Question

calculate the degree of freedom of an equilibrium gas phase containing N2O4 & NO2?
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Answer 1

Answer:

it is 1234.98

Explanation:

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Answer 2

Given:

$N_2O_4$ &$NO_2$

To find:

Degree of freedom

Step by step explanation:

Phase rule;

$F=C-P+2$

F$=$Degree of freedom,

C$=$Number of components,

P$=$Number of phase,

Equilibrium mixture of $N_2O_4_{(g)$ and $NO_2_{(g)$,

$C=1$

Because the equilibrium links the concentrations of the two species,

$P=1$

All gases are miscible,

Then substitute the values in above formula

$F=1-1+2$

$F=2$

Therefore the degree of freedom of an equilibrium gas phase containing $N_2O_2$ & $NO_2$ is  $2$.