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1.

Calculate the mass of mercury which can be liberated from `HgO` at `25^(@)C` by the treatment of excess HgO with 41.84 kJ of heat at : (a) constant pressure (b) constant volume Given : `Delta H_(f)^(@)(HgO, s) = -90.8 kJ mol^(-1)` & `M(Hg) = 200.6 g mol^(-1)`.

Answer» Correct Answer - (a) 92.435 g (b) 93.715 g
(a) `HgO rarr Hg+1/2 O_(2) " "DeltaH=90.8`
For `41.84 kJ` heat, mass of `Hg=200.6/90.8 xx41.84=-92.435 g`
(b) Calculate at constant volume
Discussion
2.

The enthalpies of neutralization of a weak acid HA & a weak acid HB by `NaOH` are -6900 Cal/equivalent & -2900 Cal/equivalent repsectively. When one equivalent of NaOH is added to a solution containing one equivalent of HB, the enthalpy change was -3900 Calories. In what ratio is the base distribute between HA & HB ?

Answer» Correct Answer - `1 : 3`
1 gram eq. of `NaOH` reacts with `HA` and `HB`
a gram eq. of `NaOH` reacts with `HA`
b gram eq. of `NaOH` react with `HB`
`a+b=1`
`6900 a+2900 b=3900`
`{:(a=0.25,,a:b),(b=0.75,,1:3):}`
Discussion
3.

Statement-I : The enthalpy of neutralization of the reaction between `HCl` and `NaOH` is `-13.7` kCal/mol. If the enthalpy of neutralization of oxalic acid `(H_(2)C_(2)O_(4))` by a strong base is `-25.4` kCal/mol, then the enthalpy change `(|Delta_(r)H|)` of the process `H_(2)C_(2)O_(4) rarr 2H^(+)+C_(2)O_(4)^(2-)` is `11.7` kCal/mol. Statement-II : `H_(2)C_(2)O_(4)` is a weak acid.A. If both Statement-I & Statement-II are True & the Statement-II is a correct explanation of the Statement-IB. If both Statement-I & Statement-II are True but Statement-II is not a correct explanation of the Statement-IC. If statement-I is True but the Statement-II is False.D. If Statement-I is False but the Satement-II is True

Answer» Correct Answer - D
`H^(+)+OH^(-) =13.7`
`2H^(+)+2OH^(-) rarr H_(2)O=27.4`
But energy released `=25.4`
`:. H_(1)C_(2)O_(4) rarr 2H^(+) +C_(2)O_(4)^(-2)" "DeltaH=2" kCal/mol."`
Discussion
4.

`NH_(3)(g) + 3Cl_(2)(g) rarr NCl_(3)(g) + 3HCl(g), " "DeltaH_(1)` `N_(2)(g)+3H_(2)(g)rarr 2NH_(3)(g), " "DeltaH_(2)` `H_(2)(g)+ Cl_(2)(g) rarr 2HCl(g) , " " DeltaH_(3)` The heat of formation of `NCl_(3)` in the terms of `DeltaH_(1), DeltaH_(2) "and" DeltaH_(3)` isA. `Delta H_(f) = - Delta H_(1) + (Delta H_(2))/(2) - (3)/(2) Delta H_(3)`B. `Delta H_(f) = Delta H_(1) + (Delta H_(2))/(2) - (3)/(2) Delta H_(3)`C. `Delta H_(f) = Delta H_(1) - (Delta H_(2))/(2) - (3)/(2) Delta H_(3)`D. None

Answer» Correct Answer - A
Discussion