2)H_(2) = [2DeltaH_(f^(@)) (CO_(2)) + DeltaH_(F^(@))(H_(2)O, l)] - Delta_(F^(@))(C_(2)H_(2))` `=[2(-394)+ (-286)] - 227` `=-1301 KJ//mol.` Now total heat released during combustion `"" 2198 x + 1301 y = 4800""(i)` `""` Combustion `=3x + 2y + z =10 ""(ii)` Total moles of `H_(2)O(l) "FORMED" = 4x + y.` moles of `C_(2)H_(4)(g)` to be prepared `=(3808)/(22.4) = 170.` Total heat absorbed during evaporation of water formation of `170` moles `C_(2)H_(4)= 4800.` `"" [(4x + y ) xx 0.875 xx 40] + (170 xx 27) = 4800` `"" 4x + y =6"(iii)` on solvent `(i), (ii)` and `(iii)` we get `x=1, y=2 "and" z=3` |
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