The bond dissociation enthalpy of {\X}_2 ∆ {\H}_{\text{bond }}^{\circ} calculated from the given data is _________ {\kJ} {\mol}{ }^{-1}. (Nearest integer) {\M}^{+} {\X}-( { s}) \rightarrow {\M}^{+}( {\g})+ {\X}^{-}( {\g}) ∆ {\H}_{\text{lattice }}^{\circ}=800 {\kJ} {\mol}^{-1} {\M}( { s}) \rightarrow {\M}( {\g}) ∆ {\H}_{\text{sub }}^{\circ}=100 {\kJ} {\mol}^{-1} {\M}( {\g}) \rightarrow {\M}^{+}( {\g})+ {\e}^{-}( {\g}) ∆ {\H}_{ {\i}}^{\circ}=500 {\kJ} {\mol}^{-1} {\X}( {\g})+ {\e}^{-}( {\g}) \rightarrow {\X}^{-}( {\g}) ∆ {\H}_{ {\eg}}^{\circ}=-300 {\kJ} {\mol}^{-1} {\M}( { s})+\frac{1}{2} {\X}_2( {\g}) \rightarrow {\M}^{+} {\X}-( { s}) ∆ {\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] [23 Jan 2025 Shift 2]

Chemical Thermodynamics Part 2

Section: Chemistry

Question : 15 of 100

Marks: +1, -0

The bond dissociation enthalpy of X2∆H‌bond ‌∘ calculated from the given data is _________ kJ‌mol‌‌−1. (Nearest integer)
‌M+X−( s)⟶M+(g)+X−(g)∆H‌lattice ‌∘=800‌kJ‌mol−1
‌M( s)⟶M(g)∆H‌sub ‌∘=100‌kJ‌mol−1
‌M(g)⟶M+(g)+e−(g)∆Hi∘=500‌kJ‌mol−1
‌X(g)+e−(g)⟶X−(g)∆Heg∘=−300‌kJ‌mol−1
‌M( s)+‌

X2(g)⟶M+X−( s)∆Hf∘=−400‌kJ‌mol−1
[Given M+X−is a pure ionic compound and X forms a diatomic molecule X2 in gaseous state]

[23 Jan 2025 Shift 2]

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