InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A `500mL` of an equilibrium mixture of gaseous `N_(2)O_(4)` and `NO_(2)` at `25^(@)C` and `753mm` of `Hg` was allowed to react with enough water to make `250mL` of solution at `25^(@)C`. Assume that all the dissolved `N_(2)O_(4)` is converted to `NO_(2)` which disproportionates in water yielding a solution of nitrous acid and nitric acid. aAsume further that disproportionation reaction goes to completion and that none of the nitrous acid disproportionates. The equilibrium constant `(K_(p))` for `N_(2)O_(4)(g)hArr 2NO_(2)(g) 0.113` at `25^(@)C`. `K_(a)` for `HNO_(2)` is `4.5 xx 10^(-4) at 25^(@)C`. a. Write balanced equation for disproportionation. b. What is the molar concentration of `NO_(2)` and `pH` of the solution? c. What is osmotic pressure of solution? d. How many grams of lime `(CaO)` would be required to neutralise the solution? |
|
Answer» a. `N_(2)O_(4) hAtt 2NO_(2)` Let `a mol` of `N_(2)O_(4)` and `bmol` of `NO_(2)` were present in equilibrium mixture. `:. (a+b) = (PV)/(RT) = (653 xx 0.5)/(760 xx 0.0821 xx 298)` `K_(p) = ((n_(NO_(2)))^(2))/((n_(N_(2)O_(4))))xx[(P)/((n_(NO_(2))+n_(N_(2)O_(4))))] ..(i)` `:. 0.113 = (b^(2))/(a) xx [(753)/(760 xx (a+b))]` `:. (b^(2))/(a(a+b)) = 0.114 ...(ii)` From equations, (i) and (ii), we get `(b^(2))/(a) = 0.114 xx 0.020 = 2.3 xx 10^(-3)` `b^(2) = 2.3 xx 10^(-3) a = 2.3 xx 10^(-3) xx (0.02 -b)` `b^(2) + 0.0023 b - 4.6 xx 10^(-5) = 0` `:. b = - (0.0023+-sqrt((0.0023)^(2)+4xx4.6xx10^(-5)xx1))/(2xx1)` `=(-0.0023+-sqrt(1.90 xx 10^(-4)))/(2)` `b = 5.73 xx 10^(-3)` `:.` By equaiton (i) `a = 0.014` b. `2NO_(2)(aq) + 2H_(2) (l) rarr NHO_(2)(aq) + H_(3)O^(o+)(aq) + NO_(3)^(Theta) (aq)` Total `NO_(2)` moles `=underset((From NO_(2)))(5.73xx10^(-3))+2xx` moles of `N_(2)O_(4)` `= 5.73 xx 10^(-3) + 2 xx 0.014` `= 0.0337` `{:(2NO_(2)(aq)+,2H_(2)O(l)rarr,HNO_(2)+,H_(3)O^(o+)(aq)+,NO_(3)^(o+)(aq)),(0.0337,,0,0,0),(0,,(0.0337)/(2),(0.0337)/(2),(0.0337)/(2)):}` `:. [HNO_(2)] = (0.337 xx 1000)/(2xx250) = 0.0674 M`, `[H^(o+)] = (0.337xx1000)/(2xx250) = 0.0674M`, Due to common ion effect `(H_(3)O^(o+)` furnished by `HNO_(3))`, the dissociation of `HNO_(2)` is suppressed. `K_(a) = ([H^(o+)][NO_(2)^(Theta)])/([HNO_(2)]) = (0.067 xx [NO_(2)^(Theta)])/(0.0674) = 4.5 xx 10^(-4)` `:. [NO_(2)^(Theta)] = 4.5 xx 10^(-4)` Also `pH =- log [H^(o+)] =- log 0.0674 = 1.17` c. `pi = iERT` (where `i =` Number of particles present in solution or it is Vant Hoff factor) `= 0.0674 xx 0.0821 xx 298 xx3` `(2NO_(2)` furnishes three particles) `= 4.95 atm` d. `Eq` of `CaO` required `= Eq of HNO_(2) + Eq` of `HNO_(3)` `= (0.0337)/(2) xx 1 + (0.0337)/(2) xx1 = 0.0337` `:.` Mole of `CaO` required `= (0.0337)/(2)or (W)/(56) = (0.0337)/(2)` `W_(CaO) = 0.924g` |
|