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The second order rate law

-dA/dt = k [A] [B]

for example if B is present in excess amount, the reaction rate reduces to pseudo first order,

A0=0.01 M

B0: 2 M

AT= 0.0 M

BT=1.99 M

so the reaction rate depends on the concentration of A and

the new reaction rate is pseudo first order;

-dA/dt = kı [A]

The third order rate law

-dA/dt = k [A2] [B]

If B is in excess amount, molarity of B0 is a very closer value to Bt ( for example ; B0= 2 M BT=1.99 M)

the reaction becomes to be pseudo second order.

-dA/dt = k [A2]

Behzat BALCI

enviromental enginneering of cukurova university

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Related Questions

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To determine the rate constant k for a pseudo-first order reaction, you can plot the natural logarithm of the concentration of reactant vs. time. The slope of the resulting line will be equal to -k. This approach is often used for reactions where one reactant is present in excess and its concentration remains constant throughout the reaction.


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