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The enthalpy changes for the following p...

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

A

`+16.8kJmol^(-1)`

B

`+244`

C

`-14.6kJmol^(-1)`

D

`-16.8mol`

Text Solution

Verified by Experts

The correct Answer is:
A
`(1)/(2)I_(2)+(1)/(2)Cl_(2)rarrICl,DeltaH=H_(R)-H_(P)`
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