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Question

The enthalpies of atomization of CH4(g) and C2H6(g) are respectively 400 kJ mol1 and 670 kJ mol1. The value of HCC would be

A
270 kJ mol1
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B
70 kJ mol1
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C
200 kJ mol1
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D
240 kJ mol1
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Solution

The correct option is B 70 kJ mol1
CH4(g)C(g)+4H(g) Hatom=400 kJ
Hatom=4×ΔHCH
HCH=4004=100 kJ mol1
CH3CH3(g)2C(g)+6H(g); Hatom=670 kJ
670=HCC+6HCH
HCC=6706×100=70 kJ mol1
hence, HCC=70 kJ mol1

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