Respuesta :
Answer: The vapor pressure of naphthalene in the flask is [tex]2.906\times 10^{-4}[/tex] atm.
Explanation:
For the conversion of naphthalene solid to naphthalene gas, the equilibrium reaction follows:
[tex]C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)[/tex]
- The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\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))})[/tex]
We are given:
[tex]\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol[/tex]
- The equation used to calculate gibbs free change is of a reaction is:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)][/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\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))})[/tex]
We are given:
[tex]\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol[/tex]
- To calculate the [tex]K_1[/tex] (at 25°C) for given value of Gibbs free energy, we use the relation:
[tex]\Delta G^o=-RT\ln K_1[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = [tex]25^oC=[273+25]K=298K[/tex]
[tex]K_1[/tex] = equilibrium constant at 25°C = ?
Putting values in above equation, we get:
[tex]22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}[/tex]
- To calculate the equilibrium constant at 35°C, we use the equation given by Arrhenius, which is:
[tex]\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]K_2[/tex] = Equilibrium constant at 35°C = ?
[tex]K_1[/tex] = Equilibrium constant at 25°C = [tex]1.14\times 10^{-4}[/tex]
[tex]\Delta H[/tex] = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J
R = Gas constant = [tex]8.314J/K mol[/tex]
[tex]T_1[/tex] = Initial temperature = [tex]25^oC=[273+25]K=298K[/tex]
[tex]T_2[/tex] = Final temperature = [tex]35^oC=[273+35]K=308K[/tex]
Putting values in above equation, we get:
[tex]\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}[/tex]
- To calculate the partial pressure of naphthalene at 35°C, we use the expression of [tex]K_p[/tex], which is:
[tex]K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)[/tex]
Partial pressure of solid is taken as 1 at equilibrium.
So, the value of [tex]K_2[/tex] will be equal to [tex]K_p[/tex]
[tex]p_{C_{10}H_8}=2.906\times 10^{-4}[/tex]
Hence, the partial pressure of naphthalene at 35°C is [tex]2.906\times 10^{-4}[/tex] atm.
