Week 11 · Chapter 3 (B): surface area and concentration, the law of mass action, the equilibrium constant Kc and what its size means (short, calculation-light text: two and a half book pages, add worked examples and prediction tasks) · 3 lessons
1
Surface area and concentration: more collisions
Chemistry Sec 3 Term 1, Chapter 3: pp. 59–61
GoalExplain, using collisions, why a larger surface area and a higher concentration make a reaction faster.
New words
surface area · the total area of a solid that is in contact with the other reactantcollision · a meeting of two particles; a reaction can happen only when particles meetconcentration · the amount of dissolved substance in each litre of solution, in mol/L
Explanation
A reaction needs particles to meet. In a solid only the particles at the surface can meet the other reactant, so grinding a lump into powder exposes far more particles without changing the mass. Take zinc and dilute hydrochloric acid on paper: a gram of zinc powder is used up much sooner than a gram of zinc block, yet both give the same amount of hydrogen in the end, because the amount of zinc is the same. Concentration works in the same way for dissolved particles. One litre of a 0.40 mol/L solution holds twice as many particles as one litre of a 0.20 mol/L solution, so each particle meets partners more often and the rate rises. A larger surface and a higher concentration therefore change how fast a reaction gets there, not how much a complete reaction finally gives. Both belong to the factors of week 10, and both come from the same idea: more collisions in each second.
One cube of volume 1 cm³ cut into smaller cubes
Pieces
Side of one piece
Surface of one piece
Total surface
1
1 cm
6 cm²
6 cm²
8
0.5 cm
1.5 cm²
12 cm²
64
0.25 cm
0.375 cm²
24 cm²
1000
0.1 cm
0.06 cm²
60 cm²
Worked examples
Cutting a cube. A cube of side 1 cm has six faces of 1 cm², so its surface is 6 cm² and its volume is 1 cm³. Cut it into 8 equal cubes: each has side 0.5 cm and surface 6 × 0.25 = 1.5 cm², so the total surface is 8 × 1.5 = 12 cm², while the volume is still 8 × 0.125 = 1 cm³. Cut into 1000 cubes of side 0.1 cm and the total surface is 1000 × 0.06 = 60 cm². The mass is unchanged, and the exposed surface is ten times larger.
Same amount, different speed (paper only). 6.5 g of zinc (Zn = 65) is 6.5 ÷ 65 = 0.10 mol. In Zn + 2HCl → ZnCl₂ + H₂ the ratio Zn : H₂ is 1 : 1, so 0.10 mol of H₂ forms, which is 0.10 × 22.4 = 2.24 L at STP, whether the zinc is powder or one block, as long as the acid is in excess. The powder reaches 2.24 L sooner, so its curve is steeper, but both curves end at the same height. The times in the picture are invented.
Concentration means particles in the same space. 50 mL of a 0.20 mol/L solution holds 0.20 × 0.050 = 0.010 mol of solute. 50 mL of a 0.40 mol/L solution holds 0.40 × 0.050 = 0.020 mol, twice as many particles in the same volume, so about twice as many meetings each second. This is also why a rate falls as a reaction goes on: the reactants are used up, their concentration falls, and the collisions become less frequent (the curves of week 10).
Class activity
Vinegar strength test, with a grown-up. Put one level teaspoon of baking soda in each of two clear cups. Into cup A pour 4 tablespoons of kitchen vinegar. Into cup B pour 2 tablespoons of vinegar mixed with 2 tablespoons of cool tap water. Pour both at the same moment and count the bubbles you see rising in ten seconds. See what happens: the full-strength vinegar usually fizzes faster, because each spoonful holds more acid particles. Leave the cups open, never taste them, and wash up. Then write which factor you changed.
Practice
1
A 2 cm cube (surface 24 cm²) is cut into eight cubes of side 1 cm. Find the new total surface and say how many times larger it is.
Each small cube has 6 × 1 = 6 cm², so the total is 8 × 6 = 48 cm², which is 48 ÷ 24 = 2 times larger.
2
13 g of zinc (Zn = 65) reacts with excess dilute acid, once as powder and once as one block. What volume of hydrogen forms at STP in each case, and which finishes first?
n = 13 ÷ 65 = 0.20 mol, and the ratio Zn : H₂ is 1 : 1, so V = 0.20 × 22.4 = 4.48 L in both cases. The powder finishes first because its surface is larger.
3
Use collisions to explain why the rate of a reaction falls as the reactants are used up.
The concentration of the reactants falls, so there are fewer particles in each litre, fewer collisions each second, and a lower rate.
4
How many moles of solute are in 250 mL of a 0.40 mol/L solution, and in 250 mL of a 0.80 mol/L solution? Which reacts faster with the same partner, and why?
0.40 × 0.250 = 0.10 mol and 0.80 × 0.250 = 0.20 mol. The 0.80 mol/L solution is faster: twice the particles in the same volume mean more collisions each second.
