Rate equations (A-level only) — AQA A-Level Chemistry
Test yourself on Rate equations (A-level only) with AQA A-Level practice questions.
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Rate equations (A-level only) explained
A rate equation has the form rate = k[A]ᵐ[B]ⁿ, where m and n are the orders with respect to each reactant and take only the values 0, 1 or 2 at this level.
Read the full explanation
The order tells you how the rate responds. Zero order means the concentration has no effect at all, so that species does not appear in the rate equation; first order means doubling the concentration doubles the rate; second order means doubling it quadruples the rate, since the rate depends on the square. Adding the individual orders gives the overall order, which fixes the units of k: mol dm⁻³ s⁻¹ for zero order overall, s⁻¹ for first, mol⁻¹ dm³ s⁻¹ for second and mol⁻² dm⁶ s⁻¹ for third.
These equations and the gas constant, R, will be given when required.
The Arrhenius equation, in which the rate constant k equals A multiplied by an exponential factor containing −Eₐ/RT, and its logarithmic form ln k = −Eₐ/RT + ln A, are both supplied in the exam along with the gas constant R, so no recall is needed; what is tested is the handling. The equation says that k, and so the rate, rises steeply as the temperature rises or as the activation energy falls, because Eₐ and T sit inside that exponent. The logarithmic form is the usable one: plotting ln k against 1/T gives a straight line of gradient −Eₐ/R and intercept ln A, so the activation energy comes from multiplying the gradient by −R. Keep the units consistent, since R is in J K⁻¹ mol⁻¹, so an activation energy found this way is in J mol⁻¹ and usually needs dividing by 1000.
Your focus
- Deduce the order with respect to a reactant from the effect of a concentration change on the rate.
- Write a rate equation from stated orders, omitting any species of zero order.
- Calculate the value of the rate constant from a rate and a set of concentrations and derive its units.
Show all 8 objectives
- Explain why the orders cannot be read from the balanced equation for the reaction.
- Use the given logarithmic form of the Arrhenius equation to determine an activation energy from a gradient.
- Plot ln k against 1/T from tabulated data and draw an appropriate line of best fit.
- Convert an activation energy obtained with R in J K⁻¹ mol⁻¹ into kJ mol⁻¹.
- Predict the effect on k of raising the temperature or of lowering the activation energy, using the form of the equation.
Rate equations (A-level only) exam tips
Marking Points
- one mark for the rate equation written with the correct order as an index on each concentration
- one mark for omitting any zero order species from the rate equation
- one mark for the overall order as the sum of the individual orders
- one mark for the units of k, obtained by rearranging the rate equation and cancelling
- one mark for selecting the correct given form of the equation for the data supplied
- one mark for a plot of ln k against 1/T with a line of best fit
- one mark for a gradient value read from the drawn line of best fit, within the accepted range
- one mark for Eₐ obtained as the gradient multiplied by −R, converted to kJ mol⁻¹
Examiner Tips
- 💡Derive the units of k each time by substituting mol dm⁻³ for each concentration and cancelling; do not rely on memory.
- 💡Check a suspicious order against the data: if doubling the concentration quadruples the rate, the order is 2.
- 💡Show the rearranged expression for k before substituting numbers, since it is credited separately.
- 💡Locate the equation and the value of R on the data sheet before you begin, then substitute carefully.
- 💡An activation energy must come out positive; a negative answer means the sign of the gradient was mishandled.
- 💡Take gradient points from opposite ends of the drawn line to reduce the uncertainty.
Common Mistakes
- taking the orders from the balancing numbers in the chemical equation
- writing a zero order species in the rate equation with an index of zero rather than leaving it out
- quoting the units of k as mol dm⁻³ s⁻¹ whatever the overall order
- inverting the rearrangement, so k comes out as concentration divided by rate
- reporting an order such as 1.5 or 3 from data that should be read as 1 or 2
- plotting k against T and expecting a straight line
- forgetting the minus sign, so a negative gradient yields a negative activation energy
- leaving Eₐ in J mol⁻¹ when the answer is required in kJ mol⁻¹
- reading the gradient from two plotted data points rather than from the line of best fit
- using a temperature in degrees Celsius inside 1/T