Sulphuric Acid — ICSE Class 10 Chemistry Important Questions

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

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

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

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

Answer: (c) White.

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

Answer: (b) Oxidising agent.

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

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

The dehydrating property is the ability of concentrated sulphuric acid to remove the chemically combined elements of water (hydrogen and oxygen in the ratio $2:1$2:1) from a compound.

Example - charring of sugar:
$C_{12}H_{22}O_{11}\xrightarrow{H_2SO_4}12C+11H_2O$C_12H_22O_11→ (H_2SO_4)12C+11H_2O

White sugar turns into a spongy black mass of carbon.

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:
$BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl$BaCl_2+H_2SO_4→ BaSO_4+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+H_2SO_4(dil.)→ ZnSO_4+H_2

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

(iii) With sodium carbonate (liberates carbon dioxide):
$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+CO_2\uparrow$Na_2CO_3+H_2SO_4→ Na_2SO_4+H_2O+CO_2

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

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

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

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

Sulphur is oxidised to $SO_2$SO_2; the acid is again reduced to $SO_2$SO_2. 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 $SO_2$SO_2 (burning sulphur or roasting sulphide ores):
$S+O_2\rightarrow SO_2$S+O_2→ SO_2

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

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

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

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

(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+H_2SO_4→ (Δ)NaHSO_4+HCl

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

(c) Observation: the blue crystals turn white. Concentrated sulphuric acid, being a dehydrating agent, removes the water of crystallisation:
$CuSO_4\cdot5H_2O\xrightarrow{H_2SO_4}CuSO_4+5H_2O$CuSO_4·5H_2O→ (H_2SO_4)CuSO_4+5H_2O

leaving white anhydrous copper(II) sulphate.

(i) The dehydrating property of concentrated sulphuric acid.

(ii) $C_{12}H_{22}O_{11}\xrightarrow{H_2SO_4}12C+11H_2O$C_12H_22O_11→ (H_2SO_4)12C+11H_2O

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

(iv) The same dehydrating property is shown when conc. $H_2SO_4$H_2SO_4 turns blue hydrated copper(II) sulphate crystals ($CuSO_4\cdot5H_2O$CuSO_4·5H_2O) white by removing the water of crystallisation. (Alternatively, it chars formic acid to $CO$CO or oxalic acid to $CO$CO and $CO_2$CO_2.)