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$color(white)(0, 2, 1/2, 1, 0)$
Well, consider the following half-lives:
$t_"1/2" = ([A]_0)/(2k)$ (zero order)
$t_"1/2" = (ln2)/k$ (first order)
$t_"1/2" = 1/(k[A]_0)$ (second order)
$t_"1/2" = 3/(2k[A]_0^2)$ (third order)In all of these, only the zero order half-life is directly proportional to the concentration of
$A$ . Hence, the reaction is zero order with respect to$A$ .
You can figure out a lot from the ...
$r(t) = k[A]^n$ Knowing that doubling the concentration leads to quadrupling the rate, i.e.
$color(red)(2)[A] -> color(red)(4)r(t)$ , we have that
$color(red)(4)cdotr(t) = k(color(red)(2)[A])^n = color(red)(2)^n cdot k[A]^n$ Thus, we have that
$2^n = 4$ . What must$bbn$ be? The reaction order with respect to$A$ is of this$n$ th order.
Same process as
$(2)$ . Now we claim that:
$color(red)(1.41)cdotr(t) = k(color(red)(2)[A])^n = color(red)(2)^n cdot k[A]^n$ But
$1.41 ~~ sqrt2$ . Hence, we have that$sqrt2 = 2^n$ . What is$bbn$ this time? (What is$2$ raised to that correlates with a square root?) The reaction is of this$n$ th order with respect to$A$ .
It discusses the first and second half-lives and claims that they are equal. This just says that the half-life (of THIS order) does not depend on what concentration you start at.
Look above. Which order half-life does not depend on
$[A]_0$ ? Hint: it's not a prime number.
This says that the rate of reaction does not depend on the current
$[A]$ . That is,$A$ has no influence on the rate. That can only be the case if$A$ is zero order, i.e. the rate depends only on the rate constant:
$r(t) = k[A]^0 = k$
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