Chemistry › Maths skills › Graphs: gradients, tangents and intercepts
Graphs: gradients, tangents and intercepts
A rate is a gradient, a half-life is read off a curve, and the temperature a calorimeter would have reached is found by extending a line back to where it started. Three readings, one ruler.
A gradient is a rate
The gradient of a graph of y against x is the change in y divided by the change in x, taken over as long a stretch of the line as the ruler allows. Its units are the units of y divided by the units of x, so the gradient of a concentration–time graph is in mol dm⁻³ s⁻¹, which is a rate. When the graph is a straight line the rate is constant and one gradient is the whole story: a zero-order reaction gives a straight concentration–time line whose gradient is −k.
On a straight line y = mx + c the intercept on the y axis is c, the value of y when x is zero. It is read where the line crosses the axis, which means the axis has to start at zero; an axis that starts at 2.5 to spread the points out has no intercept on it, and the value has to be calculated from the gradient and one point instead.
Most graphs in rates work are curves, and a curve has a different gradient at every point. The gradient at one point is the gradient of the tangent there, the straight line that touches the curve at that point and nowhere nearby. The tangent is drawn by eye with a ruler, then its gradient is taken from two points on the tangent far apart, not from points on the curve.
The initial rate from a tangent
The initial rate is the gradient at time zero, where the curve is steepest and the concentrations are the ones mixed. A tangent at t = 0 starts at the first point and leaves the curve at once; drawn too shallow it reads as a chord, which underestimates the rate. Read two points from the tangent, ideally where it crosses gridlines, and take the gradient between them.
The initial rate from a concentration–time graph
A graph of [A] against time starts at 0.500 mol dm⁻³. A tangent drawn at t = 0 passes through (0 s, 0.500 mol dm⁻³) and (40 s, 0.100 mol dm⁻³). Find the initial rate.
Show the working
Gradient of the tangent = (0.100 − 0.500) ÷ (40 − 0) = −0.400 ÷ 40 = −0.0100 mol dm⁻³ s⁻¹.
The gradient is negative because A is used up. A rate is quoted as a positive number, so the initial rate = 1.00 × 10⁻² mol dm⁻³ s⁻¹.
At 20 s the tangent would be shallower and the rate smaller, which is what a curve that flattens means: the rate falls as the reactant is used up, and only the tangent at t = 0 gives the rate at the starting concentration that the rate equation is written for.
A rate from a tangent is accepted by a mark scheme to a tolerance, usually ±10%, because two students draw the same tangent differently. The gradient mark goes with the working: two points shown, a subtraction, a division and the units.
Half-lives from a curve
The half-life is the time for a concentration to fall to half its value, and it is read straight off a concentration–time curve: find the time at which [A] has halved, then the time at which it has halved again. If the two intervals are the same, the half-life is constant and the reaction is first order in A, and k = ln 2 ÷ t½. A half-life that lengthens as the concentration falls is second order; one that shortens is zero order, where the line is straight and a half-life is not a useful idea.
Order and k from successive half-lives
On a concentration–time curve, [A] falls from 0.800 mol dm⁻³ to 0.400 mol dm⁻³ in 60 s, and from 0.400 mol dm⁻³ to 0.200 mol dm⁻³ between 60 s and 120 s. Find the order with respect to A and the rate constant.
Show the working
First half-life: 0.800 → 0.400, 60 s. Second half-life: 0.400 → 0.200, 120 − 60 = 60 s. The half-life is constant, so the reaction is first order in A.
k = ln 2 ÷ t½ = 0.693 ÷ 60 = 1.16 × 10⁻² s⁻¹.
Checking: a third half-life should end at 180 s with [A] = 0.100 mol dm⁻³. If the curve reads 0.100 there, the order is confirmed; if it reads higher, the half-life is lengthening and the order is two.
Extrapolating a cooling curve
A reaction in a plastic cup loses heat while it is happening, so the highest temperature the thermometer shows is lower than the temperature the mixture would have reached with no loss. The fix is a graph of temperature against time: readings before mixing give a flat line, readings after the peak give a falling straight line, and that falling line is extended back to the time of mixing. The gap between the two lines at that time is ΔT.
The extrapolation assumes the rate of heat loss after the peak is the rate that applied throughout, which is close enough for a cup over a few minutes. On the graph it looks like a line drawn through the cooling points, continued with the ruler past the data to the mixing time, with a vertical read off there. A mark scheme awards the extrapolated value, not the highest reading, and expects the line to run through the cooling points, not the peak.
If the temperature is read every 30 s and the mixture was added at 3.0 min, the cooling line, say falling 0.2 °C a minute from 25.8 °C at 4.0 min, reaches 26.0 °C at 3.0 min. With a starting temperature of 19.0 °C, ΔT = 7.0 °C rather than the 6.8 °C the peak reading gives, a difference of 3% that goes straight into the enthalpy change.
TRY IT: Four readings from graphs
(a) A tangent at t = 0 passes through (0 s, 1.20 mol dm⁻³) and (50 s, 0.60 mol dm⁻³). The initial rate.
(b) Successive half-lives of 100 s, 100 s and 100 s. The order and k.
(c) A cooling curve falling 0.4 °C a minute reads 31.6 °C at 5.0 min; the reactants were mixed at 2.0 min, starting at 21.0 °C. ΔT.
(d) A concentration–time graph is a straight line of gradient −2.0 × 10⁻³ mol dm⁻³ s⁻¹. The order and k.
Check your answer
(a) (0.60 − 1.20) ÷ 50 = −0.012, so the rate is 1.2 × 10⁻² mol dm⁻³ s⁻¹.
(b) Constant half-life, so first order; k = 0.693 ÷ 100 = 6.93 × 10⁻³ s⁻¹.
(c) Three minutes of cooling at 0.4 °C a minute is 1.2 °C, so the extrapolated temperature at 2.0 min is 32.8 °C and ΔT = 32.8 − 21.0 = 11.8 °C.
(d) A straight line means zero order, and k is the size of the gradient, 2.0 × 10⁻³ mol dm⁻³ s⁻¹.
The lessons that use this: Collision theory, activation energy and the rate of reaction · Finding orders from initial rates, half-lives and graphs · Measuring the rate of a reaction · Measuring an enthalpy change: calorimetry. Every skill the boards number, and where each is taught: maths skills.