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. `33.5 kJ mol^(-1)`
B. `16.8 kJ mol^(-1)`
C. `- 14.6 kJ mol^(-1)`
D. `- 211.3 kJ mol^(-1)`

Answer

B

Explanation

The requried thermochemical equation is
`(1)/(2) I_(2) (s) + (1)/(2) Cl_(2) (g) rarr Icl (g) , Delta_(r) H^(@) = ?`
To calculate its enthalpy change, we apply Hess's law of constant heat summation:
`{:((i), I_(2) (s) rarr I_(2) (g),,,Delta H^(@) = 62.8 kJ mol^(-1)),((ii),I_(2) (g) rarr 2 I (g),,,Delta H^(@) = 151 kJ mol^(-1)),((iii),Cl_(2) (g) rarr 2 Cl (g),,,Delta H^(@) = 242.3 kJ mol^(-1)),((iv),ICl (g) rarr I (g) + Cl (g),,,Delta H^(@) = 211.3 kJ mol^(-1)):}`
To get the desired thermochemical equation, we multiply Eqs. (i),(ii) and (iii) by `1//2` and add them term-by-term. Finally, we subract Eq. (iv) from them:
`(1)/(2) (i) + (1)/(2) (ii) + (1)/(2) (iii) - (iv)`
Thus,
`Delta_(r) H = [(1)/(2) (62.8) + (1)/(2) (151) + (1)/(2) (242.3)] - (211.3)`
`= 16.75 kJ`