InterviewSolution
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InterviewSolution
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When 100 mL of 1.0 M HCl was mixed 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase `5.7^(@)C` was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant `(-57.0 kJmol^(-1))`, this experiment could be used to measure the calorimeter constant. In a second experiment (Expt.2) 100 mL of 2.0 M acetic acid `(K_(a)=2.0xx10^(-5))` was mixed with 100 mL of 1.0 M NaOH (under indentical conditions to Expt. 1) where a temperature rise of `5.6^(@)C` was measured (Consider heat capacity of all solutions as `4.2 Jg^(-1)K^(-1)` and density of al solutions as 1.0 gm`L^(-1)`) Enthalpy of dissociation (in `KJmol^(-1)`) of acetic acid obtained from the Expt.2 isA. `1.0`B. `10.0`C. `24.5`D. `51.4` |
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Answer» Correct Answer - A Let the heat capacity of insulated beaker be C Mass of aqueous content in expt. `1=(100+100)xx1=200g` implies Total heat capacity = `(C+200xx4.2)J//K` Moles of acid, base neutralised in expt. `1=0.1xx1=0.1` implies Heat released in expt. `1=0.1xx57=5.7 KJ` `implies5.7xx1000=(C+200xx4.2)xxDeltaT` `5.7xx1000=(C+200xx4.2)xx5.7` `implies (C+200xx4.2)=1000` In second experiment, `n_(CH_(3)COOH)=0.2, n_(NaOH)=0.1` Total mass of aqueous content = 200 g implies Total heat capacity = (C+200`xx`4.2)=1000 implies Heat released = `1000xx5.6=5600 J` Overall, only 0.1 mol of `CH_(3)COOH` undergo neutralization `implies DeltaH_("neutralization")" of "CH_(3)COOH=(-5600)/(0.1)` `=-56000 J//mol=-56 KJ//mol` `impliesDeltaH_("ionization")" of "CH_(3)COOH=57-56=1KJ//mol` |
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