What is the bond enthalpy of `Xe-F` bond ? `XeF_(4)(g)rarrXe^(+)(g)+F^(-)(g)+F_(2)(g)+F(g)," "Delta_(r)H=292"kcal/mol"` Given : Ionization energy of `Xe=279"kcal/mol"` `B.E.(F-F) = 30"kcal/mol"`, Electron affinity of F = 85 kcal/mol
What is the bond enthalpy of Xe-F bond if Ionization energy of Xe = 279 kcal/mol B.E.(F-F) = 38 kcal/mol, electron affinity of F = 85 kcal//mol
What is the DeltaH of the following affinity. Mg_(g)+2F_(g) rarr Mg_((g))^(2+)+2F_((g))^(-) . If the electron affinity of F_(g) = -328 kJ " mol"^(-1) and first ionisation energy of Mg = 737.7kJ " mol"^(-1) and second ionisation energy of Mg = 1451kJ " mol"^(-1)
Calculate the enthalpy change for the following reaction: XeF_(4) rarr Xe^(o+) +F^(Theta) +F_(2) +F . The average Xe-F bond energy is 34 kcal mol^(-1) , first IE of Xe is 279 kcal mol^(-1), EA of F is 85 kcal mol^(-1) and bond dissociation enegry of F_(2) is 38 kcal mol^(-1)
In XeF_(4) the bond angle F-Xe-F is
NARENDRA AWASTHI-THERMODYNAMICS-Level 3 - Match The Column