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  • In certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours and so on. The dimensions of the rate constant are:Opt

In certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours and so on. The dimensions of the rate constant are:

Option: 1

\mathrm{{hour }^{-1}}


Option: 2

\mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}


Option: 3

\mathrm{L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}


Option: 4

\mathrm{mol} \mathrm{~s}^{-1}


Answers (1)

best_answer

The given reaction is a zero-order reaction because on plotting the percent concentration of reactant left after decomposition of reactants vs time, we will obtain a straight line.

For a zero-order reaction, concentration versus time graph is a straight line with a negative slope.

Consider a zero order reaction, Rate, \mathrm{R=K[A]^{\circ}}
\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, \quad A=[A]_0 \\ }

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

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

\mathrm{ {[A]_0 \quad\left[A_t=K t\right]} \\ }

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

Its in \mathrm{y=-m x+c } form.
Hence,
Order of the given reaction is zero
Rate = Rate constant = \mathrm{\frac{d x}{d t} }
units of \mathrm{\mathrm{K}=\mathrm{mol} ~L^{-1} s^1 }

Posted by

mansi

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