For a certain reaction A to P,the half-life f | vedclass

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(NONE) For a reaction of order $n$,the half-life is related to the initial concentration by the relation: $t_{1/2} \propto [A_0]^{1-n}$.Given data:$1$

Vedclass · Accepted Answer

For $[A_0] = 1 \ M$,$t_{1/2} = 25 \ s$

Vedclass · Answer

(NONE) For a reaction of order $n$,the half-life is related to the initial concentration by the relation: $t_{1/2} \propto [A_0]^{1-n}$.Given data:$1$) $[A_0]_1 = 0.1 \ M$,$(t_{1/2})_1 = 100 \ s$$2$) $[A_0]_2 = 0.025 \ M$,$(t_{1/2})_2 = 50 \ s$Taking the ratio: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{[A_0]_1}{[A_0]_2} \right)^{1-n}$$\frac{100}{50} = \left( \frac{0.1}{0.025} \right)^{1-n}$$2 = (4)^{1-n}$$2 = (2^2)^{1-n} = 2^{2-2n}$Equating exponents: $1 = 2 - 2n \implies 2n = 1 \implies n = 0.5$.Since $n = 0.5$,the reaction is of order $0.5$. None of the provided options $A$ or $B$ are correct. For $n=0.5$,the unit of $k$ is $M^{1-0.5} \ s^{-1} = M^{0.5} \ s^{-1}$. Thus,$D$ is incorrect. Option $C$ is also incorrect based on the calculated order.

$[A_0]$	$0.1 \ M$	$0.025 \ M$
$t_{1/2} \ (s)$	$100 \ s$	$50 \ s$

Expt	$\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$	$[NO] \ (mol \ L^{-1})$	$[H_2] \ (mol \ L^{-1})$
$1$	$4.8 \times 10^{-5}$	$1 \times 10^{-2}$	$1 \times 10^{-3}$
$2$	$43.2 \times 10^{-5}$	$3 \times 10^{-2}$	$1 \times 10^{-3}$
$3$	$86.4 \times 10^{-5}$	$3 \times 10^{-2}$	$2 \times 10^{-3}$

For a certain reaction $A \to P$,the half-life for different initial concentrations of $A$ is mentioned below:
$[A_0]$ $0.1 \ M$ $0.025 \ M$
$t_{1/2} \ (s)$ $100 \ s$ $50 \ s$

Which of the following option$(s)$ is/are correct?

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For a certain reaction $A \to P$,the half-life for different initial concentrations of $A$ is mentioned below:$[A_0]$$0.1 \ M$$0.025 \ M$$t_{1/2} \ (s)$$100 \ s$$50 \ s$Which of the following option$(s)$ is/are correct?