The integrated rate laws for zero-, first-, and second-order reaction may be arr

Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line, y = mx + b. Order Integrated Rate Law Graph Slope 0 { m [A] =} -kt+[{ m A}]_0 { m [A]~vs.}~t -k 1 ln[{ m A}] = -kt+ln[{ m A}]_0 { m ln[A]~vs.}~t -k 2 { m {1 over [A]} =}~kt+{1 over [{ m A}]_0} { m {1 over [A]}~vs.}~t k The reactant concentration in a zero-order reaction was 0.100 M after 115 s and 4.00

Explanation / Answer

follow thfor the reaction... A ---> B + C... rate = -d[A] / dt = k x [A]^n where.. [ ] means concentration. d[A] / dt means change in concentration of A over time - is because concentration of A is decreasing k is the rate constant. [A] is the concentration of A at any given time n is the order of the reactant and if n = 0, then

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