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    In the vapour state, hydrogen and iodine undergo the following reaction.
    H2(g) + I2(g) (reversible arrows) 2HI(g)

    reaction 2.1Write an equation, including state symbols, for the bond enthalpy of
    I - I.

    The answer is I-I(g) -> 2I(g)

    Why is it not I2-> I +I because doesn't bond enthalpy show the bonds breaking?

    Thanks.
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    (Original post by coconut64)
    In the vapour state, hydrogen and iodine undergo the following reaction.
    H2(g) + I2(g) (reversible arrows) 2HI(g)

    reaction 2.1Write an equation, including state symbols, for the bond enthalpy of
    I - I.

    The answer is I-I(g) -> 2I(g)

    Why is it not I2-> I +I because doesn't bond enthalpy show the bonds breaking?

    Thanks.
    Same thing? I think they may have written I2 as I-I to emphasise the bond, and 2I is the way to write I + I.
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    (Original post by EricPiphany)
    Same thing? I think they may have written I2 as I-I to emphasise the bond, and 2I is the way to write I + I.
    So is this saying that the 2I bond is broken to form I + I? And why do they have to be gaseous?? Thanks.
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    (Original post by coconut64)
    So is this saying that the 2I bond is broken to form I + I? And why do they have to be gaseous?? Thanks.
    Iodine gas, which is two Iodine atoms joined with a covalent bond is written as \mathrm{I}_2. 2I implies two separate iodine atoms (or radicals).
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    (Original post by EricPiphany)
    Iodine gas, which is two Iodine atoms joined with a covalent bond is written as I_2. 2I implies two separate iodine atoms (or radicals).
    So this equation is suggesting that after bond breaking, two iodine atoms are formed?
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    (Original post by coconut64)
    So this equation is suggesting that after bond breaking, two iodine atoms are formed?
    Yep.
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    (Original post by EricPiphany)
    Yep.
    And elements in the bond enthalpy equation are always gaseous?
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    (Original post by coconut64)
    And elements in the bond enthalpy equation are always gaseous?
    Yep.
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    (Original post by EricPiphany)
    Yep.
    Aha okay thank you.
 
 
 
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