MyQuestionIcon

1

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

Heat of Formation

The enthalpy ...

Question

The enthalpy of formation of reaction :
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) i s Δ_{r} H^{o} = - 572 k J m o l^{- 1}$
What will be the enthalpy of formation of $H_{2} O (l)$ ?

A

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

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

B

286 k J m o l^{- 1}

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

C

- 1144 k J m o l^{- 1}

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

D

1144 k J m o l^{- 1}

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

Open in App

Solution

The correct option is A $- 286 k J m o l^{- 1}$
Given: $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) Δ_{r} H^{o} = - 572 k J m o l^{- 1}$ .

Enthalpy of formation is the enthalpy change of the reaction when one mole of the compound is formed from its element, i.e. we aim at
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ_{r} H^{o} = ?$

this can be obtained if we divide given reaction by 2.
so, $Δ_{f} H^{o} (H_{2} O (l)) = \frac{Δ_{r} H^{o}}{2} = \frac{- 572}{2} = - 286 k J m o l^{- 1}$

hence, enthalpy of formation of $H_{2} O (l)$ is $- 286 k J m o l^{- 1}$ .

Suggest Corrections

thumbs-up

0

Similar questions

Q. The enthalpy of formation of reaction :

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) i s Δ_{r} H^{o} = - 572 k J m o l^{- 1}

What will be the enthalpy of formation of

H_{2} O (l)

?

Q. Calculate the standard enthalpy of formation of

C H_{3} O H

from the following data.

C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l); Δ_{r} H^{⊖} = - 726 k J m o l^{- 1}

C (g) + O_{2} (g) \to C O_{2} (g); Δ_{c} H^{⊖} = - 393 k J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ_{f} H^{⊖} = - 286 k J m o l^{- 1}

Q. For the reaction:

4 N H_{3} (g) + 3 O_{2} (g) \to 2 N_{2} (g) + 6 H_{2} O (l)

298 K

. Given that heat of formation of

N H_{3} (g

H_{2} O (l)

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

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

respectively. Calculate heat of reaction at

298 K

.

Calculate the standard enthalpy of formation of CH₃OH₍_l₎ from the following data:

CH₃OH₍_l₎ + O₂₍_g₎ CO₂₍_g₎ + 2H₂O₍_l₎ ; Δ_rH^θ = –726 kJ mol^–1

C₍_g₎ + O₂₍_g₎ CO₂₍_g₎ ; Δ_cH^θ = –393 kJ mol^–1

H₂₍_g₎ +O₂₍_g₎ H₂O₍_l₎ ; Δ_fH^θ = –286 kJ mol^–1.

Q. Calculate the enthalpy change for the following reaction:

{C H}_{4} (g) + 2 O_{2} (g) ⟶ {C O}_{2} (g) + 2 H_{2} O (l)

Given, enthalpies of formation of

{C H}_{4}, {C O}_{2}

H_{2} O

74.8 k J {m o l}^{- 1}, - 393.5 k J {m o l}^{- 1}, - 286 k J {m o l}^{- 1}

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Heat of Formation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon