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  • How long does it take for reactant concentration to decrease from <img alt="\mathrm{4 \times 10^{-4} molL^{-1} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B4%20%5Ctimes%

How long does it take for reactant concentration to decrease from \mathrm{4 \times 10^{-4} molL^{-1} }  to  \mathrm{2 \times 10^{-4} molL^{-1} }  in a zero-order reaction. 
Rate constant of the reaction, \mathrm{k=5\times 10^{-8} mol L^{-1} s^{-1} }

Option: 1

7500 s


Option: 2

2500 s


Option: 3

4000 s


Option: 4

5000 s


Answers (1)

best_answer

Let us consider a zero order reaction:
A \rightarrow B

Rate Law for Zero Order Reaction is given as:
Rate, \mathrm{R=k[A]^0}

\mathrm{-\frac{d[A]}{d t} =k[A]^0 \\ }

\mathrm{-\frac{d[A]}{d t} =k \times 1 \\ }

\mathrm{-d[A] =k d t }

Taking limit on both side

\mathrm{-\int_{[a]_0}^{[A]_t} d[A]=k \int_0^t d t }

When


\mathrm{\text { time }=0, A=[A]_o \\ }

\mathrm{ \text { time }=t, A=[A]_t \\ }

\mathrm{ -[A]_{[A]_o}^{[A]_t}=K[t]_0^t \\ }

\mathrm{ {[A]_0-[A]_t=k t} \\ }

\mathrm{ {[A]_t=[A]_0-k t \ldots \ldots \text { eqn }(1)} \\ }
\mathrm{ \text { where }[A]_0 \text { is the initial concentration of Reactant A. } \\ }

\mathrm{K=\frac{[A]_0-[A]_t}{t} }

\mathrm{ t=\frac{A_0-A_t}{k} \\ }

\mathrm{t=\frac{\left(4 \times 10^{-4}\right)-\left(2 \times 10^{-4}\right)}{5 \times 10^{-8}} \\ }

\mathrm{ \text { Time, } t=4000 \mathrm{~s} }
\end{aligned}

Posted by

Pankaj

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