(a)(i) Draw the energy profile diagram for the reaction
H\(_{2(g)}\) + I\(_{2(g)}\) ---> 2HI\(_{(g)}\) \(\Delta\) = —13 kJmol\(^3\)
(ii) If the concentration of HI increases from 0 to 0.001 mol dm\(^3}\) in 50 seconds, what is the rate of the reaction?
(b) State the type of salt represented by each of the following compounds:
(i) K\(_4\)Fe(CN)\(_6\) (ii) (NH\(_4\))\(_2\)Fe(SO\(_4\))\(_2\)6H\(_2\)O (iii) Mg(OH)NO\(_3\) (iv) NaH\(_2\)PO\(_4\).
(c) Explain, giving equations, the following observation: When carbon (IV) oxide is passed into lime water, it turns milky initially but turns clear with excess carbon (IV) oxide.
(d)(i) Give one use for each of the following compounds: CaCO\(_3\), CaSO\(_4\), NaHCO\(_3\).
(ii) State a drying agent for each of the following gases: i. NH\(_3\), II. HCI Ill. SO\(_4\).
(iii) Write an equation to illustrate the reaction of ammonia as a reducing agent.
(e) An industrial raw material has the following composition by mass:
Iron = 28.1%
Chlorine = 35.7%
Water of crystallization = 36.2%
Calculate the formula for the material. [ H = 1, 0 = 16, Cl = 35.5, Fe = 56 ].
(a)(i) Energy profile for H2(g) + I2(g) → 2HI(g), ΔH = -13 kJ mol-1
The reaction is exothermic, so the products lie below the reactants. The curve rises from the reactant level over an activation-energy hump and falls to a lower product level; the drop from reactants to products equals \( \Delta H = -13\ \text{kJ mol}^{-1} \). See the sketch.
(ii) Rate of reaction
\[ \text{Rate} = \frac{\Delta[\text{HI}]}{\Delta t} = \frac{0.001}{50} = \mathbf{2 \times 10^{-5}\ mol\,dm^{-3}\,s^{-1}} \]
(b) Types of salt
- (i) K4Fe(CN)6 - complex salt.
- (ii) (NH4)2Fe(SO4)2·6H2O - double salt.
- (iii) Mg(OH)NO3 - basic salt.
- (iv) NaH2PO4 - acid salt.
(c) CO2 passed into lime water
Initially a white precipitate of insoluble calcium carbonate forms, making the lime water milky:
\[ Ca(OH)_2 + CO_2 \to CaCO_3\downarrow + H_2O \]
With excess CO2 the precipitate dissolves to form soluble calcium hydrogencarbonate, so the mixture becomes clear:
\[ CaCO_3 + CO_2 + H_2O \to Ca(HCO_3)_2 \]
(d)(i) Uses
- CaCO3 - manufacture of cement/quicklime (or as a flux in iron extraction).
- CaSO4 - making plaster of Paris/blackboard chalk.
- NaHCO3 - baking powder (or as an antacid).
(ii) Drying agents:
- I. NH3 - calcium oxide, CaO (a basic gas cannot be dried with acidic agents).
- II. HCl - concentrated sulphuric acid.
- III. SO2 - concentrated sulphuric acid.
(iii) Ammonia as a reducing agent:
\[ 2NH_3 + 3CuO \to 3Cu + N_2 + 3H_2O \]
(e) Formula of the material
| Fe | Cl | H2O |
|---|
| % mass | 28.1 | 35.7 | 36.2 |
| ÷ molar mass | 28.1/56 = 0.502 | 35.7/35.5 = 1.006 | 36.2/18 = 2.011 |
| ÷ smallest (0.502) | 1 | 2 | 4 |
Formula \( = \mathbf{FeCl_2\cdot 4H_2O} \) (check: relative mass \(56+71+72 = 199\); Fe \(=28.1\%\), Cl \(=35.7\%\), H2O \(=36.2\%\)).
(a)(i) Energy profile for H2(g) + I2(g) → 2HI(g), ΔH = -13 kJ mol-1
The reaction is exothermic, so the products lie below the reactants. The curve rises from the reactant level over an activation-energy hump and falls to a lower product level; the drop from reactants to products equals \( \Delta H = -13\ \text{kJ mol}^{-1} \). See the sketch.
(ii) Rate of reaction
\[ \text{Rate} = \frac{\Delta[\text{HI}]}{\Delta t} = \frac{0.001}{50} = \mathbf{2 \times 10^{-5}\ mol\,dm^{-3}\,s^{-1}} \]
(b) Types of salt
- (i) K4Fe(CN)6 - complex salt.
- (ii) (NH4)2Fe(SO4)2·6H2O - double salt.
- (iii) Mg(OH)NO3 - basic salt.
- (iv) NaH2PO4 - acid salt.
(c) CO2 passed into lime water
Initially a white precipitate of insoluble calcium carbonate forms, making the lime water milky:
\[ Ca(OH)_2 + CO_2 \to CaCO_3\downarrow + H_2O \]
With excess CO2 the precipitate dissolves to form soluble calcium hydrogencarbonate, so the mixture becomes clear:
\[ CaCO_3 + CO_2 + H_2O \to Ca(HCO_3)_2 \]
(d)(i) Uses
- CaCO3 - manufacture of cement/quicklime (or as a flux in iron extraction).
- CaSO4 - making plaster of Paris/blackboard chalk.
- NaHCO3 - baking powder (or as an antacid).
(ii) Drying agents:
- I. NH3 - calcium oxide, CaO (a basic gas cannot be dried with acidic agents).
- II. HCl - concentrated sulphuric acid.
- III. SO2 - concentrated sulphuric acid.
(iii) Ammonia as a reducing agent:
\[ 2NH_3 + 3CuO \to 3Cu + N_2 + 3H_2O \]
(e) Formula of the material
| Fe | Cl | H2O |
|---|
| % mass | 28.1 | 35.7 | 36.2 |
| ÷ molar mass | 28.1/56 = 0.502 | 35.7/35.5 = 1.006 | 36.2/18 = 2.011 |
| ÷ smallest (0.502) | 1 | 2 | 4 |
Formula \( = \mathbf{FeCl_2\cdot 4H_2O} \) (check: relative mass \(56+71+72 = 199\); Fe \(=28.1\%\), Cl \(=35.7\%\), H2O \(=36.2\%\)).
