At 345 K,the half-life for the decomposition | vedclass

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(B) For a gaseous reaction,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ by the relation: $t_{1/2} \propto \frac{1}{(P_0)^{n-1}}$,w

Vedclass · Accepted Answer

$0$

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(B) For a gaseous reaction,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ by the relation: $t_{1/2} \propto \frac{1}{(P_0)^{n-1}}$,where $n$ is the order of the reaction.Given:$P_1 = 55.5 \ kPa$,$t_1 = 340 \ s$$P_2 = 27.8 \ kPa$,$t_2 = 170 \ s$Using the ratio formula:$\frac{t_1}{t_2} = \left(\frac{P_2}{P_1}\right)^{n-1}$Substituting the values:$\frac{340}{170} = \left(\frac{27.8}{55.5}\right)^{n-1}$$2 = \left(\frac{1}{2}\right)^{n-1}$$2^1 = (2^{-1})^{n-1}$$2^1 = 2^{-(n-1)}$$1 = -(n-1)$$1 = -n + 1$$n = 0$Thus,the order of the reaction is $0$.

At $345 \ K$,the half-life for the decomposition of a sample of a gaseous compound initially at $55.5 \ kPa$ was $340 \ s$. When the pressure was $27.8 \ kPa$,the half-life was found to be $170 \ s$. The order of the reaction is $......$ [integer answer]

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