A general reaction A_{(g)} to text{product} o | vedclass

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Question

(A) For a gaseous reaction,the concentration of $A$ is given by $[A] = \frac{n_A}{V}$.Taking the derivative with respect to time $t$,we get $-\frac{d[

Vedclass · Accepted Answer

$-\frac{d[A]}{dt} = -\frac{1}{V} \frac{dn_A}{dt} = -\frac{1}{RT} \frac{dP_A}{dt}$

Vedclass · Answer

(A) For a gaseous reaction,the concentration of $A$ is given by $[A] = \frac{n_A}{V}$.Taking the derivative with respect to time $t$,we get $-\frac{d[A]}{dt} = -\frac{1}{V} \frac{dn_A}{dt}$.From the ideal gas law,$P_A V = n_A RT$,so $P_A = \frac{n_A}{V} RT = [A] RT$.Therefore,$[A] = \frac{P_A}{RT}$.Substituting this into the rate expression,we get $-\frac{d[A]}{dt} = -\frac{d}{dt} \left( \frac{P_A}{RT} \right) = -\frac{1}{RT} \frac{dP_A}{dt}$ (assuming $T$ is constant).Thus,the correct relation is $-\frac{d[A]}{dt} = -\frac{1}{V} \frac{dn_A}{dt} = -\frac{1}{RT} \frac{dP_A}{dt}$.

$A$ general reaction $A_{(g)} \to \text{product}$ occurs in a container of volume $V$ at temperature $T$. Which of the following is correct?

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