Enthalpy of vaporisation of a liquid is - H -30 kJ mol -1 and entropy of vaporisation is - S -75 J K -1 mol -1- Find the boiling temperature at 1 atm-A- 400 KB- 300 KC- 150 KD- 425 K

The correct option is A 400 KGiven that, Δ H_vaps = 30 kJ mol^ 1 = 30 × 10^3 J mol^ 1. Δ S_vaps = 75 J mol^ 1 We know that, Δ S = Δ H_vapT_BP ∴Δ H = T Δ S ...

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Standard XII

Chemistry

Ionization Enthalpy

Enthalpy of v...

Question

Enthalpy of vaporisation of a liquid is $Δ H = 30 k J m o l^{- 1}$ and entropy of vaporisation is $Δ S = 75 J K^{- 1} m o l^{- 1}$ . Find the boiling temperature at 1 atm.

400 K

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300 K

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150 K

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425 K

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Solution

The correct option is A $400 K$
Given that,
$Δ H_{v a p s} = 30 k J m o l^{- 1} = 30 \times 10^{3} J m o l^{- 1} .$
$Δ S_{v a p s} = 75 J m o l^{- 1}$
We know that, $Δ S = \frac{Δ H_{v a p}}{T_{B P}} ∴ Δ H = T Δ S \Rightarrow 30 \times 10^{3} = T \times 75. T = 400 K$

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Enthalpy of vaporisation of a liquid is ΔH=30 kJ mol−1 and entropy of vaporisation is ΔS=75 J K−1 mol−1. Find the boiling temperature at 1 atm.

The correct option is A 400 KGiven that, ΔHvaps=30 kJ mol−1=30×103 J mol−1. ΔSvaps=75 J mol−1 We know that, ΔS=ΔHvapTBP∴ΔH=TΔS⇒30×103=T×75.T=400 K

Enthalpy of vaporisation of a liquid is $Δ H = 30 k J m o l^{- 1}$ and entropy of vaporisation is $Δ S = 75 J K^{- 1} m o l^{- 1}$ . Find the boiling temperature at 1 atm.

The correct option is A $400 K$
Given that,
$Δ H_{v a p s} = 30 k J m o l^{- 1} = 30 \times 10^{3} J m o l^{- 1} .$
$Δ S_{v a p s} = 75 J m o l^{- 1}$
We know that, $Δ S = \frac{Δ H_{v a p}}{T_{B P}} ∴ Δ H = T Δ S \Rightarrow 30 \times 10^{3} = T \times 75. T = 400 K$