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2 years ago

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Naphthalene, C10H8, melts at 80.2Â°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8Â°C and 5.3 kPa at 119.3Â°C, use th

e Clausiusâ Clapeyron equation to calculate
(a) the enthalpy of vaporization,

(b) the normal boiling point,

(c) the enthalpy of vaporization at the boiling point

1 answer:

sweet-ann [11.9K]2 years ago

3 0

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

P₁ = 1.3 kPa

P₂ = 5.3 kPa

T₁ = 85.8°C = 358.96 K

T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)

ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)

ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)

1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]

1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]

1/T₂ = 2.049 * 10⁻³ K⁻¹

T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

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