Rate of Reaction Worked ExamplesIGCSE Chemistry 0620

(a)(i) Larger chips of the same mass have a smaller surface area, so fewer acid particles can reach the solid: the rate decreases.

(ii) A higher temperature always increases the rate. (iii) Diluting the acid lowers its concentration, so the rate decreases.

(b) For gases, a higher pressure squeezes the same particles into a smaller volume, so the rate increases.

(c) A catalyst increases the rate of a reaction and is unchanged (chemically, and in mass) at the end of the reaction. (d) Biological catalysts are enzymes.

(a) The curve levels off at $80\text{ cm}^3$. (b) It first becomes horizontal at $70$ s, so the reaction stops at $70$ s.

(c) Read the volume at $20$ s: $50\text{ cm}^3$. Mean rate = volume of gas ÷ time.

(d) At $40$ s the volume is $70\text{ cm}^3$, so $20\text{ cm}^3$ is made in the next $20$ s.

The rate falls because the reactants are being used up: the magnesium gets smaller and the acid becomes less concentrated, so there are fewer collisions per second.

(e) The acid is in excess, so the magnesium is the limiting reactant. Once it has all reacted, no more hydrogen is made and the volume stays constant.

Increasing the concentration puts more acid particles in each unit of volume. The zinc surface is hit more often, so the collision frequency (collisions per second) increases. Answer: B.

A describes the effect of raising the temperature: the temperature is unchanged, so each particle has the same kinetic energy. C describes a catalyst — only a catalyst lowers the activation energy.

D describes a change in surface area: the zinc granules are the same, so the surface area is the same.

(a) Marking point 1: at a lower temperature the particles have less kinetic energy (they move more slowly).

Marking point 2: the particles therefore collide less frequently — the collision frequency decreases (fewer collisions per second).

Marking point 3: a smaller proportion of the collisions have energy equal to or greater than the activation energy, $E_a$, so a smaller fraction of collisions are successful.

(b) Marking point 1: at a lower pressure there are fewer particles per unit volume (the gas molecules are further apart). Marking point 2: so the collision frequency decreases. The energy of the particles is unchanged because the temperature is constant.

(a) Activation energy is measured from the reactants up to the top of the curve. (i) Without the catalyst: P. (ii) With the catalyst (lower, dashed curve): Q. (iii) The enthalpy change is the energy difference from reactants to products: R.

(b) Exothermic: the products are at a lower energy than the reactants, so energy is released to the surroundings and $\Delta H$ is negative.

(c) Activation energy, $E_a$, is the minimum energy that colliding particles must have in order to react.

(d) The catalyst provides a pathway with a lower activation energy. A greater proportion of collisions now have energy equal to or greater than $E_a$, so there are more successful collisions per second.

The reactant and product levels are unchanged, so $\Delta H$ stays the same.

(a) Carbon dioxide gas is produced and escapes from the flask, so the mass of the flask and contents decreases.

(b) Curve A is steeper and finishes after $3$ minutes instead of $8$, so it is the powder. The powder has a larger surface area, so more particles are exposed to the acid and the collision frequency is greater.

(c) Find the moles of calcium carbonate. $M_r(\text{CaCO}_3) = 40 + 12 + 3 \times 16 = 100$.

The equation shows $1$ mol $\text{CaCO}_3$ gives $1$ mol $\text{CO}_2$. $M_r(\text{CO}_2) = 12 + 2 \times 16 = 44$.

Both curves level off at $1.10$ g because the same mass of calcium carbonate (the limiting reactant) is used — surface area changes the rate, not the amount of product. (d) The cotton wool lets the carbon dioxide escape (so the mass falls) but stops acid spray being lost, which would make the mass loss too large.

(a) Read the minute hand first, then the seconds hand, and convert everything to seconds. At $20\,^{\circ}\text{C}$: $1$ min $36$ s $= 60 + 36 = 96$ s. At $30\,^{\circ}\text{C}$: $50$ s. At $40\,^{\circ}\text{C}$: $26$ s.

(b) Evaluate each fraction as a decimal.

