For a certain reaction A_{(g)} to B_{(g)},the | vedclass

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(D) We know that the relationship between half-life $(t_{1/2})$ and initial pressure $(P_{A_0})$ for a reaction of order $n$ is given by: $t_{1/2} \pr

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$0.5$

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(D) We know that the relationship between half-life $(t_{1/2})$ and initial pressure $(P_{A_0})$ for a reaction of order $n$ is given by: $t_{1/2} \propto (P_{A_0})^{1-n}$Therefore,$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{(P_{A_0})_1}{(P_{A_0})_2} ight)^{1-n}$Given: $(t_{1/2})_1 = 100 	ext{ s}$,$(P_{A_0})_1 = 0.1 	ext{ atm}$$(t_{1/2})_2 = 50 	ext{ s}$,$(P_{A_0})_2 = 0.025 	ext{ atm}$Substituting the values:$\frac{100}{50} = \left( \frac{0.1}{0.025} ight)^{1-n}$$2 = (4)^{1-n}$$2^1 = (2^2)^{1-n}$$2^1 = 2^{2(1-n)}$Comparing the powers: $1 = 2(1-n)$$0.5 = 1 - n$$n = 1 - 0.5 = 0.5$Thus,the order of the reaction is $0.5$.

$P_{A_0} \text{ (atm)}$	$0.1$	$0.025$
$t_{1/2} \text{ (sec)}$	$100$	$50$

For a certain reaction $A_{(g)} \to B_{(g)}$,the half-life for different initial pressures of $A$ is given below:
$P_{A_0} \text{ (atm)}$ $0.1$ $0.025$
$t_{1/2} \text{ (sec)}$ $100$ $50$

The correct statement about the order of reaction is:

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For a certain reaction $A_{(g)} \to B_{(g)}$,the half-life for different initial pressures of $A$ is given below:$P_{A_0} \text{ (atm)}$$0.1$$0.025$$t_{1/2} \text{ (sec)}$$100$$50$The correct statement about the order of reaction is: