What is the enthalpy of vaporisation of liquid water in kJ mol-1 if- i H2 g -12O2 g - H2Ol- 285-77 kJmol-1 ii H2 g -12O2 g - H2Ov- 241-84 kJ mol-1

The correct option is A + 43.93 ( i ) H_2( g ) +12O_2( g )⟶ H_2O(l)+ 285.77 kJmol^ 1 ( ii ) H_2( g ) +12O_2( g )⟶ H_2O(v)+ 241.84 kJ mol^ 1 Su ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Difference between Internal Energy and Enthalpy

What is the e...

Question

What is the enthalpy of vaporisation of liquid water in $k J m o l^{- 1}$ if:

( i ) $H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l) + 285.77 k J m o l^{- 1}$
( ii ) $H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (v) + 241.84 k J m o l^{- 1}$

+ 43.93

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

- 43.93

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

+ 527.61

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

- 527.61

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

Open in App

Solution

Suggest Corrections

Similar questions

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}

H_{2 (g)} + 1 / 2 O_{2 (g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 285.77 k J / m o l e

H_{2 (g)} + 1 / 2 O_{2 (g)} ⟶ {H_{2} O}_{(l)}; Δ H = - 241.84 k J / m o l e

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = x

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = y

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

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

C a C l_{2}

C a (s)

121 k J m o l^{- 1}

C l_{2} (g)

2 C l (g)

242.8 k J m o l^{- 1}

C a (g)

C a^{2 +} (g)

2422 k J m o l^{- 1}

C l (g)

C l^{- 1} (g)

- 355 k J m o l^{- 1}

△_{f} H

- 795 k J m o l^{- 1}

Join BYJU'S Learning Program