AQA GCSE Chemistry contents

Unit 6 · The Rate & Extent of Chemical Change

Calculating Rates of Reaction

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Mean rate of reaction

  • The rate of a reaction is found by measuring how much reactant is used or how much product is formed in a given time
mean rate=quantity of reactant usedtime takenmean rate=quantity of product formedtime taken \text{mean rate} = \frac{\text{quantity of reactant used}}{\text{time taken}} \qquad \text{mean rate} = \frac{\text{quantity of product formed}}{\text{time taken}}
  • The quantity is a mass in g or a volume in cm³, so the units of rate are g/s or cm³/s
  • (HT) The quantity can also be in moles, giving a rate in mol/s
  • Worked example: 48 cm³ of hydrogen is collected in 60 s, so the mean rate is 4860=0.80\frac{48}{60} = 0.80 cm³/s
  • Worked example (HT): 0.0048 mol of gas in 60 s: 8.0×10−58.0 \times 10^{-5} mol/s

Rate graphs

  • A graph of product formed against time starts steep and levels off: the reaction is fastest at the start, slows as the reactants are used up, and stops when one reactant runs out
  • The steeper the line, the faster the rate. When the line is flat the reaction has finished
  • A graph of reactant remaining against time falls steeply and then flattens

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the rate of a reactiontime (s)volume of gas (cm³)both level off at thesame final volume — thesame amount of reactantwas usedhigher concentrationlower concentrationriserundraw a tangent to find the rateat one moment:rate = rise ÷ runThe gradient of the curve is the rate. It is steepest at the start, when the concentration ishighest, and falls to zero when the reaction finishes. Mean rate = total gas ÷ total time.
Volume of gas against time for the same reaction at two concentrations: the higher concentration curve steeper and levelling off earlier at the same final volume; a tangent drawn at one point on a curve with its rise and run marked

Rate at a particular time

  • The rate at one moment is the gradient of the curve at that point. Draw a tangent, a straight line that just touches the curve there, and find its slope
  • (HT) Gradient of the tangent:
rate=change in ychange in x \text{rate} = \frac{\text{change in } y}{\text{change in } x}
  • Worked example (HT): a tangent at 45 s passes through (20 s, 10 cm³) and (70 s, 60 cm³). Rate =60−1070−20=5050=1.0= \frac{60 - 10}{70 - 20} = \frac{50}{50} = 1.0 cm³/s
  • Use a large triangle on the tangent, so small reading errors matter less