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Consider the following set of reactions: `CHCl_(2)COOH+NaOH to CHCl_(2)COONa+H_(2)O " " DeltaH_(1)=-12830 Cal` `HCl+NaOH to NaCl to NaCl+H_(2)O " " DeltaH_(2)=-13680 Cal`. `NH_(4)OH+HCl to NH_(4)Cl+H_(2)O " " DeltaH_(3)=-12270 Cal`. Select the correct option(s) :A. Enthalpy of neutralization of `CHCl_(2)COOH` by `NH_(4)OH` is `-11420 Cal.`B. Enthalpy of dissociation of `NH_(4)OH=1410 Cal`C. Enthalpy of dissociatiion of `NH_(4)OH=1410 Cal `.D. Enthalpy change for the reaction `H_(2)O to H^(+)+OH^(-) is -13680 Cal ` |
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Answer» (A),( C) `DeltaH=DeltaH_(1)+DeltaH_(3)-DeltaH_(2)` `=-(12830+12270-13680)` `=-11420 Cal` (B) `DeltaH_(2)-DeltaH_(1)=13680-12830` `=850 Cal` (C) `DeltaH_(2)-Delta_(3)=13680-12270` `=1410 Cal` (D) `H_(2)O to H^(+)+OH^(-)` `=+13680K` |
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