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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40.5 kJ

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

62.3 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.25 kJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

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:

Join BYJU'S Learning Program

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