The bond dissociation energies for `Cl_(2),I_(2)` and Icl are 242.3, 151 and 211.3 kJ/mol respectively. The enthalpy of sublimation of iodine is 62.8 kJ/mol. What is the standard enthalpy of formation of `ICl_((s))` ?

Options

A. 211.3 kJ/mol
B. 4019Cal/mol
C. 16.8 kJ/mol
D. 33.5 kJ/mol

Answer

B, C

Explanation

Reaction for formation of `ICl_((g))` is `(1)/(2) I_(2(g)) + (1)/(2) Cl_(2(g)) rarr ICl_((g)), Delta H_(1)` ….(i) get equation for `Delta H_(1)` in terms of Bond energy we have to use
`I_(2(g)) rarr I_(2(g)), Delta H_("sublimation")` ..(2)
eq (1) - eq (2) `xx (1)/(2)` gives
`(1)/(2) I_(2(g)) + (1)/(2) Cl_(2(g)) rArr ICl_((g)) Delta H_(3)`
`Delta H_(3) = Delta H_(1) - (Delta H_(2))/(2)`
`= (1)/(2) ("B.E of " I_(2)) + (1)/(2) ("B. E of " Cl_(2)) -` B. E of ICl
`rArr Delta H_(1) - (62.8)/(2) = (1)/(2) (242.3) + (1)/(2) (151) - 211.3`
`Delta H_(1) = 16.75 KJ = 4019` Cal