The enthalpy changes for the following processes are listed below: Cl2 (g) = 2Cl (g), 242.3 kJ mol-1 I2 (g) = 21 (g), 151. kJ mol-1 ICl (g) = I (g) + Cl (g), 211.3 kJ mol-1 I2(s) = I2 (g). 62.76 kJ mol-1 Given that the standard states for iodine and chlorine are I2 (s) and Cl2 (g), the standard enthalpy of formation of ICl (g) is :

Question

The enthalpy changes for the following processes are listed below: Cl2 (g) = 2Cl (g), 242.3 kJ mol-1 I2 (g) = 21 (g), 151. kJ mol-1 ICl (g) = I (g) + Cl (g), 211.3 kJ mol-1 I2(s) = I2 (g). 62.76 kJ mol-1 Given that the standard states for iodine and chlorine are I2 (s) and Cl2 (g), the standard enthalpy of formation of ICl (g) is : (A) -14.6 KJ mol-1 (B) -16.8 KJ mol -1 (C) +16.8 KJ mol -1 (D) +244.8 KJ mol-1

Tardigrade Admin · Accepted Answer

(1/2) l2 (s)+(1/2) Cl2 (g) → ICl (g) Δ H=[(1/2)Δ H s→ g+(1/2)Δ Hdiss. (Cl2)+(1/2)Δ Hdiss (I2)]-Δ HICI =((1/2)×62.76+(1/2)×242.3+(1/2)×151.0)-211.3 =228.03-211.3 Δ H=16.73

$- 14.6 K J m o l^{- 1}$

$- 16.8 K J m o l^{- 1}$

$+ 16.8 K J m o l^{- 1}$

$+ 244.8 K J m o l^{- 1}$

−14.6KJmol−1

−16.8KJmol−1

+16.8KJmol−1

+244.8KJmol−1

21​l2​(s)+21​Cl2​(g)→ICl(g) ΔH=[21​ΔHs→g​+21​ΔHdiss.​(Cl2​)+21​ΔHdiss​(I2​)]−ΔHICI​ =(21​×62.76+21​×242.3+21​×151.0)−211.3 =228.03−211.3 ΔH=16.73

$- 14.6 K J m o l^{- 1}$

$- 16.8 K J m o l^{- 1}$

$+ 16.8 K J m o l^{- 1}$

$+ 244.8 K J m o l^{- 1}$