Chemistry, asked by theman9494, 1 year ago

First order reaction
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Answered by sehangshu22
1

Explanation:

to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:

rate=−Δ[A]Δt=k[A](14.5.1)

If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).

The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:

[A]=[A]0e−kt(14.5.2)

where [A]0 is the initial concentration of reactant A at t = 0; k is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation 14.5.2 predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation 14.5.2 and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t:

ln[A]=ln[A]0−kt(14.5.3)

Because Equation 14.5.3 has the form of the algebraic equation for a straight line, y = mx + b, with y = \ln[A] and b = \ln[A]0, a plot of \ln[A] versus t for a first-order reaction should give a straight line with a slope of −k and an intercept of \ln[A]0. Either the differential rate law (Equation 14.5.1 ) or the integrated rate law (Equation 14.5.3 ) can be used to determine whether a particular reaction is first order.

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