(a)(i) Draw the energy profile diagram for the reaction \(\mathrm{H_{2(g)} + I_{2(g)} \longrightarrow 2HI_{(g)}}\) \(\Delta =\) —13 kJmol\(^3\) (ii) If the ...
Question 1 Report
(a)(i) Draw the energy profile diagram for the reaction
\(\mathrm{H_{2(g)} + I_{2(g)} \longrightarrow 2HI_{(g)}}\) \(\Delta =\) —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) \(\mathrm{K_4Fe(CN)_6}\) (ii) \(\mathrm{(NH_4)_2Fe(SO_4)_2 6H_2O}\) (iii) \(\mathrm{Mg(OH)NO_3}\) (iv) \(\mathrm{NaH_2PO_4}\).
(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: \(\mathrm{CaCO_3}\), \(\mathrm{CaSO_4}\), \(\mathrm{NaHCO_3}\).
(ii) State a drying agent for each of the following gases: i. \(\mathrm{NH_3}\), II. HCI Ill. \(\mathrm{SO_4}\).
(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 ].
Answer Details
(a)(i) Energy profile diagram
The reaction is exothermic, since \(\Delta H=-13\ \text{kJ mol}^{-1}\). Therefore, the products are at a lower energy level than the reactants.
(ii) Rate of reaction
\[\text{Rate}=\frac{\text{change in concentration of HI}}{\text{time taken}}=\frac{0.001-0}{50}=2.0\times10^{-5}\ \text{mol dm}^{-3}\text{s}^{-1}.\]
(b) Types of salts
- \(\mathrm{K_4Fe(CN)_6}\): complex salt.
- \(\mathrm{(NH_4)_2Fe(SO_4)_2\cdot6H_2O}\): double salt.
- \(\mathrm{Mg(OH)NO_3}\): basic salt.
- \(\mathrm{NaH_2PO_4}\): acid salt.
(c) Action of carbon(IV) oxide on lime water
Initially, carbon(IV) oxide reacts with lime water to form insoluble calcium trioxocarbonate(IV), \(\mathrm{CaCO_3}\), which is a white precipitate. This makes the lime water milky.
\[\mathrm{Ca(OH)_2(aq)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)}\]
On passing excess carbon(IV) oxide, the calcium trioxocarbonate(IV) dissolves to form soluble calcium hydrogentrioxocarbonate(IV). The solution therefore becomes clear.
\[\mathrm{CaCO_3(s)+CO_2(g)+H_2O(l)\rightarrow Ca(HCO_3)_2(aq)}\]
(d)(i) Uses
- \(\mathrm{CaCO_3}\): manufacture of cement.
- \(\mathrm{CaSO_4}\): manufacture of plaster of Paris for casts.
- \(\mathrm{NaHCO_3}\): used in baking powder.
(d)(ii) Suitable drying agents
- \(\mathrm{NH_3}\): calcium oxide, \(\mathrm{CaO}\).
- \(\mathrm{HCl}\): concentrated tetraoxosulphate(VI) acid, \(\mathrm{H_2SO_4}\).
- \(\mathrm{SO_2}\): concentrated tetraoxosulphate(VI) acid, \(\mathrm{H_2SO_4}\).
(d)(iii) Ammonia as a reducing agent
Ammonia reduces copper(II) oxide to copper:
\[\mathrm{2NH_3(g)+3CuO(s)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)}\]
(e) Formula of the industrial raw material
| Constituent | Mass in 100 g | Moles | Mole ratio |
|---|---|---|---|
| Fe | 28.1 g | \(28.1/56=0.502\) | \(0.502/0.502=1\) |
| Cl | 35.7 g | \(35.7/35.5=1.006\) | \(1.006/0.502=2\) |
| \(\mathrm{H_2O}\) | 36.2 g | \(36.2/18=2.011\) | \(2.011/0.502=4\) |
Hence, the mole ratio is \(\mathrm{Fe:Cl:H_2O}=1:2:4\).
\[\boxed{\mathrm{FeCl_2\cdot4H_2O}}\]
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