Calculate the enthalpy change (∆H) of the following reaction2C2H2(g)+5O2(g)⟶4CO2(g)+2H2O(g) given average bond enthalpies of various bonds, i.e., C-H, C ≡C, O = O, C = O, O - H as 414, 814 499,724 and 640 kJ mol-1 respectively.

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-2600 kJ mol-1

+2573 kJ mol-1

-2573 kJ mol-1

+2600 kJ mol-1

answer is C.

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Detailed Solution

ΔHf=2ϵC=C+2ϵC−H+5∈O=o−42∈C=O−22∈O−HΔHf=2[814+2×414]+5[499]−[8×724]−4[640]=3284+2495−5792−2560=−2573kJ/mole

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