16.5-16.7 - Rate Graphs And Reaction Orders

From A Graph To A Rate

A reaction mixture is monitored every few seconds. The concentration of reactant A falls smoothly, but the graph is not a straight line. Does the reaction have one rate, or does its rate change as the reaction proceeds?

A rate is a change per unit time. On a graph, that means a gradient:

The sign and unit come from the graph:

If the graph is straight, the rate is constant over that interval. If the graph is curved, the rate at a particular time is found from a tangent at that time. This matters chemically because reactant concentrations change during a run, so the collision frequency and rate often change too.

Use this route:

1. Identify the graph and axes.
2. Decide whether the question asks for a mean rate, an initial rate, or a rate at a time.
3. For a curve, draw a tangent at the required point.
4. Choose two well-spaced points on the tangent.
5. Calculate the gradient, $\Delta y/\Delta x$.
6. Use the graph-axis units in the final rate.

Worked example: rate from a concentration-time graph. A tangent to a concentration-time curve at $40 \mathrm{s}$ passes through two convenient points:

$(20\,\mathrm{s},\ 0.62\,\mathrm{mol\,dm^{-3}})$ and $(70\,\mathrm{s},\ 0.26\,\mathrm{mol\,dm^{-3}})$

Calculate the rate of disappearance of A at $40 \mathrm{s}$.

The graph gradient is negative because the concentration of A is decreasing. The rate of disappearance of A is the positive magnitude:

The chemical interpretation is that, at about $40 \mathrm{s}$, A is being used up at $7.2\times 10^{-3} \mathrm{mol} \mathrm{dm}^{-3}$ each second. That is an instantaneous rate, not the mean rate for the whole experiment.

First-Order Half-Life From Graphs

A falling concentration can be compared in fractions as well as absolute changes. Half-life measures the time required for one such fractional change.

For a first-order reaction, successive half-lives are equal at constant temperature.

[DIAGRAM: asset_name: First-order half-life and tangent graph; asset_slug: edexcel_a_level_chemistry_l078_first_order_half_life_tangent; recommended_method: deterministic_chart; description: 16:9 monochrome concentration-time graph for a first-order reaction showing $[\mathrm{A}]$ falling from 0.80 to 0.40 to 0.20 to 0.10 mol dm^-3 at equal 30 s intervals, with dashed half-life construction lines and a tangent at 45 s labelled as instantaneous rate.]

Read a half-life from a concentration-time graph by choosing a concentration and finding the time taken to reach half that concentration. Then repeat from the new concentration. For a first-order reaction, these time intervals should be the same within the accuracy of the graph.

For the graph shown:

• $0.80 \to 0.40\,\mathrm{mol\,dm^{-3}}$ takes 30 s
• $0.40 \to 0.20\,\mathrm{mol\,dm^{-3}}$ takes 30 s
• $0.20 \to 0.10\,\mathrm{mol\,dm^{-3}}$ takes 30 s

So:

$t_{1/2} = 30\,\mathrm{s}$

Volume-time graphs need one extra thought. A product-volume graph rises towards a final volume. If the gas volume is directly proportional to product formed, and the gas comes from the limiting reactant being used up, then one half-life is the time taken for the product volume to reach halfway to the final volume. The second half-life is the time taken to go from halfway to three-quarters of the final volume, because half of the remaining reactant has reacted.

For a final gas volume of $80 \mathrm{cm}^{3}$:

• after 1 half-life: product volume $= 40\,\mathrm{cm^{3}}$
• after 2 half-lives: product volume $= 60\,\mathrm{cm^{3}}$
• after 3 half-lives: product volume $= 70\,\mathrm{cm^{3}}$

That sequence is not $40, 80, 120$. The product volume approaches a maximum because the limiting reactant is being used up.

Worked example: half-life from product volume. A first-order decomposition produces gas and reaches a final volume of $64.0 \mathrm{cm}^{3}$. The graph shows $32.0 \mathrm{cm}^{3}$ at $18 \mathrm{s}$ and $48.0 \mathrm{cm}^{3}$ at $36 \mathrm{s}$.

The first value, $32.0 \mathrm{cm}^{3}$, is half of the final volume, so it marks one half-life. The second value, $48.0 \mathrm{cm}^{3}$, is three-quarters of the final volume, so it marks two half-lives.

first half-life $= 18 - 0 = 18\,\mathrm{s}$; second half-life $= 36 - 18 = 18\,\mathrm{s}$

The half-lives are equal, so the graph is consistent with first-order behaviour in the limiting reactant.

Orders From Concentration-Time Graphs

A concentration-time graph shows how much reactant remains as time passes. To deduce order from this graph, look for the rate pattern, not just whether the line is curved.

[DIAGRAM: asset_name: Graph patterns for reaction order; asset_slug: edexcel_a_level_chemistry_l078_order_graph_patterns; recommended_method: deterministic_chart; description: 16:9 monochrome two-panel chart comparing zero, first and second order evidence; left panel shows concentration-time patterns with zero order as a straight decreasing line, first order as exponential decay with constant half-life markers, and second order as a curve with increasing half-lives; right panel shows rate-concentration patterns with zero order horizontal, first order straight through the origin, and second order upward-curving.]

Zero order is the easiest concentration-time graph to spot: a straight decreasing line means the same amount of A is used up each second. The rate does not fall as $[\mathrm{A}]$ falls, so A is zero order under those conditions.

These single-reactant patterns assume other rate-law factors are constant or have been accounted for. If two reactants change substantially during the same run, a curve alone may not isolate the order with respect to one of them.

For a simple second-order decay, successive half-lives double as the concentration halves. Increasing half-lives alone suggest order greater than one; the doubling pattern identifies second order within this model.

First order and second order are both curves, so "it curves" is not enough evidence. For first order, equal half-lives are the key pattern. For second order, the graph flattens in a way that gives longer successive half-lives as concentration decreases.