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

A student carried the following reaction at the laboratory:

A+B → products

After performing the experiment, the student has plotted the following 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  

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:

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

$\frac{1}{{\left[\mathrm{A}\right]}_{\mathrm{t}}}=\mathrm{kt}+\frac{1}{{\left[\mathrm{A}\right]}_0}$

• 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

  •  $\mathrm{gradient}=\frac{∆\mathrm{y}}{∆\mathrm{x}}$

  • $\mathrm{gradient}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

  • $\mathrm{gradient}=\frac{37.5-2.5}{80-0}$

  • gradient = 0.434

• Step 5: Calculate the units for the rate constant

  •  $\mathrm{rate}=\mathrm{k}{\left[\mathrm{A}\right]}^2$

  • $\mathrm{units}\mathrm{of}\mathrm{rate}=\left(\mathrm{units}\mathrm{of}\mathrm{k}\right)\times {\left(\mathrm{units}\mathrm{of}\mathrm{concentration}\mathrm{of}\mathrm{A}\right)}^2$

  • $\mathrm{units}\mathrm{of}\mathrm{k}=\frac{\mathrm{units}\mathrm{of}\mathrm{rate}}{{\left(\mathrm{units}\mathrm{of}\mathrm{concentration}\mathrm{of}\mathrm{A}\right)}^2}$

  • $\mathrm{units}\mathrm{of}\mathrm{k}=\frac{{\mathrm{Ms}}^{-1}}{{\left(\mathrm{M}\right)}^2}$

  • $\mathrm{units}\mathrm{of}\mathrm{k}=\frac{{\mathrm{Ms}}^{-1}}{\left(\mathrm{M}\right)\left(\mathrm{M}\right)}$

  • $\mathrm{units}\mathrm{of}\mathrm{k}=\frac{s^{-1}}{\mathrm{M}}$
    

  • 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