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. `-16.8 kJ "mol"^(-1)`
B. `+ 16.8 kJ "mol"^(-1)`
C. `+ 244.8 kJ "mol"^(-1)`
D. `-14.6 kJ "mol"^(-1)`

Answer

B

Explanation

(i) `CI_(2)(g) to 2CI(g), DeltaH = 242.3 kJ mol^(-1)`
`I_(2)(s) to I_(2)(g), DeltaH = 62.76 kJ "mol"^(-1)`
(iii)`I_(2)(g) to 2I(g), DeltaH = 151.10 kJ "mol"^(-1)`
(iv) `ICI(g) to I(g) + CI(g). DeltaH = 211.3 kJ "mol"^(-1)`
The required equation is
`1/2I_(2)(s) + 1/2CI_(2)(g) to ICI(g)`
Add eq. (i) and (ii)
(v) `CI_(2)(g) + I_(2)(s) to 2CI(g) + I_(2)(g) DeltaH = 305.06 kJ`
Add eq. (iii) and eq.(v)
(vi) `CI_(2)(g) + I_(2)(s) to 2CI(g) + 2I(g) DeltaH = 456.16 kJ`
Divide eq.(vi) by 2
(vii)` 1/2CI_(2)(g) + 1/2 I_(2)(s) to CI(g) + I(g) DeltaH = 228.08 kJ`
Subtract eq. (iv)
`ICI(g) to I(g) + CI(g) DeltaH = 211.3 kJ`
`1/2CI_(2)(g) + 1/2I_(2)(s) to ICI(g) DeltaH = 16.78 kJ`