Asample of solid naphthalene is introduced into an evacuated flask. use the data below to calculate the equilibrium vapor pressure of naphthalene (c10h8) in the flask at 35°c. hf (25c) gf (25c) c10h8(s) 78.5 kj/mol 201.6 kj/mol c10h8(g) 150.6 kj/mol 224.1 kj/mol

The vapor pressure of naphthalene in the flask is atm.

Explanation:

For the conversion of naphthalene solid to naphthalene gas, the equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

where,

= Gibbs free energy = 22.5 kJ/mol = 22500 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant =

T = temperature =

= equilibrium constant at 25°C = ?

Putting values in above equation, we get:

where,

= Equilibrium constant at 35°C = ?

= Equilibrium constant at 25°C =

= Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant =

= Initial temperature =

= Final temperature =

Putting values in above equation, we get:

Partial pressure of solid is taken as 1 at equilibrium.

So, the value of will be equal to

Hence, the partial pressure of naphthalene at 35°C is atm.

the answer to your question is c