Question

A sample 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

Answer 1

Answer: The vapor pressure of naphthalene in the flask is $2.906\times 10^{-4}$ atm.

Explanation:

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

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

• The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

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

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

We are given:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

• The equation used to calculate gibbs free change is of a reaction is:  

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

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

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

We are given:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

• To calculate the $K_1$ (at 25°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C = ?

Putting values in above equation, we get:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

• To calculate the equilibrium constant at 35°C, we use the equation given by Arrhenius, which is:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C = ?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

Putting values in above equation, we get:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

• To calculate the partial pressure of naphthalene at 35°C, we use the expression of $K_p$, which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

Partial pressure of solid is taken as 1 at equilibrium.

So, the value of $K_2$ will be equal to $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Hence, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.