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solong [7]
3 years ago
12

Naphthalene, C10H8, 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 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
Chemistry
1 answer:
sweet-ann [11.9K]3 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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Answer:

\boxed {\boxed {\sf B. \ 22 \ grams}}

Explanation:

We need to use the formula for heat of vaporization.

Q=H_{vap}*m

Identify the variables.

  • The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.
  • Q is the energy, in this problem, 50,000 Joules.
  • m is the mass, which is unknown.

H_{vap}=2260 \ J/g\\Q=50,000 \ J \\

Substitute the values into the formula.

50,000 \ J=2260 \ J/g*m

We want to find the mass. We must isolate the variable, m.

m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.

\frac{50,000 \ J}{2260 \ J/g} =\frac{2260 \ J/g*m}{2260 \ J/g}

\frac{50,000 \ J}{2260 \ J/g} =m

Divide. Note that the Joules (J) will cancel each other out.

\frac{50,000 \ }{2260 \ g} =m

22.1238938 \ g =m

Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.

22 \ g \approx m

The mass is about 22 grams, so choice B is correct.

3 0
3 years ago
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