Q.

The enthalpy changes for the following processes are listed below

Cl2(g)=2Cl(g);                 242.3kJmol1I2(g)=2I(g);                      151.0kJmol1ICl(g)=I(g)+Cl(g);        211.3kJmol1I2(g)=I2(g);                        62.76kJmol1

Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is :

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a

+16.8 kJ mol1

b

14.6 kJ mol1

c

16.8 kJ mol1

d

+244.8 kJ mol1

answer is B.

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Detailed Solution

12I212Cl2(g)ICl(g)ΔH1Cl=12ΔHI2(s)I2(g)+12ΔHTI+12ΔHClclΔHYCl=12×62.76+12×1.51.0+12×242.3[2.11.3]=16.83 KJ/mol

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