AQA · A-Level · Chemistry
Course version: 7405 · Version from Sept 2016
These equations and the gas constant, R, will be given when required.
Full specification wording
Rate equations (A-level only): 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…
Checked against the AQA 7405 specification (opens in a new tab) on .
- Point 2 of 2
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
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.
What earns the marks
- Selecting the correct given form of the equation for the data supplied.
- A plot of ln k against 1/T with a line of best fit.
- A gradient value read from the drawn line of best fit, within the accepted range.
- Eₐ obtained as the gradient multiplied by −R, converted to kJ mol⁻¹.
Common mistakes and how to put them right
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.
Exam tips
- 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.
Try the lesson
The start of the Rate equations (A-level only): These equations and the gas constant, R, will be given when required. lesson, made from checked teaching notes for this specification point.
Have a go first
The Arrhenius equation is given as k = Ae^{-E_a/RT}. Which of the following is the correct rearrangement to calculate the activation energy, E_a?
Listen
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. A strong answer earns marks for these points. selecting the correct given form of the equation for the data supplied. a plot of ln k against 1/T with a line of best fit. a gradient value read from the drawn line of best fit, within the accepted range. Eₐ obtained as the gradient multiplied by −R, converted to kJ mol⁻¹. Exam tip: Locate the equation and the value of R on the data sheet before you begin, then substitute carefully.
See it
Put it in order
Put these steps in the order they happen.
- 1. Divide by 1000 to give kJ mol⁻¹
- 2. Eₐ in J mol⁻¹
- 3. Draw line of best fit
- 4. Eₐ = gradient × −R
- 5. Tabulated data: k and T
- 6. Plot ln k against 1/T
- 7. Read gradient from the line of best fit
Key words
Say what each word means before you turn the card.
Arrhenius equation
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logarithmic form
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ln k against 1/T
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gradient and activation energy
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See every definition
- Arrhenius equation
- An equation in which the rate constant k equals A multiplied by an exponential factor containing −Eₐ/RT.
- logarithmic form
- The equation ln k = −Eₐ/RT + ln A, which is the usable form of the Arrhenius equation.
- ln k against 1/T
- A plot that gives a straight line of gradient −Eₐ/R and intercept ln A.
- gradient and activation energy
- The activation energy comes from multiplying the gradient by −R.
Label the diagram
Some boxes have lost their names. Pick the right name for each number.
1Box 1
2Box 2
3Box 3
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