Question #4007

(Chemistry) In a bimolecular reaction A + B → M, a moles per liter of a substance A and

b moles per liter of a substance B are combined. Under constant temperature the rate of

reaction is

y′= k(a − y)(b − y) (Law of mass action);

that is, y′ is proportional to the product of the concentration of the substances that are

reacting, where y(t) is the number of moles per liter which have reacted after time t. Solve

this ODE, assuming a and b not equal

b moles per liter of a substance B are combined. Under constant temperature the rate of

reaction is

y′= k(a − y)(b − y) (Law of mass action);

that is, y′ is proportional to the product of the concentration of the substances that are

reacting, where y(t) is the number of moles per liter which have reacted after time t. Solve

this ODE, assuming a and b not equal

Expert's answer

It is second order rate laws because the rate is proportional to the product of two concentrations. By elementary integration of these differential equations Integrated Rate Laws can be obtained

(1/(a-b)) ln((a-Y)/(b-Y))

Where a and b are the initial concentrations of A and B (assuming a not equal to b), and Y is the extent of reaction at time t. Note that the latter can also be written:

(a-x)/(b-x) = (a/b)exp[(a-b)kt].

(1/(a-b)) ln((a-Y)/(b-Y))

Where a and b are the initial concentrations of A and B (assuming a not equal to b), and Y is the extent of reaction at time t. Note that the latter can also be written:

(a-x)/(b-x) = (a/b)exp[(a-b)kt].

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