Question

For the reaction N2O4 2NO2,if the average molecular mass of the mixture after time t of the reaction is 69.find the percentage dissassociation of N2O5

Answer 1

Explanation:

Answer: Percentage dissociation of N_2O_5N

2

O

5

for the given reaction is 15%.

Explanation: We are given a chemical equation:

N_2O_4\rightleftharpoons 2NO_2N

2

O

4

⇌2NO

2

at t = 0 1 0

at t=t_{eq}t=t

eq

1-\alpha1−α 2\alpha2α

Normal molecular mass of N_2O_5N

2

O

5

= 92 g/mol

Observed molecular mass of N_2O_5N

2

O

5

= 80 g/mol

Van't Hoff factor for dissociation is given by:

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

Observed Molecular mass

Normal Molecular mass

For Association, i < 1

For dissociation, i > 1

Putting values in above equation, we get:

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

80

92

=1.15

Expression for Van't Hoff equation is given by:

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

Calculated colligative property

Observed colligative property

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

1

1−α+2α

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

1

1+α

where,

i = Van't Hoff factor

\alphaα = Degree of dissociation or association.

\begin{lgathered}1.15=1-\alpha \\\alpha = 0.15\end{lgathered}

1.15=1−α

α=0.15

Percentage of dissociation for N_2O_4N

2

O

4

is given by:

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

2

O

4

=α×100=0.15×100=15%