(c) $0.0200 \div 0.0104 \approx 1.9$ and $0.0385 \div 0.0200 \approx 1.9$: the relative rate roughly doubles for every $10\,^{\circ}\text{C}$ rise.

(d) Any two of: the mass of calcium carbonate; its particle size (powder); the volume of acid; the concentration of acid; the volume of gas collected ($40\text{ cm}^3$).

(a) Gas that escapes is not collected, so the volume recorded is smaller than it should be.

(b) Put the zinc in a small tube standing inside the flask (or on one side of a divided flask), with the stopper already in place. Start the reaction by tilting the flask so the zinc falls into the acid, and start the stop-watch at the same moment.

(c) The hydrogen made pushes air out of the flask into the cylinder. When the reaction stops, the flask is full of hydrogen instead of air, so the volume of air collected is equal to the volume of hydrogen left in the flask — the total collected equals the volume of hydrogen made.

(d) Carbon dioxide is slightly soluble in water, so some dissolves and the volume measured is too low. A gas syringe is better.

At the lower temperature the particles have less kinetic energy, so they move more slowly and collide less frequently.

The activation energy is a property of the reaction pathway — it is unchanged by temperature (only a catalyst changes it). Answer: B.

A and D wrongly change the activation energy. C misses the drop in collision frequency — slower particles collide less often. Both curves end at $80\text{ cm}^3$ because the same amounts of reactants are used.

(a) Moles of each reactant ($A_r$ of Mg $= 24$; convert $50.0\text{ cm}^3$ to $0.0500\text{ dm}^3$).

$0.00250$ mol Mg needs $2 \times 0.00250 = 0.00500$ mol HCl. There is $0.0500$ mol, so the acid is in excess and magnesium is limiting. $1$ mol Mg gives $1$ mol $\text{H}_2$, and $1$ mol of gas occupies $24\,000\text{ cm}^3$ at r.t.p.

(b)(i) Powder has a larger surface area: the curve is steeper at the start and levels off sooner, at the same $60\text{ cm}^3$ (same moles of Mg).

(ii) Half the magnesium: $0.00125$ mol $\text{H}_2$, so the curve levels off at $0.00125 \times 24\,000 = 30\text{ cm}^3$. The start is similar or slightly less steep (less magnesium surface), and the curve levels off sooner.

(iii) $n(\text{HCl}) = 0.0500 \times 0.500 = 0.0250$ mol — still more than the $0.00500$ mol needed, so magnesium is still limiting and the final volume is still $60\text{ cm}^3$. The acid is less concentrated, so the curve is less steep and takes longer to level off.

(a) Mean rate = total volume ÷ total time.

(b) Convert the volume of oxygen to moles using $24\,000\text{ cm}^3$ per mole at r.t.p.

The equation gives $2$ mol $\text{H}_2\text{O}_2$ → $1$ mol $\text{O}_2$, so double it.

Divide by the volume in $\text{dm}^3$: $25.0\text{ cm}^3 = 0.0250\text{ dm}^3$.

(c) Filter off the solid at the end, wash it with water and dry it, then weigh it: the mass is still $0.50$ g. It can be added to fresh hydrogen peroxide and still speeds up the reaction. (d) Oxygen relights a glowing splint.

Start with (b): multiply concentration by time for the results that fit the pattern: $0.050 \times 120 = 6.0$, $0.100 \times 60 = 6.0$, $0.200 \times 30 = 6.0$, $0.250 \times 24 = 6.0$. The product is constant, so time is inversely proportional to concentration and rate is proportional to concentration — doubling the concentration doubles the rate.

(a) For $0.150\text{ mol/dm}^3$ the pattern predicts $6.0 \div 0.150 = 40$ s, but $55$ s was recorded: this is the anomalous result. The reaction was slower than expected — for example the acid was colder, the powder had larger grains, or the stop-watch was started late.

(c) Use the constant product.

(d) A higher concentration means more acid particles per unit volume, so the frequency of collisions with the calcium carbonate increases. The energy of each collision is unchanged.