Graphs: plotting, gradients, intercepts and tangents — reaction order from data and plotting good graphs

Overview

Finding reaction orders from concentration-time and rate-concentration graphs and initial-rate tables, then the rules for plotting good graphs: axes, scales, lines of best fit, anomalies and gradients.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   How is a concentration–time graph used to decide whether a reaction is zero order or first order with respect to a reactant?

   Model answer

   A zero-order reactant gives a straight line of constant negative gradient, since the rate does not depend on its concentration. A first-order reactant gives a curve whose successive half-lives are equal, so the time taken for the concentration to fall from 0.80 to 0.40 mol dm⁻³ equals the time taken to fall from 0.40 to 0.20 mol dm⁻³.

2. 2

   On a concentration–time graph, the concentration of A falls from 0.800 to 0.400 mol dm⁻³ in 50 s, then from 0.400 to 0.200 mol dm⁻³ in a further 50 s. Deduce the order with respect to A and calculate the rate constant.

   Model answer

   The half-life is constant at 50 s and independent of concentration, so the reaction is first order in A. Using t½ = ln2 ÷ k, k = 0.693 ÷ 50 = 1.39 × 10⁻² s⁻¹.

3. 3

   How does the shape of a rate–concentration graph reveal the order with respect to a reactant?

   Model answer

   A horizontal line shows the rate is independent of concentration, so the reaction is zero order. A straight line through the origin shows rate ∝ [A], which is first order. An upward curve in which the rate quadruples when the concentration doubles indicates second order, and plotting rate against [A]² would then give a straight line through the origin.

4. 4

   Why must a rate–concentration graph for a first-order reactant pass through the origin?

   Model answer

   Rate = k[A], so when the concentration of A is zero the rate is zero. A straight line of constant gradient that misses the origin would indicate a rate that persists in the absence of A, which the rate equation does not allow, and usually signals a systematic error in the data.

5. 5

   Describe how initial-rate data are used to derive a rate expression.

   Model answer

   Several experiments are performed in which the concentration of one reactant is changed while all others are held constant. Comparing the initial rates shows how the rate responds: no change indicates zero order, a proportional change indicates first order, and a fourfold change on doubling indicates second order. The orders are then assembled into rate = k[A]ᵐ[B]ⁿ and k is found by substituting one set of values.

6. 6

   Three experiments give: [A] 0.100, [B] 0.100, rate 2.00 × 10⁻⁴; [A] 0.200, [B] 0.100, rate 8.00 × 10⁻⁴; [A] 0.100, [B] 0.200, rate 4.00 × 10⁻⁴ mol dm⁻³ s⁻¹. Deduce the rate equation and calculate k with units.

   Model answer

   Between the first two, [A] doubles at constant [B] and the rate quadruples, so the order in A is 2. Between the first and third, [B] doubles at constant [A] and the rate doubles, so the order in B is 1. The rate equation is rate = k[A]²[B]. Substituting the first set, k = (2.00 × 10⁻⁴) ÷ ((0.100)² × 0.100) = (2.00 × 10⁻⁴) ÷ (1.00 × 10⁻³) = 0.200 mol⁻² dm⁶ s⁻¹.

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