The enthalpy of vapourisation of liquid water using the data H2g - 1-2 O2g - H2O l- - H - - 285-77 KJ mol-1 H2g - 1-2 O2g - H2Og- - H - - 241-84 KJ mol-1

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Standard XII

Chemistry

Enthalpy

The enthalpy ...

Question

The enthalpy of vapourisation of liquid water using the data
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 285.77 K J m o l^{- 1}$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = - 241.84 K J m o l^{- 1}$

+43.93 KJ mol

^{- 1}

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-43.93 KJ mol

^{- 1}

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+527.61 KJ mol

^{- 1}

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-527.61 KJ mol

^{- 1}

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Solution

The correct option is A +43.93 KJ mol $^{- 1}$
The thermochemical reactions are as given below.
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 285.77 K J m o l^{- 1}$ ......(1)
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = - 241.84 K J m o l^{- 1}$ ......(2)
The reaction (1) is reversed and added to the reaction (2) to obtain the reaction for the vapouisation of liquid water which is $H_{2} O (l) \to H_{2} O (g)$
Hence, the heat of vapourisation of liquid water is $- 241.84 + 285.77 = 43.93 K J m o l^{- 1}$

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Similar questions

Q. Consider the following reactions.

(a)

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1}

m o l^{- 1}

(b)

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m o l^{- 1}

(c)

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Enthalpy of formation of

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(b) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - x_{2} k J m o l^{- 1}$

(d) $C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l) Δ H = x_{4} k J m o l^{- 1}$

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are

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The enthalpy of vapourisation of liquid water using the dataH2(g)+12O2(g)→H2O(l);ΔH=−285.77KJmol−1H2(g)+12O2(g)→H2O(g);ΔH=−241.84KJmol−1

The enthalpy of vapourisation of liquid water using the data
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 285.77 K J m o l^{- 1}$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = - 241.84 K J m o l^{- 1}$