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Question

The enthalpy changes for the following precesses are listed below :


Cl2(g)2Cl(g), 242.3 kJmol1
I2(g)2I(g), 151.0 kJmol1
ICl(g)I(g)+Cl(g), 211.3 kJmol1
I2(s)I2(g), 62.76 kJmol1

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

A
14.6 kJmol1
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B
16.8 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 B +16.8 kJmol1
12I2(s)+12Cl2(g)ICl(g)
H=[12Hsg+12Hdiss(Cl2)+12Hdiss(I2)]HICl
=(12×62.76+12×242.3+12×151.0)211.3
=228.03211.3
H=16.73 kJmol1

Hence the correct option is C.

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