Question

At equilibrium N2o4=2no2 the observed molecular weight of n2o4 is 80 g at 350 k the %of dissociation of n2o4(g) at350 k is

Answer 1

Answer: Percentage dissociation of $N_2O_5$ for the given reaction is 15%.

Explanation: We are given a chemical equation:

               $N_2O_4\rightleftharpoons 2NO_2$

at t = 0           1           0

at $t=t_{eq}$    $1-\alpha$        $2\alpha$

Normal molecular mass of $N_2O_5$ = 92 g/mol

Observed molecular mass of $N_2O_5$ = 80 g/mol

Van't Hoff factor for dissociation is given by:

$i=\frac{\text{Normal Molecular mass}}{\text{Observed Molecular mass}}$

For Association, i < 1

For dissociation, i > 1

Putting values in above equation, we get:

$i=\frac{92}{80}=1.15$

Expression for Van't Hoff equation is given by:

$i=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$

$i=\frac{1-\alpha +2\alpha }{1}$

$i=\frac{1+\alpha }{1}$

where,

i = Van't Hoff factor

$\alpha$ = Degree of dissociation or association.

$1.15=1-\alpha \\\alpha = 0.15$

Percentage of dissociation for $N_2O_4$ is given by:

$\text{Percentage dissociation of }N_2O_4=\alpha \times 100=0.15\times 100=15\%$

Answer 2

Answer:

15%

here's the answer and hope this helps you.