Izaz Ul Islam

Record the experimental data: measurements of the concentration of reactant [A] or product [P] versus time t.

Plot a graph: For a pseudo-first-order reaction, if the reaction is first order with respect to reactant A, the graph should be linear when plotted:

ln[A] versus t. Check for linearity: If the graph of ln[A] versus time is linear, the reaction is probably pseudo-first order.

If not, try other models (e.g., second order, zero, etc.). Determine the parameters of the model:

The slope of the line gives the reaction rate k, since:

ln[A]=−kt+ln[A]0, where [A]0 is the initial concentration.

Let's say you have the data:

Time, min Concentration [A], mol/L

0 0.10

5 0.082

10 0.067

15 0.055

Plot ln[A] against time:

Time, min ln[A]

0 ln0.10≈−2.30

5 ln0.082≈−2.50

10 ln0.067≈−2.71

15 ln0.055≈−2.90

If these points form an approximately straight line, then the reaction is pseudo-first order. Based on the slope of the line, you can find the rate k: k=−slope of the line

If the graph is truly linear, this confirms the model.

There is no answer to your question since you don't provide any information about your experimental data. Plz, attach the table with your raw data.

Yuri Mirgorod Honorable Prof. thanks a lot for your time and detail answer to the question.

Yurii V Geletii Honorable Prof. thanks a lot for your time and response. After detail discussion with some expert in the field and literature study the issue has been resolved.

Of course! If your dataset fits the first-order kinetics, it will also fit the second-order kinetics. In that case, the sum of the parameters before both 'exp' (in the 2-exp equation) will equal the same parameter in the '1-exp' equation. You can easily test it on any example dataset:

If:

(1) f(x) = A + B*exp(-k1*x)

(2) f(x) = A + B1*exp(-k2a*x) + B2*exp(-k2b*x)

and the data behaviour according to the eq. (1), then:

k2a = k2b

and

B1+B2 = B

If the dataset can be fitted only with the '2-exp' equation, you will see, that the '1-exp' equation doesn't work.