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 used time taken mean rate = quantity of product formed time 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}} mean rate = time taken quantity of reactant used mean rate = time taken quantity of product formed
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 48 60 = 0.80 \frac{48}{60} = 0.80 60 48 = 0.80 cm³/s
Worked example (HT): 0.0048 mol of gas in 60 s: 8.0 × 10 − 5 8.0 \times 10^{-5} 8.0 × 1 0 − 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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Following the rate of a reaction time (s) volume of gas (cm³) both level off at the same final volume — the same amount of reactant was used higher concentration lower concentration rise run draw a tangent to find the rate at one moment: rate = rise ÷ run The gradient of the curve is the rate. It is steepest at the start, when the concentration is highest, 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 y change in x \text{rate} = \frac{\text{change in } y}{\text{change in } x} rate = change in x change in y
Worked example (HT): a tangent at 45 s passes through (20 s, 10 cm³) and
(70 s, 60 cm³). Rate = 60 − 10 70 − 20 = 50 50 = 1.0 = \frac{60 - 10}{70 - 20} = \frac{50}{50} = 1.0 = 70 − 20 60 − 10 = 50 50 = 1.0
cm³/s
Use a large triangle on the tangent, so small reading errors matter
less
Mean rate over a period (total change divided by total time) is not
the rate at a time (gradient of the tangent). If the question says
"at 45 s", it wants the tangent. Reading the value at 45 s and dividing by
45 gives the mean rate and scores only the units or significant figure
mark
Draw the tangent at the time asked , mark the two points you read,
and write the values into the gradient equation so the examiner can see
them. Counting squares scores nothing
Read the scales carefully: check what one small square is worth on
each axis. And 1 cm³ = 0.001 dm³