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Question

The enthalpy changes for the following processes are listed below:
Cl2(g)2Cl(g),242.3kJmol1
I2(g)2I(g),151.0kJmol1
ICI(g)I(g)+Cl(g),211.3kLmol1
I2(s)I2(g),62.75kJmol1
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICI(g) is:

A
16.8kJmol1
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B
+16.8kJmol1
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C
+244.8kJmol1
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D
14.6kJmol1
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Solution

The correct option is D 16.8kJmol1
Formation enthalpy of ICl is-
12I2(s)+12Cl2(g)ICl(g)
So Multiply the 1st and 4th, 2nd reaction with 12
and reverse the third reaction , finally add them -
=(121.15+31.375)+75.5)211.3=16.8 KJ mol1

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