Heat of atomisation of NH3 and N2H4 are xkcalmol−1 and ykcalmol−1 respectively. What will be average bond energy of N−N?
A
−3y−4x3kcalmol−1
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B
3y−4x3kcalmol−1
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C
3y+4x3kcalmol−1
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D
None of the above
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Solution
The correct option is B3y−4x3kcalmol−1 Given reactions are: NH3→N(g)+3H(g)ΔHatom=xkcalmol−1=ΔH1=3×BEN−H......1 N2H4→2N(g)+4H(g)ΔHatom=ykcalmol−1=ΔH2=4×BEN−H+BEN−N......2 from equation 1 we have BEN−H=x3 put BEN−H in equation 2 to get the value of BEN−N ΔatomH=ykcalmol−1=ΔH2=4×x3+BEN−N BEN−N=y−4x3 BEN−N=3y−4x3kcalmol−1 hence, the average bond energy of N−N is 3y−4x3kcalmol−1