The enthalpy change for the following process are listed belowCl2(g)→2Cl(g);ΔH1=242.3KJmol-1I2(g)→2I(g);ΔH2=151.0KJmol-1ICl(g)→I(g)+Cl(g);ΔH3=211.3KJmol-1I2(s)→I2(g);ΔH4=62.76KJmol-1If standard states of iodine and chlorine are  I2(s) and Cl2(g)  the standard enthalpy of formation for IClg

1) I2(s)+Cl2(g)→2ICl(g) ;ΔH=?We are to find enthalpy change of the above process.Cl2(g)→2Cl(g);ΔH1I2(s)→I(g);ΔH4 I2(g)→2I(g);ΔH22I(g)+2CI(g)→2ICl(g);2ΔH3Cl2(g)+I2(s)→2ICl(g) ΔH=ΔH1+ΔH4+ΔH2−2ΔH3 =242.3+62.76+151−2×211.3 =33.46 For 2 moles ---- 33.46        1 mole ------ ? ΔHmole=33.462=16.83KJ/mol