Half-lives of a first-order and a zero-order | vedclass

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(B) For a first-order reaction,the half-life is $(t_{1/2})_{1^{st}} = 0.693 / k_1$.For a zero-order reaction,the half-life is $(t_{1/2})_{zero} = C_0

Vedclass · Accepted Answer

$2 	imes 0.693$

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(B) For a first-order reaction,the half-life is $(t_{1/2})_{1^{st}} = 0.693 / k_1$.For a zero-order reaction,the half-life is $(t_{1/2})_{zero} = C_0 / (2k_0)$.Given $(t_{1/2})_{1^{st}} = (t_{1/2})_{zero}$,we have $0.693 / k_1 = C_0 / (2k_0)$,which implies $k_1 / k_0 = (2 	imes 0.693) / C_0$.The initial rate of the first-order reaction is $r_1 = k_1 [C_0]$.The initial rate of the zero-order reaction is $r_0 = k_0 [C_0]^0 = k_0$.The ratio of the initial rates is $r_1 / r_0 = (k_1 [C_0]) / k_0 = (k_1 / k_0) 	imes C_0$.Substituting the value of $k_1 / k_0$,we get $r_1 / r_0 = [(2 	imes 0.693) / C_0] 	imes C_0 = 2 	imes 0.693$.

$a$. Rate constant for first order reaction	$i$. $mol \ lit^{-1} \sec^{-1}$
$b$. Molarity	$ii$. $\frac{k \times 1000}{M}$
$c$. Rate constant for zero order reaction	$iii$. $second^{-1}$
$d$. Limiting molar conductivity	$iv$. $\frac{\text{moles of solute}}{\text{Volume of solution (lit)}}$

Half-lives of a first-order and a zero-order reaction are the same. Assume that the initial concentration of the reactant is the same for both reactions,then the ratio of the initial rates of the first-order reaction to that of the zero-order reaction is:

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