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Question
Calculate the enthalpy change (△H) of the following reaction
2C2H2(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.
Calculate the enthalpy change (△H) of the following reaction
2C2H2(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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Solution
The correct option is C −2573 kJ mol−1
Given- Bond enthalpies of various bonds
εC−H=414 kJ mol−1
εC≡C=814 kJ mol−1
εO=O=499 kJ mol−1
εC=O=724 kJ mol−1
εO−H=640 kJ mol−1
△Hf=2[εC≡C+2εC−H]+5εO=O−4[2εC=O]−2[2εO−H]
△Hf=2[814+2×414]+5[499]−[8×724]−4[640]
=3284+2495−5792−2560=−2573 kJ/mol.
hence, △Hf=−2573 kJ/mol.
Given- Bond enthalpies of various bondsεC−H=414 kJ mol−1
εC≡C=814 kJ mol−1
εO=O=499 kJ mol−1
εC=O=724 kJ mol−1
εO−H=640 kJ mol−1
△Hf=2[εC≡C+2εC−H]+5εO=O−4[2εC=O]−2[2εO−H]
△Hf=2[814+2×414]+5[499]−[8×724]−4[640]
=3284+2495−5792−2560=−2573 kJ/mol.
hence, △Hf=−2573 kJ/mol.
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