The bond dissociation enthalpy of calculated from the given data is _____ kJ . (Nearest integer) [Given: is a pure ionic compound and X forms a diatomic molecule in gaseous state] [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 bond dissociation enthalpy of $X_{2}$ $Δ H_{bond}^{\circ}$ calculated from the given data is _____ kJ ${mol}^{- 1}$ . (Nearest integer)

$M^{+} X^{-} (s) \to M^{+} (g) + X^{-} (g) Δ H_{lattice}^{\circ} = 800 {kJ mol}^{- 1}$

$M (s) \to M (g) Δ H_{sub}^{\circ} = 100 {kJ mol}^{- 1}$

$M (g) \to M^{+} (g) + e^{-} (g) Δ H_{i}^{\circ} = 500 {kJ mol}^{- 1}$

$X (g) + e^{-} (g) \to X^{-} (g) Δ H_{e g}^{\circ} = - 300 {kJ mol}^{- 1}$

$M (s) + \frac{1}{2} X_{2} (g) \to M^{+} X^{-} (s) Δ H_{f}^{\circ} = - 400 {kJ mol}^{- 1}$

[Given: $M^{+} X^{-}$ is a pure ionic compound and X forms a diatomic molecule $X_{2}$ in gaseous state] [2025]

Ans.
(200)

$Δ H_{f}^{°} = Δ H_{sub}^{°} + \frac{1}{2} Δ H_{bond}^{°} + Δ H_{i}^{°} + Δ H_{eg}^{°} - Δ H_{lattice}^{°}$

$- 400 = 100 + \frac{1}{2} Δ H_{b o n d}^{°} + 500 - 300 - 800$

$Δ H_{b o n d}^{°} = 200 kJ/mol$

Want Us to Email you About Special Offers & Updates?