The concentration of R in the reaction R righ | vedclass

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(B) For a zero-order reaction,the rate constant $k$ is given by the expression $k = \frac{[R]_0 - [R]_t}{t}$.Using the data points:$1$. For $t = 0.05

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(B) For a zero-order reaction,the rate constant $k$ is given by the expression $k = \frac{[R]_0 - [R]_t}{t}$.Using the data points:$1$. For $t = 0.05 	ext{ min}$,$[R] = 0.75 	ext{ M}$:$k_1 = \frac{1.0 - 0.75}{0.05} = \frac{0.25}{0.05} = 5 	ext{ M min}^{-1}$.$2$. For $t = 0.12 	ext{ min}$,$[R] = 0.40 	ext{ M}$:$k_2 = \frac{1.0 - 0.40}{0.12} = \frac{0.60}{0.12} = 5 	ext{ M min}^{-1}$.$3$. For $t = 0.18 	ext{ min}$,$[R] = 0.10 	ext{ M}$:$k_3 = \frac{1.0 - 0.10}{0.18} = \frac{0.90}{0.18} = 5 	ext{ M min}^{-1}$.Since the rate constant $k$ is constant for all time intervals,the reaction follows zero-order kinetics.

$[R] \text{ (molar)}$	$1.0$	$0.75$	$0.40$	$0.10$
$t \text{ (min.)}$	$0.0$	$0.05$	$0.12$	$0.18$

The concentration of $R$ in the reaction $R \rightarrow P$ was measured as a function of time and the following data is obtained:

$[R] \text{ (molar)}$ $1.0$ $0.75$ $0.40$ $0.10$

$t \text{ (min.)}$ $0.0$ $0.05$ $0.12$ $0.18$

The order of the reaction is:

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The concentration of $R$ in the reaction $R \rightarrow P$ was measured as a function of time and the following data is obtained: $[R] \text{ (molar)}$ $1.0$ $0.75$ $0.40$ $0.10$ $t \text{ (min.)}$ $0.0$ $0.05$ $0.12$ $0.18$ The order of the reaction is: