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If Delta(f)H^(@)(C(3)H(8)(g))=-85kJ mol...

If `Delta_(f)H^(@)``_(C_(3)H_(8)(g))=-85kJ moll^(-1),`
`Delta_(f)H^(@)``_(C_(3)H_(8)(g))=-104kJ mol^(-1),`
`DeltaH` for `C_((s))rarr C_((g))` is `718kJ mol^(-1)` and heat of formation of `H-` atom is `218kJ mol^(-1)`, then `:`

A

`e_(C-H)=414kJ`

B

`e_(C-C)=345kJ`

C

`e_(H-H)=218kJ`

D

`e_(H-H)=436kJ`

Text Solution

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The correct Answer is:
A, B, D
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Given C_(2)H_(2)(g)+H_(2)(g)rarrC_(2)H_(4)(g): DeltaH^(@)=-175 " kJ mol"^(-1) DeltaH_(f(C_(2)H_(4),g))^(@)=50 " kJ mol"^(-1), DeltaH_(f(H_(2)O,l))^(@)=-280 " kJ mol"^(-1), DeltaH_(f(CO_(2)g))^(@)=-390 " kJ mol"^(-1) If DeltaH^(@) is enthalpy of combustion (in kJ "mol"^(-1) ) of C_(2)H_(2) (g), then calculate the value of |(DeltaH^(@))/(257)|

Knowledge Check

  • If DeltaH_(f(C_(2)H_(6)))^(0)(g) = -85 KJH mol^(-1) , DeltaH_(f(C_(3)H_(8)))^(0) (g) = -104 KJ mol^(-1), DeltaH^(0) for C (s) rarr C(g) is 718 KJ mol^(-1) and heat of formation of H-atom is 218 KJmol^(-1) then :

    A
    `DeltaH_(C-C) = 345` KJ
    B
    `DeltaH_(C-H) = 414` KJ
    C
    `DeltaH_(H-H)=436 KJ`
    D
    `DeltaH_(H-H) = 436` KJ

  • Calculate the enthalpy of formation of methanol (DeltaH_(f) (CH_(3) OH) (I) ) in KJ mol^(-1) using the data : DeltaH_("vap") (CH_(3)OH)(I) is 38 KJ mol^(-1) The heat of formation of gaseous atoms from elements in their standard states are H = 218 KJ mol^(-1) C = 715 KJ mol^(-1) O = 249 KJ mol^(-1) Average bond energy data is C- H= 415 KJ mol^(-1) C-O = 356 KJ mol^(-1) O -H = 463 KJ mol^(-1)

    A
    `-266`
    B
    266
    C
    313
    D
    `-313`

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