The enthaplpy changes state for the following processes are listed below:
`Cl_(2)(g)=2Cl(g)` : `242.3KJmol^(-1)`
`I_(2)(g)=2I(g)` , `151.0KJ mol^(-1)`
`ICl(g)=I(g)+Cl(g)` : `211.3KJ mol^(-1)`
`I_(2)(s)=l_(2)(g)` , `62.76KJ mol^(-1)`
Given that the standard states for iodine chlorine are `I_(2)(s)` and `Cl_(2)(g)` , the standard enthalpy of formation for `ICl(g)` is:

Options

A. `+244.8KJmol^(-1)`
B. `-14.6KJmol^(-1)`
C. `-16.8KJmol^(-1)`
D. `+16.8KJmol^(-1)`

Answer

D

Explanation

`I_(2)+Cl_(2)rarr2ICl`
`DeltaH=e_(I_(2)srarrg)+e_(I-I)+e_(Cl-Cl)-2xxe_(I-Cl)`
`=62.76+151.0+242.3-2xx211.3`
`=33.46KJ` for 2 moles of `ICl`
:. `DeltaH//mol=(33.46)/(2)=16.73KJmol^(-1)`