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Question

The enthalpy changes for the following processes are listed below:
Cl2(g)2Cl(g);ΔHo=242.3kJmol1
I2(g)2I(g);ΔHo=151.0kJmol1
ICl(g)I(g)+Cl(g);ΔHo=211.3kJmol1
I2(s)I2(g);ΔHo=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:

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

The correct option is C 16.8 kJmol1
12I212Cl2(g)ICl(g)
ΔH1Cl=[12ΔHI2(s)I2(g)+12ΔHII+12ΔHClCl]
[ΔHICl]
=[12×62.76+12×1.51.0+12×242.3][2.11.3]
=16.83KJ/mol
option (C) is correct.

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