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The standard' enthalpy'of formation of d...

The standard' enthalpy'of formation of diamond is not taken as zero, whereas that for graphite is take as zero. Give reason.

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Standard molar enthalpy of formation of CO_(2) is equal to

Knowledge Check

  • Standard molar enthalpy of formation of 'CO_2', is equal to

    A
    Zero
    B
    standard molar enthalpy of combustion
    C
    the sum of standard molar enthalpies of formation of 'CO' and 'O_2'
    D
    the standard molar enthalpy of combustion of carbon (graphite)

  • Standard molar enthalpy of formation of CO_(2) is equal to

    A
    Zero.
    B
    The standard molar enthalpy of combustion of gaseous carbon.
    C
    The sum of standard molar enthalpies of formation of` CO_(2) and O_(2)`
    D
    The standard molar enthalpy of combustion of carbon (graphite).

  • The standard enthalpy of formation of NH_(2) is -46 kJ mol^(-1). If the enthalpy of formation of H, from its atoms is -436 kJ mol^(-1) and that of N_(2) is - 712 kJ mol^(-1). the average bond enthalpy of N-H bond in NH_(3) is:

    A
    `- 1056 kJ mol ^(-1)`
    B
    `- 1102 kJmol ^(-1) `
    C
    `- 9 64 kJ mol ^(-1)`
    D
    `+ 352 kJ mol ^(-1)`

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    The standard enthalpy of formation of NH_(3) is 46 kJ mol^(-1). If the enthalpy of formation of H_(2) from its atc is 436 kJ mol^(-1) and that of N_(2) is -712 kJ mol^(-1) the average bond enthalpy of N-H bond in NH_(3) is:

    Give justification, The first lonisation enthalpy of carbon is greater than that of boron, whereas the reverse is correct for the second ionisation enthalpy.

    Unicellular organisms are immortal, whereas multicellular organisms are not. Give reason.

    Silianes get easily hydrolysed whereas alkanes do not. Give reasons.

    Assertion : The enthalpy of both graphite and diamond is taken to be zero, being elementary substances. Reason : The enthalpy of formation of an elementary substance in any state is taken as zero.

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