Given: (I) H2(g) + (1/2)O2(g) → H2O(l); Δ H°298K = -285.9kJ mol-1 (II) H2(g) + (1/2)O2(g) → H2O(g); Δ H° 298K = -244.8kJ mol-1 The molar enthalpy of vapourisation of water will be :

Question

Given: (I) H2(g) + (1/2)O2(g) → H2O(l); Δ H°298K = -285.9kJ mol-1 (II) H2(g) + (1/2)O2(g) → H2O(g); Δ H° 298K = -244.8kJ mol-1 The molar enthalpy of vapourisation of water will be : (A) 241.8 kJ mol-1 (B) 22.0 kJ mol-1 (C) 44.1 kJ mol-1 (D) 527.7 kJ mol-1

Tardigrade Admin · Accepted Answer

Given H2(g) + (1/2)O2(g) → H2O(l); Δ H° = -285.9kJ mol-1 ...(1) H2(g) + (1/2)O2(g) → H2O(g); Δ H° = -244.8kJ mol-1 ...(2) We have to calculate H2O(l)→ H2O(g) ; Δ H° = ? On substracting eqn. (2) from eqn. (1) we get H2O(l)→ H2O(g) ; Δ H° = -241.8-(-285.9) = 44.1 kJ mol-1

Q. Given :
$(I) H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l);$
$Δ H_{298 K}^{\circ} = - 285.9 k J m o l^{- 1}$
$(II) H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g);$
$Δ H_{298 K}^{\circ} = - 244.8 k J m o l^{- 1}$
The molar enthalpy of vapourisation of water will be :

$241.8 k J m o l^{- 1}$

$22.0 k J m o l^{- 1}$

$44.1 k J m o l^{- 1}$

$527.7 k J m o l^{- 1}$

Q. Given : (I)H2​(g)+21​O2​(g)→H2​O(l); ΔH298K∘​=−285.9kJmol−1 (II)H2​(g)+21​O2​(g)→H2​O(g); ΔH298K∘​=−244.8kJmol−1 The molar enthalpy of vapourisation of water will be :

241.8kJmol−1

22.0kJmol−1

44.1kJmol−1

527.7kJmol−1

Q. Given :
$(I) H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l);$
$Δ H_{298 K}^{\circ} = - 285.9 k J m o l^{- 1}$
$(II) H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g);$
$Δ H_{298 K}^{\circ} = - 244.8 k J m o l^{- 1}$
The molar enthalpy of vapourisation of water will be :

$241.8 k J m o l^{- 1}$

$22.0 k J m o l^{- 1}$

$44.1 k J m o l^{- 1}$

$527.7 k J m o l^{- 1}$