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Enthalpy

Define mixtur...

Question

Define mixtures.
(a) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$ ;
$Δ H_{298 K}^{o} = - 285.9 k J m o l^{- 1}$
(b) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ ;
$Δ H_{298 K}^{o} = - 241.8 k J m o l^{- 1}$
The molar enthalpy of vapourisation of water will be:

A

241.8

k J

m o l^{- 1}

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B

22.0

k J

m o l^{- 1}

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C

44.1

k J

m o l^{- 1}

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D

527.7

k J

m o l^{- 1}

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Solution

The correct option is C $44.1$ $k J$ $m o l^{- 1}$
Given:
(a) ${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 285.9 k J / m o l$
${H_{2} O}_{(l)} ⟶ {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)}; Δ H = 285.9 k J / m o l . . . . . (1)$
(b) ${H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2} O}_{(g)}; Δ H = - 241.8 k J / m o l . . . . . (2)$
Adding ${e q}^{n} (1) & (2)$ , we have
${H_{2} O}_{(l)} + {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} ⟶ {H_{2}}_{(g)} + \frac{1}{2} {O_{2}}_{(g)} + {H_{2} O}_{(g)}; Δ H = [285.9 + (- 241.8)] k J / m o l$
${H_{2} O}_{(l)} ⟶ {H_{2} O}_{(g)}; Δ H = 44.1 k J / m o l$
Hence the molar enthalpy of vapourisation of water is $44.1 k J / m o l$ .

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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

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