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Propane has the structure H_(3)C - CH_(2) - CH_(3) . Calculate the change inenthalpy for the reaction : C_(3) H_(8) (g) + 5O_(2)(g) rarr 3 CO_(2)(g) + 4H_(2)O(g) Given that average bond enthalpiesare : {:(C-C,C-H,C=0,O=O,O-H),(347,414,714,498,464kJ mol^(-1)):} |
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Answer» Solution :`Delta_(r)H= [B.E. (H-underset(H)underset(|)overset(H) overset(|)(C ) - underset(H) underset(|) overset(H) overset(|) (C) - underset(H)underset(|) overset(H)overset(|)(C)-H)+5 xx B.E. ( O= O) ]= [3 xx B.E. ( O= C=O) + 4B.E. ( H- O-H)]` `= [2 B.E. ( C-C) + 8 B.E. (C-H) + 5 B.E. ( O=O) ] = [ 6 xx B.E. ( C= O )+8B.E. ( O- H) ]` `= 2 ( 347) + 8 ( 414) + 5 ( 498) ]- [ 6 ( 741) +8 ( 464) ] kJ MOL^(-1)` ` = [694+ 3312+2490 ]-[4446 +3712]= - 1662k J mol^(-1)` |
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