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# Bond enthalpy watch

1. 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.
2. (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.
3. (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.
4. (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 . 2I implies two separate iodine atoms (or radicals).
5. (Original post by EricPiphany)
Iodine gas, which is two Iodine atoms joined with a covalent bond is written as . 2I implies two separate iodine atoms (or radicals).
So this equation is suggesting that after bond breaking, two iodine atoms are formed?
6. (Original post by coconut64)
So this equation is suggesting that after bond breaking, two iodine atoms are formed?
Yep.
7. (Original post by EricPiphany)
Yep.
And elements in the bond enthalpy equation are always gaseous?
8. (Original post by coconut64)
And elements in the bond enthalpy equation are always gaseous?
Yep.
9. (Original post by EricPiphany)
Yep.
Aha okay thank you.

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