The formation enthalpies, for H(g) and O(g) are 220.0 and 250.0 kJ , respectively, at 298.15 K, and for (g) is kJ at the same temperature. The average bond enthalpy of the bond in water at 298.15 K is ______ kJ (nearest integer). [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The formation enthalpies, $Δ H_{f}^{\circ}$ for H(g) and O(g) are 220.0 and 250.0 kJ ${mol}^{- 1}$ , respectively, at 298.15 K, and $Δ H_{f}^{\circ}$ for $H_{2} O$ (g) is $- 242.0$ kJ ${mol}^{- 1}$ at the same temperature. The average bond enthalpy of the $O - H$ bond in water at 298.15 K is ______ kJ ${mol}^{- 1}$ (nearest integer). [2025]

Ans.
(466)

$H_{2} O (g) \to 2 H (g) + O (g), Δ H_{r}$

$Δ H_{r} = Δ H_{B.E} (reactants) - Δ H_{B.E} (products)$

$= Δ H_{f} (products) - Δ H_{f} (reactants)$

$[2 \times B.E (O-H)] - [2 \times B.E (H) + B.E (O)]$

$= [2 \times Δ H_{f} (H) + Δ H_{f} (O)] - [Δ H_{f} (H_{2} O)]$

$[2 \times B.E (O-H)] - [0] = [2 \times 220 + 250] - [- 242]$

$B.E (O-H) = 466 {kJ mol}^{- 1}$

Want Us to Email you About Special Offers & Updates?