The enthalpy of vapourisation of water from the following equations is-H 2 g -1-2 O 2 g - H 2 O l - H -286 kJH 2 g -1-2 O 2 g - - H 2 O g - H -245-5 kJA- 1-25 kJB- 40-5 kJC- 6-02 kJD- 62-3 kJ

The correct option is B 40.5 kJH_2(g) + 12O_2(g) → H_2O(l), H = 286 kJ ...(i) H_2(g) + 12 O_2 (g), → H_2O (g), H = 245.5 kJ ...(ii) The desired equation ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy

The enthalpy ...

Question

The enthalpy of vapourisation of water from the following equations is:
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), △ H = - 286 kJ$
$H_{2} (g) + \frac{1}{2} O_{2} (g), \to H_{2} O (g), △ H = - 245.5 kJ$

6.02 kJ

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40.5 kJ

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62.3 kJ

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1.25 kJ

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Solution

The correct option is B $40.5 kJ$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), △ H = - 286 kJ . . . (i)$
$H_{2} (g) + \frac{1}{2} O_{2} (g), \to H_{2} O (g), △ H = - 245.5 kJ . . . (i i)$

The desired equation is $H_{2} O (l) \to H_{2} O (g), △ H =$ ?

On subtracting equation (i) from (ii),
$H_{2} O (l) \to H_{2} O (g)$
$(Equation (ii) - Equation (i))$
$△ H = - 245.5 kJ - (- 286 kJ) = 40.5 kJ$

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Similar questions

Q. The enthalpy of vapourisation of water from the following equations is:

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), △ H = - 286 kJ

H_{2} (g) + \frac{1}{2} O_{2} (g), \to H_{2} O (g), △ H = - 245.5 kJ

Q. 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}

Q. 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:

Q. The enthalpy of reaction,

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H_{1}

and that of following reaction is

Δ H_{2}

H_{2_{(g)}} + \frac{1}{2} O_{2} (g) \to H_{2} O_{(l)}

Then :

Q. The enthalpy of the reaction,

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(} g)

Δ H_{1}

and that of

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to {H_{2} O}_{(l)}

Δ H_{2}

then:

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The enthalpy of vapourisation of water from the following equations is: H2(g)+12O2(g)→H2O(l),△H=−286 kJ H2(g)+12O2(g),→H2O(g),△H=−245.5 kJ

The correct option is B 40.5 kJH2(g)+12O2(g)→H2O(l),△H=−286 kJ...(i) H2(g)+12O2(g),→H2O(g),△H=−245.5 kJ...(ii) The desired equation is H2O(l)→H2O(g),△H= ? On subtracting equation (i) from (ii), H2O(l)→H2O(g) (Equation (ii) - Equation (i)) △H=−245.5 kJ−(−286 kJ)=40.5 kJ

The enthalpy of vapourisation of water from the following equations is:
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), △ H = - 286 kJ$
$H_{2} (g) + \frac{1}{2} O_{2} (g), \to H_{2} O (g), △ H = - 245.5 kJ$