Enthalpy of vapourisation for water is 186.5 KJ mole^-1. the entropy change during vapourisation is 1) 0.5 kj k^-1 mole^-1 2)1.0 kj k^-1 mole^-1 3) 1.5 kj k^-1 mole^-1 4)2.0 kj k^-1 mole^-1

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Q. Enthalpy of vaporisation for water is

186.5

m o l e^{- 1}

. The entropy change during vaporisation is ______ KJ

m o l e^{- 1}

Q. Calculate the change in entropy for fusion of 1 mole of ice. The melting point of ice is 273 K and molar enthalpy of fusion for ice = 6.0 kJ

m o l e^{- 1}

Assuming that water vapour is an ideal gas, the internal energy change (∆U) when 1 mol of water is vapourised at 1 bar pressure and 100 $^{\circ}$ C, (given : molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol $^{- 1}$ and R = 8.3 J mol $^{- 1}$ K $^{- 1}$ ) will be

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); {Δ H}_{295 K} = - 285.8 k J .

The molar enthalpy of vapourisation of water at 1 atm and

25^{o} C

is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at

25^{o} C

is:

The molar entropies of $H I (g), H (g)$ and $I (g)$ at $298 K$ are $206.5, 114.6$ , and $180.7 J m o l e^{- 1} K^{- 1}$ respectively. Using the $Δ G^{0}$ given below,Calculate the bond energy of HI.

$H (g) \to H (g) + I (g)$ ; $Δ G^{0} = 271.8 k J$

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