Physics, 05.11.2019 03:31 paytonhuff8580

Naphthalene, c 10 h 8 , melts at 80.2°c. if the vapour pressure of the liquid is 1.3 kpa at 85.8°c and 5.3 kpa at 119.3°c, use the clausius–clapeyron equation to calculate (a) the enthalpy of vaporization, (b) the normal boiling point, and (c) the enthalpy of vaporization at the boiling point.

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Answer from: timothyashburn8

Answer :

(a) The value of $\Delta H_{vap}$ is 48.6 kJ/mol

(b) The the normal boiling point is 489.2 K

Explanation :

(a) To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

P_1 = vapor pressure at temperature 85.8^oC = 1.3 kPa

P_2 = vapor pressure at temperature 119.3^oC = 5.3 kPa

$\Delta H_{vap}$ = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 85.8^oC=[85.8+273]K=358.8K

T_2 = final temperature = 119.3^oC=[119.3+273]K=392.3K

Putting values in above equation, we get:

$\ln(\frac{5.3kPa}{1.3kPa})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{358.5}-\frac{1}{392.3}]\\\\\Delta H_{vap}=48616.4J/mol=48.6kJ/mol$

Therefore, the value of $\Delta H_{vap}$ is 48.6 kJ/mol

(b) The clausius claypron equation is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

P_1 = vapor pressure at temperature 85.8^oC = 1.3 kPa

P_2 = vapor pressure at temperature normal boiling point = 101.3 kPa

$\Delta H_{vap}$ = Enthalpy of vaporization = 48.6 kJ/mol

R = Gas constant = $8.314\times 10^{-3}kJ/mol.K$

T_1 = initial temperature = 85.8^oC=[85.8+273]K=358.8K

T_2 = final temperature = ?

Putting values in above equation, we get:

$\ln(\frac{101.3kPa}{1.3kPa})=\frac{48.6kJ/mol}{8.314\times 10^{-3}kJ/mol.K}[\frac{1}{358.5}-\frac{1}{T_2}]\\\\T_2=489.2K$

Therefore, the normal boiling point is 489.2 K

$\Delta S_{vap}=\frac{\Delta H_{vap}}{T_b}$

where,

$\Delta S_{vap}$ = entropy of vaporization = ?

$\Delta H_{vap}$ = enthalpy of vaporization = 48.6 kJ/mol

T_b = boiling point = 489.2 K

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

$\Delta S_{vap}=\frac{48.6kJ/mol}{489.2K}=99.3J/K$

Therefore, the entropy of vaporization at the boiling point is 99.3 J/K

Answer from: Quest

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Answer from: Quest

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from wikipedia

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Naphthalene, c 10 h 8 , melts at 80.2°c. if the vapour pressure of the liquid is 1.3 kpa at 85.8°c a...

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