Sulphuric Acid — ICSE Class 10 Chemistry Important Questions

Answer: (c) Vanadium pentoxide ($V_2O_5$V₂O₅).

$2SO_2+O_2\xrightleftharpoons[V_2O_5]{}2SO_3$2SO₂+O₂[V₂O₅]2SO₃ is the key catalytic step of the Contact process.

Answer: (b) Direct absorption in water forms a dense corrosive mist of sulphuric acid.

So $SO_3$SO₃ is absorbed in conc. $H_2SO_4$H₂SO₄ to form oleum ($H_2S_2O_7$H₂S₂O₇), which is then diluted.

Answer: (c) White.

Conc. $H_2SO_4$H₂SO₄ is a dehydrating agent: $CuSO_4\cdot5H_2O\xrightarrow{H_2SO_4}CuSO_4+5H_2O$CuSO₄·5H₂O→ (H₂SO₄)CuSO₄+5H₂O, so blue hydrated crystals turn white anhydrous $CuSO_4$CuSO₄.

Answer: (b) Oxidising agent.

$C+2H_2SO_4(conc.)\rightarrow CO_2+2SO_2+2H_2O$C+2H₂SO₄(conc.)→ CO₂+2SO₂+2H₂O. Carbon is oxidised to $CO_2$CO₂; $H_2SO_4$H₂SO₄ is reduced to $SO_2$SO₂, so it acts as an oxidising agent.

Answer: (a) $NaCl+H_2SO_4\rightarrow NaHSO_4+HCl\uparrow$NaCl+H₂SO₄→ NaHSO₄+HCl↑. Being non-volatile, $H_2SO_4$H₂SO₄ displaces the volatile acid $HCl$HCl; R correctly explains A.

(i) $Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\uparrow$Na₂SO₃+H₂SO₄→ Na₂SO₄+H₂O+SO₂↑ (sulphur dioxide, smell of burning sulphur).

(ii) $FeS+H_2SO_4\rightarrow FeSO_4+H_2S\uparrow$FeS+H₂SO₄→ FeSO₄+H₂S↑ (hydrogen sulphide, smell of rotten eggs).

To the solution add a little dilute hydrochloric acid followed by barium chloride solution.

A white precipitate insoluble in dilute acids confirms the sulphate ion:
$BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl$BaCl₂+H₂SO₄→ BaSO₄↓+2HCl

(The dilute $HCl$HCl dissolves interfering sulphite/carbonate, whose barium salts are acid-soluble.)

(i) With zinc (active metal, liberates hydrogen):
$Zn+H_2SO_4(dil.)\rightarrow ZnSO_4+H_2\uparrow$Zn+H₂SO₄(dil.)→ ZnSO₄+H₂↑

(ii) With copper(II) oxide (a base):
$CuO+H_2SO_4\rightarrow CuSO_4+H_2O$CuO+H₂SO₄→ CuSO₄+H₂O

(iii) With sodium carbonate (liberates carbon dioxide):
$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+CO_2\uparrow$Na₂CO₃+H₂SO₄→ Na₂SO₄+H₂O+CO₂↑

In each case hot concentrated $H_2SO_4$H₂SO₄ is itself reduced to sulphur dioxide.

(i) With copper:
$Cu+2H_2SO_4(conc.)\xrightarrow{\Delta}CuSO_4+SO_2\uparrow+2H_2O$Cu+2H₂SO₄(conc.)→ (Δ)CuSO₄+SO₂↑+2H₂O

Copper is oxidised to $Cu^{2+}$Cu^2+; the acid is reduced to $SO_2$SO₂.

(ii) With sulphur:
$S+2H_2SO_4(conc.)\rightarrow 3SO_2\uparrow+2H_2O$S+2H₂SO₄(conc.)→ 3SO₂↑+2H₂O

Sulphur is oxidised to $SO_2$SO₂; the acid is again reduced to $SO_2$SO₂. In both cases the reduction product of the acid is sulphur dioxide.

The dilution of concentrated sulphuric acid with water is a highly exothermic process, releasing a large amount of heat.

If water is added to the acid, the small amount of water heats up rapidly, may boil suddenly and spurt the corrosive acid out of the container, causing burns.

Therefore the acid is added slowly, in a thin stream, to a large volume of water with constant stirring, so that the heat is distributed and safely dissipated by the large mass of water.

In the Contact process sulphuric acid is manufactured from sulphur dioxide.

Step 1 - production of $SO_2$SO₂ (burning sulphur or roasting sulphide ores):
$S+O_2\rightarrow SO_2$S+O₂→ SO₂

Step 2 - catalytic oxidation of $SO_2$SO₂ (the key reversible step):
$2SO_2+O_2\xrightleftharpoons[V_2O_5,\ 450^\circ C,\ 2\ atm]{}2SO_3$2SO₂+O₂[V₂O₅, 450° C, 2 atm]2SO₃

The purified, dry gases are passed over vanadium pentoxide at about $450^\circ C$450° C.

Step 3 - formation of oleum: $SO_3$SO₃ is absorbed in concentrated sulphuric acid (not water, to avoid an acid mist):
$SO_3+H_2SO_4\rightarrow H_2S_2O_7$SO₃+H₂SO₄→ H₂S₂O₇ (oleum / fuming sulphuric acid)

Step 4 - dilution: oleum is diluted with the calculated amount of water:
$H_2S_2O_7+H_2O\rightarrow 2H_2SO_4$H₂S₂O₇+H₂O→ 2H₂SO₄

(a) Uses of sulphuric acid:

1. Manufacture of fertilisers such as ammonium sulphate and superphosphate of lime.
2. In lead storage batteries (accumulators) as the electrolyte.
3. Refining of petroleum and pickling of metals before galvanising.
4. Manufacture of other chemicals - detergents, paints, dyes, explosives, and other acids.

(b) As a non-volatile acid (displacing a volatile acid from its salt):
$NaCl+H_2SO_4\xrightarrow{\Delta}NaHSO_4+HCl\uparrow$NaCl+H₂SO₄→ (Δ)NaHSO₄+HCl↑

Being non-volatile with a high boiling point, $H_2SO_4$H₂SO₄ drives out the more volatile $HCl$HCl.

(c) One molecule of $H_2SO_4$H₂SO₄ has two replaceable hydrogen ions, so it is dibasic and forms two kinds of salt with sodium hydroxide:

$NaOH+H_2SO_4\rightarrow NaHSO_4+H_2O$NaOH+H₂SO₄→ NaHSO₄+H₂O (acid salt, sodium hydrogen sulphate)

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$2NaOH+H₂SO₄→ Na₂SO₄+2H₂O (normal salt, sodium sulphate)

(i) The dehydrating property of concentrated sulphuric acid.

(ii) $C_{12}H_{22}O_{11}\xrightarrow{H_2SO_4}12C+11H_2O$C₁₂H₂₂O₁₁→ (H₂SO₄)12C+11H₂O

(iii) The black solid left behind is carbon (a spongy mass of carbon).

(iv) The same dehydrating property is shown when conc. $H_2SO_4$H₂SO₄ turns blue hydrated copper(II) sulphate crystals ($CuSO_4\cdot5H_2O$CuSO₄·5H₂O) white by removing the water of crystallisation. (Alternatively, it chars formic acid to $CO$CO or oxalic acid to $CO$CO and $CO_2$CO₂.)