Chemistry, 01.08.2019 01:10 aj567

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

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Answer from: jfjhbc

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.

Answer from: Quest

Heat is involved in chemical change in two ways; chemical changes either result in heat being added to or removed from an object, and chemical changes require a certain amount of heat before they can take place.

Answer from: Quest

Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o