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Calculate enthalpy change of the followi...

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.

Text Solution

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`-125 kJ//"mole"`
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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

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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
    `-125 kJ`
    C
    `-150 kJ`
    D
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  • The reagent used for the following reaction is/are: CH_(3)-CH=CH-C_(2)H_(5) to CH_(3)CHO+C_(2)H_(5)CHO

    A
    `O_(3)` and Zn stem
    B
    Baeyer's reagent
    C
    `(KMnO_(4))/H_(2)SO_(4)`
    D
    Lemieux reagent

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