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Which of the following is correct expression for an irreversible nth order reaction?

$\begin{array}{1 1} kt = \large\frac{n}{n-1}\big(\large\frac{1}{[A]^{n-1}} - \large\frac{1}{[A_0]^{n-1}}\big) \\ kt = \large\frac{1}{n}\big(\large\frac{1}{[A]^{n-1}} - \large\frac{1}{[A_0]^{n-1}}\big) \\ kt = \large\frac{1}{n-1}\big(\large\frac{1}{[A]^{n-1}} - \large\frac{1}{[A_0]^{n-1}}\big) \\ kt = (n-1)\big(\large\frac{1}{[A]^{n-1}} - \large\frac{1}{[A_0]^{n-1}}\big)\end{array}$

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Answer: $kt = \large\frac{1}{n-1}$$\big($$\large\frac{1}{[A]^{n-1}}$$ - \large\frac{1}{[A_0]^{n-1}}$$\big)$
Given: $nA \overset{k} \rightarrow \text{Products}$
Rate Equation: $-\large\frac{d[A]}{dt}$$ = k[A]^n = n\large\frac{d[P]}{dt}$
Integrating, we get: $\large\frac{1}{[A]^{n-1}}$$ - \large\frac{1}{[A_0]^{n-1}}$$ = (n-1)kt$
$\Rightarrow kt = \large\frac{1}{n-1}$$\big($$\large\frac{1}{[A]^{n-1}}$$ - \large\frac{1}{[A_0]^{n-1}}$$\big)$
answered Dec 16, 2013 by sreemathi.v
edited Jul 26, 2014 by balaji.thirumalai
 

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