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

when atmosphere interacts with biosphere they focuses on on the sources and sinks of green house gases (ghgs) air pollutants, particular matters, water and energy between the earth's surface and the atmosphere. the biosphere affects the climate. the atmosphere and biosphere are in constant interaction with each other, and all living things rely on gases in the atmosphere for life. organisms such as animals take in oxygen gas form the atmosphere to use in every cell of their body for smooth functioning of daily routine exercises.

What happens when the atmosphere interacts with biosphere

Answer from: Quest

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

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Asample of solid naphthalene is introduced into an evacuated flask. use the data below to calculate...