(a) The following reaction scheme is an illustration of the contact process. Study the scheme and answer the questions that follow.
(i) Name X and Y
(ii) Write a balanced chemical equation for each of the processes I, II, III and IV
(iii) Name the catalyst used in process II
(iv) Using Le Chatelier's principle, explain briefly why increasing the temperature would not favour the reaction in II
(v) State two uses of SO\(_2\)
(b) Consider the following equation: 2H\(_{2(g)}\) + O\(_{2(g)}\) \(\to\) 2H\(_2\)O\(_{(g)}\)
Calculate the volume of unused oxygen gas when 40 cm\(^3\) of hydrogen gas is sparked with 30cm\(^3\) of oxygen gas
(c) Calcium carbonate of mass 1.0 g was heated until there was no further change.
- Write an equation for the reaction which took place.
- Calculate the mass of the residue.
- Calculate the volume of the gas evolved at s.t.p.
- What would be the volume of the gas measured at 15 and 760 mm Hg? [C= 12.0, O = 16.0, Ca = 40.0, molar volume of a gas at s.t.p. = 22.4 dm\(^3\) ]
a)
(i) X is sulfur dioxide (SO2) and Y is sulfur trioxide (SO3).
(ii)
Process I: Sulfur (S) is burned to form sulfur dioxide (SO2).
S (s) + O2 (g) → SO2 (g)
Process II: Sulfur dioxide (SO2) is oxidized to form sulfur trioxide (SO3) using oxygen (O2) in the presence of a catalyst, vanadium(V) oxide (V2O5).
2SO2 (g) + O2 (g) → 2SO3 (g)
Process III: Sulfur trioxide (SO3) is dissolved in concentrated sulfuric acid (H2SO4) to form oleum (H2S2O7).
SO3 (g) + H2SO4 (l) → H2S2O7 (l)
Process IV: Oleum (H2S2O7) is diluted with water (H2O) to form concentrated sulfuric acid (H2SO4).
H2S2O7 (l) + H2O (l) → 2H2SO4 (l)
(iii) The catalyst used in process II is vanadium(V) oxide (V2O5).
(iv) In process II, the reaction is exothermic (it releases heat). According to Le Chatelier's principle, increasing the temperature would shift the equilibrium in the opposite direction, favoring the reactants (SO2 and O2). Therefore, increasing the temperature would not favor the forward reaction in process II.
(v) Two uses of SO2 are:
- - It is used in the production of sulfuric acid (H2SO4), which is an important industrial chemical.
- - It is used as a preservative for dried fruits, such as apricots, to prevent them from turning brown.
b)
The equation given is:
2H2 (g) + O2 (g) → 2H2O (g)
To solve the problem, we can use the mole ratio
a)
(i) X is sulfur dioxide (SO2) and Y is sulfur trioxide (SO3).
(ii)
Process I: Sulfur (S) is burned to form sulfur dioxide (SO2).
S (s) + O2 (g) → SO2 (g)
Process II: Sulfur dioxide (SO2) is oxidized to form sulfur trioxide (SO3) using oxygen (O2) in the presence of a catalyst, vanadium(V) oxide (V2O5).
2SO2 (g) + O2 (g) → 2SO3 (g)
Process III: Sulfur trioxide (SO3) is dissolved in concentrated sulfuric acid (H2SO4) to form oleum (H2S2O7).
SO3 (g) + H2SO4 (l) → H2S2O7 (l)
Process IV: Oleum (H2S2O7) is diluted with water (H2O) to form concentrated sulfuric acid (H2SO4).
H2S2O7 (l) + H2O (l) → 2H2SO4 (l)
(iii) The catalyst used in process II is vanadium(V) oxide (V2O5).
(iv) In process II, the reaction is exothermic (it releases heat). According to Le Chatelier's principle, increasing the temperature would shift the equilibrium in the opposite direction, favoring the reactants (SO2 and O2). Therefore, increasing the temperature would not favor the forward reaction in process II.
(v) Two uses of SO2 are:
- - It is used in the production of sulfuric acid (H2SO4), which is an important industrial chemical.
- - It is used as a preservative for dried fruits, such as apricots, to prevent them from turning brown.
b)
The equation given is:
2H2 (g) + O2 (g) → 2H2O (g)
To solve the problem, we can use the mole ratio
