The rate constant for a zero order reaction A | vedclass

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(C) For a zero order reaction,the integrated rate equation is given by: $k = \frac{[R]_0 - [R]_t}{t}$ Where: $k = 0.0030 \ mol \ L^{-1} \ s^{-1}$ (rat

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$8.33$

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(C) For a zero order reaction,the integrated rate equation is given by: $k = \frac{[R]_0 - [R]_t}{t}$ Where: $k = 0.0030 \ mol \ L^{-1} \ s^{-1}$ (rate constant) $[R]_0 = 0.10 \ M$ (initial concentration) $[R]_t = 0.075 \ M$ (concentration at time $t$) Rearranging the formula to solve for time $t$: $t = \frac{[R]_0 - [R]_t}{k}$ $t = \frac{0.10 \ M - 0.075 \ M}{0.0030 \ mol \ L^{-1} \ s^{-1}}$ $t = \frac{0.025}{0.0030} \ s = 8.33 \ s$

The rate constant for a zero order reaction $A \rightarrow$ products is $0.0030 \ mol \ L^{-1} \ s^{-1}$. How long will it take for the initial concentration of $A$ to fall from $0.10 \ M$ to $0.075 \ M$ (in $s$)?

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