Question

calculate the molar entropy change when (a) water and (b) benzene are evaporated at their boiling points at a pressure of 1 atm. what are the entropy changes in (i) the system, (ii) the surroundings, and (iii) the universe, in each case?

Answer #1

At the boiling temperature, temperature and the pressure is constant. The only heat of transformation is the latent heat of vapourization. latent heat of vapourization of water = 40.66 kJ/mol

S = q/T = n *dH/T = 1 *40.66 kJ/mol/373.15 = 0.12 kJ/K

S (sys) = 0.12 kJ/K. No heat is transferred to the environment. So, S(surr) =0

S (univ )= 0.12 kJ/mol + 0 = 0.12kJ/mol

-----------------------------------------------------------

Latent heat of vaporization of benzene = 30.8 kJ/mol and boiling point = 80.1 oC = 353.1 K

S (sys) = q/T = 1mol *30.8 kJ/mol/ 353.1 K =0.087 kJ/K

S (surr) = 0 as no heat is transferred to the surrounding.

S (univ) = S (sys)+ S (surr) = 0.087 kJ/mol

What is the change in the entropy when 1g of benzene, boils
reversibly at its boiling point of 80.1C and a constant pressure of
1atm? The heat of vaporization of benzene is 395 J/g

The standard molar entropy of benzene is 173.3 J/K-mol.
Calculate the change in its standard molar Gibbs energy when
benzene is heated from 25C to 45C.

For each case determine the change in entropy for
the surroundings and the universe, when 1 mole of a
diatomic gas expands from its initial volume of 5 L and 31.5 ◦C
to:
CASE (a) a final volume of 25 L reversibly and
isothermically.
CASE (b) irreversibly and isothermically against an external
pressure of 1 atm.
CASE (c) a final volume of 25 L reversibly and adiabatically.
Yes, it is zero, but you have to demonstrate it.
CASE (d) irreversibly...

For each process below, determine the change in entropy of the
system. Indicate whether the changes in the entropy of the
surroundings and of the universe will be negative or positive (5
pts each).
(a) One mole of ammonia gas decomposes to
nitrogen and hydrogen gases.
(b) 12 g of liquid water freezes.

Calculate the change in entropy of the surroundings when 1.0 mol
of water vaporizes as 1000C. The heat of vaporization of
water is 40.6 kj/mol

Calculate the change in entropy of the system when 15.0g of ice
at -12.0C is converted to water vapor at 105C at a constant
pressure of 1bar. The constant pressure molar heat capacity ofH2O
(s) and H2O (l) is 75.291J/mol K and that of H2O (g) is 33.58 J/K
mol

One mole of either carbon monoxide or benzene are completely
combusted with oxygen at constant temperature and pressure (298 K
and 1 atm) to generate CO2 and H2O. Assume all substances are ideal
gases for calculating volume changes.
a. Write out balanced combustion reactions for each
reaction.
b. Calculate the change in entropy for the system for each
reaction, using the table, below.
c. Use the enthalpies of formation to calculate the heat lost or
gained during this reaction.
d....

Trouton’s rule states that the entropy of boiling at the normal
point is 85 J/mol * K. (a) Does the data from Example 3.2 support
Trouton’s rule? (b) H2O has a heat of vaporization of 40.7 kJ/mol.
Does the Delta vapS for H2O at its normal boiling point support
Trouton’s rule? Can you explain any deviation? (c) Predict the
boiling point of cyclohexane, C6H12, if its Delta vapH is 30.1
kJ/mol. Compare your answer to the measured normal boiling point...

If the temperature of the surroundings is 5.45 °C, calculate the
entropy change (in J/K) for the system (ΔSsys),
surroundings (ΔSsur) and universe
(ΔSuniverse) when 33.6 g of gaseous diethyl ether
(C4H10O) condenses. Report your answers to
two decimal places.
Tfus(°C)
-116.30
Tvap(°C)
34.50
ΔH°fus (kJ/mol)
7.27
ΔH°vap (kJ/mol)
26.52
i need these :
1- ΔSsys =
2- ΔSsur =
3 -ΔSuniverse =

Calculate the change in the
entropy of 2.0 kg of water when it is converted from liquid to
vapor.
Heat of fusion for water is
334*10^3 (J/kg)
Boiling pint of water is 373.15 K
Heat of vaporization for water is 2256*10^3 (J/kg)

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