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Using the bond enthalpy data given below...

Using the bond enthalpy data given below, calculate the enthalpy change for the reaction
`C_(3)H_(4)(g)+H_(2)(g)rarrC_(2)H_(6)(g)`
`{:("Data":,"Bond",,"Bond enthalpy",),(,C-C,,336KJmol^(-1),),(,C=C,,606KJmol^(-1),),(,C-H,,410KJ mol^(-1),),(,H-H,,431KJ mol^(-1),):}`

Text Solution

Verified by Experts

`Delta_(r) H^(@) = sum` reactant - `sum` product
`Delta H_(r)^(@) = sum C = C + 4 sum C - H + sum H - H - sum C - C - 6 sum C - H`
`Delta H_(r)^(@) = 606.68 + 4 xx 410.87 + 431.79 - 336.81 - 6 xx 410 .87 = 2681 .95 - 2802 . 03`
`Delta H_(r)^(@) =- 120 . 08` kJ/mol
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