The enthalpy of vapourisation of water from the following equations is-H2g - 12O2g - H2Ol- H - -286 kJH2g - 12 O2 g- - H2O g- H - - 245-5 kJ

H_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 is H_2O(l) → H_2O(g), H = ? On ...

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Standard XIII

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:

Heat of formation of $C H_{4}$ is:

If given:

$C (s) + O_{2} (g) \to C O_{2} (g) △ H = - 394 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) △ H = - 284 k J$

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) △ H = - 892 k J$

Q. Given that :

2 C (s) + 2 O_{2} (g) \to 2 C O_{2} (g)

;

Δ H = - 787

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

;

Δ

= - 286

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)

;

Δ

= - 1310

kJ.

Heat of formation of acetylene is:

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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:
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), △ H = - 286 kJ$
$H_{2} (g) + \frac{1}{2} O_{2} (g), \to H_{2} O (g), △ H = - 245.5 kJ$