(a)(i) Draw the energy profile diagram for the reaction; H\(_{2(g)}\) + I\(_{2(g)}\) \(\to\) 2Hl\(_{(g)}\); \(\Delta\)H = –13 KJ mol\(^{-1}\)
(ii) If the concentration of HI\(_{(g)}\) increases from 0.000 to 0.002 mol dm\(^{-3}\) in 80 seconds, what is the rate of t reaction?
(b)(i)Give one use of each of the following compounds: I. NaHCO\(_{3}\); II. CaSO\(_{4}\); III. CaCO\(_{3}\).
(ii) State a drying agent that can be used for each of the following gases: I. SO\(_{2}\); II. HCI; Ill. NH\(_{3}\)
(c)(i) Write an equation for the complete combustion of carbon in oxygen.
(ii) Calculate the number moles of carbon (IV) oxide produced from the complete combustion of 2.5 g of carbon. [ C = 12.0, O = 16]
(iii) Mention one use of I. carbon (II) oxide; II. carbon (IV) oxide.
(d) An industrial raw material has the following composition by mass:
Iron = 28.1%; Chlorine = 35.7%; Water cf crystallization = 36.2%.
Calculate the formula for the material. [H = 1.00, O = 16.0, CI = 35.5, Fe = 56.0]
(e) Give one example of a (i) metal that is liquid at room temperature,
(ii) non-metal that is liquid room temperature.
(a)(i) The reaction is exothermic because \(\Delta H=-13\ \text{kJ mol}^{-1}\). The products are therefore at a lower energy level than the reactants.
Energy profile diagram for \(\mathrm{H_2(g)+I_2(g)\rightarrow 2HI(g)}\). (ii)
\[\text{Rate of reaction}=\frac{\text{change in concentration of HI}}{\text{time}}\]
\[=\frac{0.002-0.000}{80}=2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\]
Rate of reaction = \(2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\).
(b)(i)
\(\mathrm{NaHCO_3}\): manufacture of baking powder. \(\mathrm{CaSO_4}\): manufacture of plaster of Paris. \(\mathrm{CaCO_3}\): manufacture of cement. (ii)
\(\mathrm{SO_2}\): concentrated \(\mathrm{H_2SO_4}\). \(\mathrm{HCl}\): concentrated \(\mathrm{H_2SO_4}\). \(\mathrm{NH_3}\): calcium oxide, \(\mathrm{CaO}\). (c)(i)
\[\mathrm{C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}}\]
(ii)
From the equation, 1 mol of carbon produces 1 mol of \(\mathrm{CO_2}\).
\[n(\mathrm C)=\frac{\text{mass}}{\text{molar mass}}=\frac{2.5}{12.0}=0.2083\ \text{mol}\]
\[n(\mathrm{CO_2})=0.2083\ \text{mol}\approx\boxed{0.208\ \text{mol}}\]
(iii)
Carbon(II) oxide, \(\mathrm{CO}\): reducing agent in the extraction of metals. Carbon(IV) oxide, \(\mathrm{CO_2}\): used in fire extinguishers. (d)
Using 100 g of the material:
Constituent Mass (g) Number of moles Mole ratio Fe 28.1 \(\frac{28.1}{56.0}=0.502\) \(\frac{0.502}{0.502}=1\) Cl 35.7 \(\frac{35.7}{35.5}=1.006\) \(\frac{1.006}{0.502}=2\) \(\mathrm{H_2O}\) 36.2 \(\frac{36.2}{18.0}=2.011\) \(\frac{2.011}{0.502}=4\)
Therefore, the formula of the material is \(\boxed{\mathrm{FeCl_2\cdot4H_2O}}\).
(e)
(i) Mercury, \(\mathrm{Hg}\).
(ii) Bromine, \(\mathrm{Br_2}\).
(a)(i) The reaction is exothermic because \(\Delta H=-13\ \text{kJ mol}^{-1}\). The products are therefore at a lower energy level than the reactants.
Energy profile diagram for \(\mathrm{H_2(g)+I_2(g)\rightarrow 2HI(g)}\). (ii)
\[\text{Rate of reaction}=\frac{\text{change in concentration of HI}}{\text{time}}\]
\[=\frac{0.002-0.000}{80}=2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\]
Rate of reaction = \(2.5\times10^{-5}\ \text{mol dm}^{-3}\text{ s}^{-1}\).
(b)(i)
\(\mathrm{NaHCO_3}\): manufacture of baking powder. \(\mathrm{CaSO_4}\): manufacture of plaster of Paris. \(\mathrm{CaCO_3}\): manufacture of cement. (ii)
\(\mathrm{SO_2}\): concentrated \(\mathrm{H_2SO_4}\). \(\mathrm{HCl}\): concentrated \(\mathrm{H_2SO_4}\). \(\mathrm{NH_3}\): calcium oxide, \(\mathrm{CaO}\). (c)(i)
\[\mathrm{C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}}\]
(ii)
From the equation, 1 mol of carbon produces 1 mol of \(\mathrm{CO_2}\).
\[n(\mathrm C)=\frac{\text{mass}}{\text{molar mass}}=\frac{2.5}{12.0}=0.2083\ \text{mol}\]
\[n(\mathrm{CO_2})=0.2083\ \text{mol}\approx\boxed{0.208\ \text{mol}}\]
(iii)
Carbon(II) oxide, \(\mathrm{CO}\): reducing agent in the extraction of metals. Carbon(IV) oxide, \(\mathrm{CO_2}\): used in fire extinguishers. (d)
Using 100 g of the material:
Constituent Mass (g) Number of moles Mole ratio Fe 28.1 \(\frac{28.1}{56.0}=0.502\) \(\frac{0.502}{0.502}=1\) Cl 35.7 \(\frac{35.7}{35.5}=1.006\) \(\frac{1.006}{0.502}=2\) \(\mathrm{H_2O}\) 36.2 \(\frac{36.2}{18.0}=2.011\) \(\frac{2.011}{0.502}=4\)
Therefore, the formula of the material is \(\boxed{\mathrm{FeCl_2\cdot4H_2O}}\).
(e)
(i) Mercury, \(\mathrm{Hg}\).
(ii) Bromine, \(\mathrm{Br_2}\).
