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Calculate enthalpy change of the followi...

Calculate enthalpy change of the following reaction `:`
`H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g))`
The bond energy of `C-H,C-C,C=C,H-H` are `414,347,615` and `435k J mol^(-1)` respectively.

Text Solution

Verified by Experts

`DeltaH_(Reaction)=` Bond energy data for the formation of bond `+` Bond energy data for the dissociation of bond
`=-[1-(C-C)+6(C-H)]+[1-(C=C)+4(C-H)+1-H-H)]`
`=-1(C-C)-2(C-H)+1(C=C)+1(H-H)`
`=-347-2xx414+615+435=-125kJ`
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Calculate enthalpy change of the following reaction : CH_(2) = CH_(2)(g) + H_(2)(g) rarr CH_(3) - CH_(3)(g) The bond energy of C - H, C - C, C = C, H - H are 414, 615 and 436 kJ mol^(-1) respectively.

Calculate the enthalpy of the following reaction : H_(2)C = CH_(2)(g) + H_(2)(g) rarr H_(3)C - CH_(3)(g) The bond energies of C-H,C-C,C=C and H-H are 414,347,615 and 435 kJ mol^(-1) respectively

Knowledge Check

  • Calculate the enthalpy change of the following reaction H_2C=CH_2+H_2(g)rarrH_3C-CH_3(g) The bond energy of C-H,C-C,C=C,H-H are 414,347, 615 and 453 KJ mol^(-1) respectively.

    A
    `+125kJ`
    B
    `-12.5 kJ`
    C
    `-125 kJ`
    D
    `+12.5 kJ`

  • What is enthalpy change of the following reaction: CH_(2) = CH_(2) + H_(2)(g) to CH_(3)- CH_(3) Given, C-H, C-C, C=C, H-H and 414,347,615 and 435 kJ mole^(-1) respectively.

    A
    `-100 kJ`
    B
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    C
    `-150 kJ`
    D
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