The molar enthalpy of vaporization of hexane (CH14) is 28.9 kl/mol, and its normal boiling point is 68.73 C What is the vapor pressure

The molar enthalpy of vaporization of hexane (CH14) is 28.9 kl/mol, and its normal boiling point is 68.73 C What is the vapor pressure of hexane at 25°C?

Answers

(a) increase delta S is positive

(b) Value of delta S is 89.21 J/molK

Explanation:

Boiling point is defined as the temperature where atmospheric pressure becomes equal to the vapor pressure of a liquid.  

Entropy is the degree of randomness present within the molecules of a substance.

Further Explanation:

At constant temperature and pressure, the amount of heat energy necessary to change one mole of a substance into gaseous phase from liquid phase is known as molar enthalpy of vaporization.

As due to heating of a liquid there occurs an increase in entropy due to increase in randomness between the molecules. Therefore, entropy will increase.

As the reaction is Br_{2}(l) rightarrow Br_{2}(g)

The given data is as follows.

  T = 58.8^{o}C

      = (58.8 + 273) K

      = 331.8 K

Delta H_{vap} = 29.6 kJ/mol

As 1 kJ = 1000 J.

So, 29.6 kJ/mol = 29600 J/mol.

Relation between change in entropy and change in enthaly are as follows:

              Delta S_{vap} = frac{Delta H}{T}    

Now, putting the given values into the above formula as follows:

                    Delta S_{vap} = frac{Delta H}{T}

                      = frac{29600 J/mol}{331.8 K}

                      = 89.21 J/mol K

Since, liquid molecules are changing into gas molecules therefore, there will occur an increase in entropy.  

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Keywords:

enthalpy of vaporization, change in entropy.

286 J/K

Explanation:

The molar Gibbs free energy for the vaporization (ΔGvap) is:

ΔGvap = ΔHvap - T.ΔSvap

where,

ΔHvap: molar enthalpy of vaporization

T: absolute temperature

ΔSvap: molar entropy of the vaporization

When T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.

ΔHvap - Tb . ΔSvap = 0

ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.mol

When 1.35 mol of methanol vaporizes, the change in the entropy is:

1.35mol.frac{212J}{K.mol} =286 J/K

a) Q = 68.587,kJ, b) L_{f} = 79.778,frac{cal}{g}

Explanation:

a) The energy absorbed by the water is:

Q = (30.3,g)cdot (frac{1,mol}{18.02,g}  )cdot (40.79,frac{kJ}{mol} )

Q = 68.587,kJ

b) The molar enthalpy of fusion of ice is:

L_{f} = (6.009,frac{kJ}{mol} )cdot (frac{1000,J}{1,kJ} )cdot (frac{1,mol}{18.02,g} )cdot (frac{1,cal}{4.18,J} )

L_{f} = 79.778,frac{cal}{g}

a) increase, ΔS is positive

b) ΔS = 0.089 kJ/K

Explanation:

a) Entropy is a thermodynamic quantity which measures the degree of randomness of a system. Greater the disorder, greater will be the entropy. For different states of matter the entropy increases from solid to liquid to gas.

In the given example, when Br2(l) boils it changes into the gas phase

Br2(l) ↔ Br2(g)

Thus, entropy will increase. i.e. S(product) > S(reactant) and ΔS is positive.

b) It is given that:

Br2(l) ↔ Br2(g)    ΔH = 29.6 kJ/mol

dS = frac{qrev}{T} \\qrev = dH * moles = 29.6 kJ/mol * 1 mol = 29.6 kJ\\dS = frac{29.6}{58.8 + 273} K = 0.089 kJ/K

Answer : The amount of energy absorbed is, 81.2 kJ

Explanation :

The process involved in this problem are :

(1):H_2O(l)(0^oC)rightarrow H_2O(l)(100^oC)\\(2):H_2O(l)(100^oC)rightarrow H_2O(g)(100^oC)

The expression used will be:

Q=[mtimes c_{p,l}times (T_{final}-T_{initial})]+[mtimes Delta H_{vap}]

where,

Q = heat required for the reaction = ?

m = mass of liquid = 30.3 g

c_{p,l} = specific heat of liquid water = 4.18J/g^oC

Delta H_{vap} = enthalpy change for vaporization = 40.79kJ/mol=frac{40790J/mol}{18.02g/mol}=2263.6J/g

Now put all the given values in the above expression, we get:

Q=[30.3gtimes 4.18J/g^oCtimes (100-0)^oC]+[30.3gtimes 2263.6J/g]

Q=81252.48J=81.2kJ

Therefore, the  amount of energy absorbed is, 81.2 kJ

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

Delta S=frac{Delta H_{vap}}{T_b}

where,

Delta S = change in entropy

Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

Delta S=frac{Delta H_{vap}}{T_b}

Delta S=frac{40.5kJ/mol}{352K}

Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

Explanation:

Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane C₄H₁₀:

n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol.

∴ 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized.

Know using cross multiplication:

0.0688 mol of butane to be vaporized requires → 1.67 kJ.

1.0 mol of butane to be vaporized requires → ??? kJ.

∴ 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ.

∴ The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

According to this formula when:

ΔG = ΔH - TΔS = 0

∴ ΔS = ΔH/T

∴ ΔS = n*ΔHVap / Tvap

- when n is the number of moles = mass/molar mass 

when the mass = 24.1 g 

and the molar mass = 187.3764 g/mol

by substitution: 

∴ n = 24.1 / 187.3764g/mol

      = 0.129 moles

and ΔHvap is the molar enthalpy of vaporization is 27.49 kJ/mol

and Tvap is the temperature in Kelvin = 47.6 + 273 = 320.6 K

So by substitution, we will get the ΔS the change in entropy:

∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K

      = 11 J/K

The answer is 236.5 J/K

According to Δ G formula:

ΔG = ΔH - TΔS

when ΔG is the change in free energy (KJ)

and ΔH is the change in enthalpy (KJ)= ΔHvap * moles

                                                              = 71.8 KJ/mol * 1.11 mol
                                                             
                                                              =   79.7 KJ

and T is the absolute temperature (K)= 64 °C + 273°C = 337 K

Δ S is the change in entropy  KJ/K

by substitution:

when at equilibrium ΔG = 0 

∴ΔS = ΔH / T

       =79.7 KJ/ 337 K

     = 0.2365 KJ/K

     = 236.5 J/K

The molar enthalpy of vaporization is the amount of energy required to vaporize one mole of substance at a given temperature and pressure.

When we are given the amount of substance present, and the molar enthalpy of vaporization, we may simply use the formula:

ΔH = n * ΔH(vap) 

To find the enthalpy change occurring

ΔH = 2.15 * 71.8

The value of ΔH is 154.27 kJ.

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