hmmm asked in Science & MathematicsChemistry · 1 decade ago

Calculate the standard enthalpy of formation for diamond, given?

Calculate the standard enthalpy of formation for diamond, given that

C(graphite) + O2(g) → CO2(g) ΔH0 = –393.5 kJ/mol

C(diamond) + O2(g) → CO2(g) ΔH0 = –395.4 kJ/mol

Is it 1.9 or -1.9 and why?

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  • 1 decade ago
    Best Answer

    It has to be 1.9 kJ/mol.

    The reason being, you are looking for the enthalpy of FORMATION of DIAMOND.

    So when you apply Hess' Law, the second equation MUST be flipped so you get this equation...remember when you reverse the reaction, you must change the sign of the enthalpy (H).

    CO2(gas) -> C(diamond) + O2(gas) , H=+395.4 kJ/mol

    Then you retain your graphite equation and add these two equations together.

    C(graphite) + O2(g) -> CO2 (g) , H = -393.5 kJ/mol

    CO2(g) -> C(diamond) + O2(g) , H = +395.4 kJ/mol

    The end result when you add these two together is this equation...note that the CO2 and O2 gas drops out of the equation:

    C(graphite) --> C(diamond) , H = -393.5 + 395.4 = +1.9 KJ/mol

    Remember, you flipped the second equation and not the first because you wanted to FORM , i.e. you are looking for the enthalpy of FORMATION, of C(diamond)

  • pavick
    Lv 4
    3 years ago

    Calculating Change In Enthalpy

  • Anonymous
    4 years ago

    This Site Might Help You.

    RE:

    Calculate the standard enthalpy of formation for diamond, given?

    Calculate the standard enthalpy of formation for diamond, given that

    C(graphite) + O2(g) → CO2(g) ΔH0 = –393.5 kJ/mol

    C(diamond) + O2(g) → CO2(g) ΔH0 = –395.4 kJ/mol

    Is it 1.9 or -1.9 and why?

    Source(s): calculate standard enthalpy formation diamond given: https://tr.im/Qj6AS
  • 1 decade ago

    If you reverse the second reaction, then add the two togther, the CO2 and O2 cancel out

    C (graphite) + O2 (g) ---> CO2 (g)

    CO2 (g) ----> C (diamond) + O2 (g)

    and you get

    C(graphite) ---> C(diamond)

    By Hess's Law, you change the sign of the enthalpy for the second reaction (because you reversed it), and add it to the enthalpy of the first.

    So, the enthalpy change is -393.5 kJ/mol + (+395.4 kJ/mol)

    or +1.9 kJ/mol.

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  • 3 years ago

    C Graphite

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