InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temerature increase of `0.7^(@)`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 kJ `"mol"^(-1)`), the 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 identical conditions of Expt.1) where hte temperature rise of `5.6^(@)C` was measured. (Consider heat capacity of all solutions as `4.2 J g^(-1) K^(-1)` and density of all solutions as 1.0 m `mL^(-1)`) Enthalpy of dissociation (in kJ `"mol"^(-1)`) of acetic acid obtained from Expt. 2 isA. `1.0`B. `10.0`C. `24.5`D. `51.4` |
|
Answer» Correct Answer - A Moles of HCl or NaOH neutralized (n) `=100xx1` mmol = 0.1 mole Heat evolved = 0.1 mole `xx` 57.0 kJ `"mol"^(-1)` = 5.7 kJ = 5700 J Heat used to increase temperature of the solution (200 mL ) by `5.7^(@)` `=200 xx 4.2 xx 5.7` (sp. heat capacity of solution `= 4.2 JK^(-1)g^(-1)`) = 4788 J `:.` Heat used to increase the temperature of the calorimeter `=5700-4788 J = 912 J` ms `Delta t = 912` or ms `(5.7) = 912` or ms (calorimeter constant ) `= (912)/(5.7) = 160 J//^(@)C` In Expt. 2, heat evolved by neutralization of `CH_(3)CO OH` with NaOH `{:(=,200xx4.2xx5.6,+,160xx5.6),(,("Absorbed by solution"),,("Absorbed by calorimeter")):}` `= 4704+896 J = 5600 J` `CH_(3)CO OH` present in 100 mL of 2.0 M solution `=100xx2 ` mmol = 0.2 mol NaOH present in 100 mL of 1.0 M NaOH `=100xx1` mmol = 0.1 mol `:. CH_(3)CO OH` neutralized by NaOH = 0.1 mol Thus, heat evolved by neutralisation of 0.1 mol of `CH_(3)CO OH = 5600 J` `:.` Heat used in the dissociation of 0.1 mol of `CH_(3)CO OH =5700-5600J = 100 J ` `:.` Enthalpy of dissociation of acetic acid per mole = 1000 J = 1 kJ |
|