Question

A substance 'A' undergoes decomposition as shown $2A\left(g\right)\to 3B\left(g\right)+C\left(g\right)$. If the rate constant for the decomposition of 'A' is $4\times {10}^{-2}mi{n}^{-1}$ then identify the correct statement(s).

• A. The value of $\frac{d\left[C\right]}{dt}=2\times {10}^{-2}\left[A\right]$,where [specie] represents concentration of specie at time t
• B. The value of $\frac{d\left[C\right]}{dt}=8\times {10}^{-2}\left[A\right]$,where [specie] represents concentration of specie at time t
• C. Rate at which B is being formed is 1.5 times the rate at which A is decomposing
• D. Pressure of the gases would never be double of the original in a finite interval of time keeping volume and temperature constant

Answer 1

Answer: (A), (C), (D)

Pressure of the gases would never be double of the original in a finite interval of time keeping volume and temperature constant

(A)For a first order reaction$\left[A\right]=[A{]}_0{e}^{-kt}$
$[A{]}_0=$ conc. of reactant at $t=0$
$\left[A\right]=$ conc. of reactant at time$=t$
When reaction reaches completion, the value $\left[A\right]=0$
So, we get
$0=[A{]}_0{e}^{-kt}\text{or}{e}^{-kt}=0[\therefore [A{]}_0\ne 0]$
But $e^{-kt}=0$ only when $t=\infty$
This means a first-order reaction takes infinite time for its completion.
(B)For the given decomposition reaction:
$2A\to 3B+C$
$\frac{1}{2}\left[\frac{-d\left[A\right]}{dt}\right]=\frac{1}{3}\frac{d\left[B\right]}{dt}$
$\to \frac{d\left[B\right]}{dt}=\frac{3}{2}[-\frac{d\left[A\right]}{dt}]$
That means statement B is correct.
(C)$\frac{d\left[C\right]}{dt}=\frac{1}{2}[-\frac{d\left[A\right]}{dt}]$
Given $\left[\frac{-dA}{dt}\right]=[4\times {10}^{-2}]$
$\therefore \frac{d\left[C\right]}{dt}=\frac{1}{2}\times 4\times {10}^{-2}\left[A\right]$
Incorrect statement