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Question

The enthalpy changes for the following processes are listed below:
Cl2(g)=2Cl(g),242.3 kJ mol−1
I2(g)=2I(g),151.0 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 for ICl(g) is:

A
16.8 kH mol1
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B
+16.8 kJ mol1
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C
+244.8 kJ mol1
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D
14.6 kJ mol1
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Solution

The correct option is A 16.8 kH mol1
2HoStd=sub+atomCl2+(211.32)
2HoStd=62.76+151.0+242.32211.3
2HoStd=32.7KJmol1
HoStd=+16.8KJmol1

1077324_692232_ans_607e1a65154b47c9a67ae514ae043521.PNG

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