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When 2mole of C(2)H(6) are completely bu...

When 2mole of `C_(2)H_(6)` are completely burnt `-3129kJ` of heat is liberated. Calculate the heat of formation of `C_(2)H_(6).Delta_(f)H^(Theta)` for `CO_(2)` and `H_(2)O` are `-395` and `-286kJ`, respectively.

Text Solution

Verified by Experts

We have to find
`2C_((s))+3H_(2(g)) rarr C_(2)H_(6(g)),DeltaHimplies ....(1)`
Given ,
`C_((s))+O_(2(g))rarr,DeltaH=-395kJ ...(2)`
`H_(2(g))+(1)/(2)O_(2(g))rarr H_(2)O_((l)), DeltaH=-286kJ ...(3)`
`C_(2)H_(6(g))+(7)/(2)O_(2(g))rarr 2CO_(2(g))+3H_(2)O_((l)), DeltaH=-(3129)/(2)kJ ....(4)`
`=-1564.5kJ`
Multiplying eq. `(2)` and `eq. (3)` by 3, then adding
`2C_((s))+3H_(2(g))+(7)/(2)O_(2(g))rarr 2CO_((g))+3H_(2)O_((l)),
DeltaH=-1648kJ ...(5)`
Subtracting `eq. (4)` from `eq. (50` ,
`ul( {:(C_(2)H_(6(g)),+(7)/(2)O_(2(g)),rarr,2CO_(2(g)),+3H_(2)O_((l)),,DeltaH=-1564.5kJ),(-,-,,-,-," +"):} )`
`2C_((s))+3H_(2(g)) rarrC_(2)H_(6(g)),DeltaH=-83.5kJ`
` :. DeltaH_(f)` of `C_(2)H_(6) =-83.5kJ`
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Knowledge Check

  • The standard enthalpy of formation of octane (C_(8)H_(18)) is -250kJ //mol . Calculate the enthalpy of combustion of C_(8)H_(18) . The enthalpy of CO_(2)(g) and H_(2)O(l) are -394 kJ//mol and -286kJ//mol respectively.

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