Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Metal Halides help

Ok, so my problem is that when for example, sodium iodide reacts with concentrated sulphuric acid, there seems to be two possibilities:

NaI + H2SO4 -> NaHSO4 + HI

or

2NaI + 2H2SO4 -> Na2SO4 + 2H2O + I2 + SO2

How do you know when to put one and not the other, or does it even matter?

are the conditions for reacting the same?

OP

There is no mention of conditions in the question. According to the mark scheme, for NaCl the first equations applies but it's the second for NaI. In both cases concentrated sulphuric acid is added to the solid sodium halide.

I is a stronger reducing agent than Cl so the reaction goes further (not just to the hydrogen halide stage)

OP

Ah right, that'll be why then.

So is bromine strong enough to reduce it to SO2 and H2S as well?

no only iodine can do that as strength of reducing agents decreases down the group. with HBR the reaction is

2HBr + H2SO4 --> Br2 + SO2 + 2H2O

OP

Alright cheers, I understand it now.

Thanks again :smile:

:smile:

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