(a) State two factors which can affect the rate of a chemical reaction.
(b) 0.72g of magnesium was added to different volumes of 2 mol. per dm\(^{3}\) hydrochloric acid. The volume of liberated was as measured at room temperature and pressure. The result of the experiment was as tabulated
| vol. of 2 mol. per dm\(^{3}\) HCl used (cm\(^{3}\) |
Vol. of H\(_2\) evolved in cm\(^{3}\) (to the nearest 10cm\(^{3}\)) |
| 5 |
120 |
| 15 |
360 |
| 25 |
550 |
| 35 |
600 |
| 45 |
600 |
Use the data in the table to plot a graph of the volume of hydrogen liberated against the volume of acid used.
(c) From the graph in (b) above, determine the volume of: (i) hydrogen that would be produced if 50 cm\(^{3}\) of the acid were added to 0.72g of magnesium.
(ii) the acid which must be added to 0.72 g of magnesium to produce 480 cm\(^{3}\) of hydrogen;
(iii) the acid needed exactly to dissolve 0.72 g of magnesium completely.
(d) Explain your answer to (c)(iii).
(e) From your answers to (c) above, deduce the: (i) volume of the acid which will dissolve 1 mole of magnesium completely. (Mg = 24)
(ii) volume of hydrogen that would be liberated if 1 mole of magnesium dissolves completely in the acid;
(iii) equation for the reaction between magnesium and hydrochloric acid. Show clearly how you arrived at you answers
(a) Two factors that affect the rate of a chemical reaction: temperature and concentration of the reactants. (Surface area, a catalyst and, for gases, pressure are also acceptable.)
(b) The data to be plotted are:
| Vol. of 2 mol dm-3 HCl used (cm3) | 5 | 15 | 25 | 35 | 45 |
|---|
| Vol. of H2 evolved (cm3) | 120 | 360 | 550 | 600 | 600 |
Plot volume of H2 (y-axis) against volume of acid (x-axis). The graph rises steeply and in direct proportion at first (about 24 cm3 of H2 per 1 cm3 of acid), then bends over and becomes a horizontal line (a plateau) at 600 cm3 once all the magnesium has reacted.
(c) From the graph:
- (i) At 50 cm3 of acid the curve is already on the plateau, so the volume of hydrogen is 600 cm3 (adding more acid gives no more gas because the magnesium is used up).
- (ii) To produce 480 cm3 of hydrogen, read on the rising (proportional) part of the graph. Since \(120\ \text{cm}^3\) H2 corresponds to \(5\ \text{cm}^3\) acid, the ratio is \(24\ \text{cm}^3\) H2 per cm3 acid:
\[V_{acid} = \frac{480}{24} = \mathbf{20\ cm^3}\]
- (iii) The acid needed to dissolve the magnesium exactly is the volume at which the graph just reaches the plateau (600 cm3), which is read as 30 cm3.
(d) At 30 cm3 of acid, the acid is exactly enough to react with all 0.72 g of magnesium; this is the point where the maximum volume of hydrogen (600 cm3) is first produced. Below this volume the acid is the limiting reactant (less hydrogen). Above it, the magnesium is completely used up and is now the limiting reactant, so no extra hydrogen is formed and the graph stays flat.
(e) First find the moles of magnesium used:
\[n_{Mg} = \frac{0.72}{24} = 0.03\ \text{mol}\]
(i) From (c)(iii), 0.03 mol Mg needs 30 cm3 of the acid. For 1 mole:
\[V = \frac{30}{0.03} = 1000\ \text{cm}^3 = \mathbf{1\ dm^3}\]
(ii) From the plateau, 0.03 mol Mg liberates 600 cm3 of hydrogen. For 1 mole:
\[V_{H_2} = \frac{600}{0.03} = 20000\ \text{cm}^3 = \mathbf{20\ dm^3}\]
(iii) Deduce the mole ratios. The acid is 2 mol dm-3, so 1 dm3 contains 2 mol HCl. Therefore:
- 1 mol Mg reacts with 2 mol HCl (from (e)(i)).
- 1 mol Mg produces about 1 mol H2 (20 dm3 is close to the molar gas volume, from (e)(ii)).
This gives the ratio Mg : HCl : H2 = 1 : 2 : 1, so the equation is:
\[\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)\]
(a) Two factors that affect the rate of a chemical reaction: temperature and concentration of the reactants. (Surface area, a catalyst and, for gases, pressure are also acceptable.)
(b) The data to be plotted are:
| Vol. of 2 mol dm-3 HCl used (cm3) | 5 | 15 | 25 | 35 | 45 |
|---|
| Vol. of H2 evolved (cm3) | 120 | 360 | 550 | 600 | 600 |
Plot volume of H2 (y-axis) against volume of acid (x-axis). The graph rises steeply and in direct proportion at first (about 24 cm3 of H2 per 1 cm3 of acid), then bends over and becomes a horizontal line (a plateau) at 600 cm3 once all the magnesium has reacted.
(c) From the graph:
- (i) At 50 cm3 of acid the curve is already on the plateau, so the volume of hydrogen is 600 cm3 (adding more acid gives no more gas because the magnesium is used up).
- (ii) To produce 480 cm3 of hydrogen, read on the rising (proportional) part of the graph. Since \(120\ \text{cm}^3\) H2 corresponds to \(5\ \text{cm}^3\) acid, the ratio is \(24\ \text{cm}^3\) H2 per cm3 acid:
\[V_{acid} = \frac{480}{24} = \mathbf{20\ cm^3}\]
- (iii) The acid needed to dissolve the magnesium exactly is the volume at which the graph just reaches the plateau (600 cm3), which is read as 30 cm3.
(d) At 30 cm3 of acid, the acid is exactly enough to react with all 0.72 g of magnesium; this is the point where the maximum volume of hydrogen (600 cm3) is first produced. Below this volume the acid is the limiting reactant (less hydrogen). Above it, the magnesium is completely used up and is now the limiting reactant, so no extra hydrogen is formed and the graph stays flat.
(e) First find the moles of magnesium used:
\[n_{Mg} = \frac{0.72}{24} = 0.03\ \text{mol}\]
(i) From (c)(iii), 0.03 mol Mg needs 30 cm3 of the acid. For 1 mole:
\[V = \frac{30}{0.03} = 1000\ \text{cm}^3 = \mathbf{1\ dm^3}\]
(ii) From the plateau, 0.03 mol Mg liberates 600 cm3 of hydrogen. For 1 mole:
\[V_{H_2} = \frac{600}{0.03} = 20000\ \text{cm}^3 = \mathbf{20\ dm^3}\]
(iii) Deduce the mole ratios. The acid is 2 mol dm-3, so 1 dm3 contains 2 mol HCl. Therefore:
- 1 mol Mg reacts with 2 mol HCl (from (e)(i)).
- 1 mol Mg produces about 1 mol H2 (20 dm3 is close to the molar gas volume, from (e)(ii)).
This gives the ratio Mg : HCl : H2 = 1 : 2 : 1, so the equation is:
\[\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)\]
