For an n^{th} order reaction,the rate law is | vedclass

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(C) The rate law is $\frac{dx}{dt} = K[A]^n$. Taking the logarithm on both sides: $\log \left( \frac{dx}{dt} \right) = n \log [A] + \log K$. This equa

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Both order of reaction and $K$ (Rate constant)

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(C) The rate law is $\frac{dx}{dt} = K[A]^n$. Taking the logarithm on both sides: $\log \left( \frac{dx}{dt} \right) = n \log [A] + \log K$. This equation is in the form of a straight line $y = mx + c$,where $y = \log \left( \frac{dx}{dt} \right)$,$x = \log [A]$,slope $m = n$,and intercept $c = \log K$. Thus,by plotting $\log \left( \frac{dx}{dt} \right)$ versus $\log [A]$,the slope gives the order of reaction $(n)$ and the intercept gives the logarithm of the rate constant $(\log K)$. Therefore,both the order of reaction and the rate constant can be determined.

Rate $(mol \ L^{-1} \ s^{-1})$	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$
$0.10$	$20$	$0.5$
$0.40$	$x$	$0.5$
$0.80$	$40$	$y$

For an $n^{th}$ order reaction,the rate law is given by $\frac{dx}{dt} = K[A]^n$. Which parameter can be determined from the logarithmic plot?

Similar Questions

If the rate constant $k = 4.5 \times 10^{-7} \text{ L}^2 \text{ mol}^{-2} \text{ s}^{-1}$, then what is the order of the reaction?

For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.
Rate $(mol \ L^{-1} \ s^{-1})$ $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$
$0.10$ $20$ $0.5$
$0.40$ $x$ $0.5$
$0.80$ $40$ $y$

What is the value of $x$ and $y$?

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