Question

N2O4=2NO2 the observed molecular weight of N2O4 IS 80g/mol at 350K the percentage of dissociation of N2O4 at 350 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 $t=t_{eq}$     $1-\alpha$         $2\alpha$

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

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

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

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:

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

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

Answer 2

Answer:

15%

here's the answer and hope this helps you.