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rate = k[A], k= rate constant / of proprotion , [] =concentration. (@ = change in []concentration ,@t change in time)

a A +bB-> cC from this the rate = -1/a(@[A]/@t) =-1/b(@[B]/@t) =1/c(@[C]/@t where the small letters are the coefficents . so the rate of a reaction is equal to each other when ajusted by the coeffiects . note that the reactants are negative becasue it is decreasing as products form..

the basic difference of a fist order and a second is proportions.. fist is directly proportional id est concentration goes up by 2 the rate goes up by two.

while second is proportional to the square.

concentration goes up by two the rate goes up by 4

1. First-Order Reactions

A first order reaction (order = 1) has a rate proportional to the concentration of one of the reactants. A common example of a first-order reaction is the phenomenon of radioactive decay. The rate law is:

rate = k[A] (or B instead of A), with k having the units of sec-1

2. Second-Order Reactions

A second-order reaction (order = 2) has a rate proportional to the concentration of the square of a single reactant or the product of the concentration of two reactants:

rate = k[A]2 (or substitute B for A or k multiplied by the concentration of A times the concentration of B), with the units of the rate constant M-1sec-1

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11y ago
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11y ago

If the rate of the reaction depends upon both the reactants.. wrong!

A second order reaction yields the rate proprontional to the square of the concentration. id est the concentration goes up by 2 : the rate goes up by 4

for the former to be correct, both reactants must be of the first order becasue the over all order is determined by adding the value of [ A]^ n & [B]^ m

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Q: What is the difference between an first order reaction and second order reaction?
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