(B) Given,The reaction is $PCl_5 \longrightarrow PCl_3 + Cl_2$. Rate $= 1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. Rate constant $(k) = 3.4 \times

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(B) Given,The reaction is $PCl_5 \longrightarrow PCl_3 + Cl_2$. Rate $= 1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. Rate constant $(k) = 3.4 \times 10^{-5} \ s^{-1}$. The rate law for this first-order reaction is: $\text{Rate} = k [PCl_5]$ Substituting the given values: $1.02 \times 10^{-4} = 3.4 \times 10^{-5} \times [PCl_5]$ $[PCl_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}$ $[PCl_5] = 3 \ mol \ L^{-1}$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

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For the reaction,$PCl_5 \longrightarrow PCl_3 + Cl_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively at a given instant. The molar concentration of $PCl_5$ at that instant is

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A
$8.0 \ mol \ L^{-1}$
B
$3.0 \ mol \ L^{-1}$
C
$0.2 \ mol \ L^{-1}$
D
$2.0 \ mol \ L^{-1}$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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For the reaction,$PCl_5 \longrightarrow PCl_3 + Cl_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively at a given instant. The molar concentration of $PCl_5$ at that instant is

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