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Concentration-Time Graphs & Rate Constants (College Board AP® Chemistry)

Study Guide

Martín

Written by: Martín

Reviewed by: Stewart Hird

Concentration-Time Graphs & Rate Constants

Line equations for the Concentration-Time Graphs

  • Zero order concentration-time graph is a straight line

    • The equation for this line is:

begin mathsize 16px style left square bracket straight A right square bracket subscript straight t minus left square bracket straight A right square bracket subscript 0 equals negative kt end style

  • This equation links the concentration of A at any time (), with the initial concentration of A (), the rate constant (k), and the time (t)

    • It can be used to calculate the concentration of A at any time, just by replacing the time

  • If the line equation is rearranged, the following equation is obtained:

begin mathsize 16px style left square bracket straight A right square bracket subscript straight t equals negative kt space plus left square bracket straight A right square bracket subscript 0 end style 

  • Looking carefully into the equation, if [A]t is plotted against t, an straight line going down is obtained

    • The slope of the straight line is -k

    • The y-intercept of the line is [A]0

  • An straight line can be drawn for a first order reaction by using the integrated law for a first order reaction

    • The equation for the integrated law is shown below in two equivalent ways:

 ln [A]t - ln [A]0 = -kt

  ln [A]t  = -kt +ln [A]0

  • An straight line can be drawn for a second order reaction by using the integrated law for a second order reaction

    • The equation for the integrated law is shown below in two equivalent ways:

1 over left square bracket straight A right square bracket subscript straight t minus 1 over left square bracket straight A right square bracket subscript 0 space equals space kt

1 over left square bracket straight A right square bracket subscript straight t space equals space kt space plus space 1 over left square bracket straight A right square bracket subscript 0

Line equations for zero order, first order and second order reactions

line-equations-for-the-three-orders-of-reaction

Comparison of the axis labels, slopes and y-intercepts between the line equations for a zero order, first order and second order reaction

Worked Example

A student carried the following reaction at the laboratory:

A+B → products

After performing the experiment, the student has plotted the following graphs, 

worked-example-graphs

Using the graphs above, determine the order of reaction with respect to A

Answer:

  • The only graph from these three that is an straight line is Graph II

  • The axis from this graph are ln[A] vs time

  • By comparing the three straight line equations, the order of the reaction with respect to [A] must be 1 because it is the only one in which ln[A] is present

ln [A]t = -kt + ln [A]0

  • Therefore, it is a first order reaction with respect to A  

The importance of the slope in concentration-time graphs

  • The slope in concentration-time graphs determine the rate constant of the reaction

    • The formula for the slope/gradient is:

gradient space equals space fraction numerator increment straight y over denominator increment straight x end fraction

  • Therefore, if the overall order of reaction is determined, the rate constant can be obtained by calculating the slope of the concentration-time graphs

    • For a zero-order reaction, plotting the concentration of reactant vs time

[A]t versus  t

  • For a first-order reaction, using a plot natural logarithm of the concentration of reactant vs time

ln [A]t versus  t

  • For a second-order reaction, plotting the inverse of the concentration of reactant vs time

begin mathsize 14px style 1 over left square bracket straight A right square bracket subscript straight t space versus space space straight t end style

Worked Example

A student carried the following reaction at the laboratory:

A+B → products

The student knows it is a zero order with respect to B. She has plotted the following graph:

worked-example-rate-constant-determination

Using the graph above, determine the rate constant and write the general rate equation for the reaction

Answer:

  • Step 1: Determine the order of reaction respect to A

    • The axis from this graph are 1/[A] vs time

    • By comparing the three straight line equations, the order of the reaction with respect to [A] must be 2 because it is the only one in which 1/[A] is present

1 over left square bracket straight A right square bracket subscript straight t space equals space kt space plus space 1 over left square bracket straight A right square bracket subscript 0

  • Therefore, it is a second order reaction with respect to A

  • Step 2: Determine the overall order of reaction

    • Since the statement establishes that it is a zero order reaction respect to B, the overall order is 2

    • Therefore, the general rate equation should look like this:

rate =k [A]2   

  • Step 3: Choose two points and write down their coordinates

    • Point 1 (0, 2.5)

    • Point 2 (80, 37.5)

  • Step 4: Calculate the slope/gradient using the coordinates

    •  begin mathsize 14px style gradient space equals space fraction numerator increment straight y over denominator increment straight x end fraction end style

    • begin mathsize 14px style gradient space equals space fraction numerator straight y subscript 2 minus straight y subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction end style

    • gradient space equals fraction numerator 37.5 minus 2.5 over denominator 80 minus 0 end fraction space

    • gradient = 0.434

  • Step 5: Calculate the units for the rate constant

    •  begin mathsize 16px style rate space equals straight k space left square bracket straight A right square bracket squared end style

    • units space of space rate space equals left parenthesis units space of space straight k right parenthesis space cross times space left parenthesis units space of space concentration space of space straight A right parenthesis squared

    • begin mathsize 14px style units space of space straight k space equals fraction numerator units space of space rate over denominator left parenthesis units space of space concentration space of space straight A right parenthesis squared end fraction space end style

    • begin mathsize 14px style units space of space straight k space equals Ms to the power of negative 1 end exponent over left parenthesis straight M right parenthesis squared end style

    • units space of space straight k space equals fraction numerator Ms to the power of negative 1 end exponent over denominator left parenthesis straight M right parenthesis stretchy left parenthesis straight M stretchy right parenthesis end fraction

    • units space of space straight k space equals s to the power of negative 1 end exponent over straight M

    • units of k = M-1 s-1

  • Step 6: Write down the final rate equation by replacing the calculated value for k

    •  rate = k [A]2

    • rate = (0.434 M-1 s-1) [A]2

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    Martín

    Author: Martín

    Expertise: Chemistry Content Creator

    Martín, a dedicated chemistry teacher and tutor, excels in guiding students through IB, AP, and IGCSE Chemistry. As an IB Chemistry student, he came from hands-on preparation, focusing on practical exam techniques and rigorous practice. While at Universidad San Francisco de Quito, his academic journey sparked a passion for computational and physical chemistry. Martín specializes in chemistry, and he knows that SaveMyExams is the right place if he wants to have a positive impact all around the world.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.