For a reaction: 3X to 4Y,calculate the rate o | vedclass

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Question

(C) The rate law for the reaction $3X 	o 4Y$ is given by $-\frac{1}{3} \frac{d[X]}{dt} = k[X]^n$.For an $n$-th order reaction where $n 
eq 1$,the in

Vedclass · Accepted Answer

$\frac{1}{9\sqrt{3}} 	imes 10^{-4} \ M \ \min^{-1}$

Vedclass · Answer

(C) The rate law for the reaction $3X 	o 4Y$ is given by $-\frac{1}{3} \frac{d[X]}{dt} = k[X]^n$.For an $n$-th order reaction where $n 
eq 1$,the integrated rate law is $\frac{1}{[X]^{n-1}} - \frac{1}{[X]_0^{n-1}} = 3(n-1)kt$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \frac{1}{[X]^3}$,we identify $n-1 = 3$,so $n = 4$.The slope of the graph is $3(n-1)k = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$.Substituting $n-1 = 3$,we get $3(3)k = \frac{1}{\sqrt{3}}$,which implies $9k = \frac{1}{\sqrt{3}}$,or $k = \frac{1}{9\sqrt{3}}$.The rate of reaction $r$ is given by $r = k[X]^n = \frac{1}{9\sqrt{3}} 	imes (0.1)^4$.$r = \frac{1}{9\sqrt{3}} 	imes 10^{-4} \ M \ \min^{-1}$.

For a reaction: $3X \to 4Y$,calculate the rate of reaction based on the following graph when the concentration of $X$ is $0.1 \ M$.

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