138 g of N_(2)O_(4(g)) is placed in 8.2 L container at 300 K. The equilibrium vapour density of mixture was found to be 30.67. The (R = 0.082 L atm mol^(-1)K^(-1))

138 g of N_(2)O_(4(g)) is placed in 8.2 L container at 300 K. The equi

1.	138 g of N_(2)O_(4(g)) is placed in 8.2 L container at 300 K. The equilibrium vapour density of mixture was found to be 30.67. The (R = 0.082 L atm mol^(-1)K^(-1))
Answer» the total pressure at equilibrium = 4.5 atm the degree of DISSOCIATION of `N_(2)O_(5)=0.25` the total number of moles at equilibrium is 1.5 `K_(p)` of `N_(2)O_(4)iff2NO_(2(g))` will be 6 atm Solution :`N_(2)O_(4)iff2NO_(2)` `{:("INITIAL moles",1,0),("Eq. moles",(1-alpha),2ALPHA):}` Total no. of moles at equilibrium = `1-alpha+2alpha=1+alpha` `alpha=(1)/(n-1)((D-d)/(d))` where, D = Theoretical vapour density `=("MOL. mass")/(2)=(92)/(2)=46` d = Observed vapour density n = No. of moles of products formed from the dissociation of 1 MOLE of `N_(2)O_(4)` `therefore alpha=(1)/(2-1)((46-30.67)/(30.67))=0.4998~=0.5` Thus, total no. of moles at equilibrium = `1+0.5=1.5` Total pressure `=(1.5xx0.082xx300)/(8.2)=4.5` atm So, `K_(p)=(4alpha^(2))/(1-alpha^(2))xx4.5=6` atm