CE Level Chemistry Qu : about conc. sulpuric acid

其實照(1)既eqt.睇黎,都係想form chlorine gas姐,
但係chlorine gas 既oxdizing power 係勁過sulphuric acid,
so 係form唔到架。

Concentrated sulphuric acid is a high-boiling-point acid (or simply, a non-volatile acid) which can produce volatile acids when reacting with the salts of the volatile acids. Therefore, concentrated sulphuric acid would react with sodium chloride, sodium bromide and sodium iodide to give hydrogen chloride gas, hydrogen bromide gas and hydrogen iodide gas respectively.

However, concentrated sulphuric acid is an oxidizing agent. The reducing power of hydrogen chloride gas is too weak to react with concentrated sulphuric acid, but the reducing power of hydrogen bromide gas and that of hydrogen iodide gas are strong enough to reduce sulphuric acid to give bromine and iodine respectively.

(1) is incorrect.
Sodium chloride reacts with concentrated sulphuric acid to give hydrogen chloride gas.
NaCl + H2SO4 → NaHSO4 + HCl
There is no reaction between the HCl and the excess concentrated H2SO4.

(2) is correct.
Sodium bromide reacts with concentrated sulphuric acid to give hydrogen bromide gas.
NaBr + H2SO4 → NaHSO4 + HBr
HBr is a moderately strong reducing agent which can undergo redox react with the excess concentrated H2SO4. The HBr is oxidized to Br2, while the concentrated H2SO4 is reduced to SO2 gas.
2HBr + H2SO4 → 2H2O + SO2 + Br2

(3) is correct.
Sodium iodide reacts with concentrated sulphuric acid to give hydrogen iodide gas.
NaI + H2SO4 → NaHSO4 + HI
HI is a stronger reducing agent than HBr. The HI formed can undergo redox react with the excess concentrated H2SO4. The HI is oxidized to I2, while the concentrated H2SO4 is reduced to H2S gas.
8HI + H2SO4 → H2S + 4H2O + 4I2