Question

For a gaseous phase $1^{st}$ order reaction: $A\left(g\right)\to B\left(g\right)+2C\left(g\right)$ (rate constant $K={10}^{-2}tim{e}^{-1}$), iin a closed vessel of $2$ litre containing $5$ mole of $A\left(g\right)$ at ${27}^{o}C$, which of the following is incorrect?

• A. Rate of appearance of $C\left(g\right)$ is $5\times {10}^{-2}mol{L}^{-1}{t}^{-1}$
• B. Rate of disappearance of $A\left(g\right)$ is $6.15\times {10}^{-1}atm{t}^{-1}$
• C. Rate of disappearance of $A\left(g\right)$ is $5.0\times {10}^{-2}mol{t}^{-1}$
• D. Rate of appearance of $B\left(g\right)$ is $5\times {10}^{-2}mol{L}^{-1}{t}^{-1}$

Answer 1

The correct option is D Rate of appearance of $B\left(g\right)$ is $5\times {10}^{-2}mol{L}^{-1}{t}^{-1}$

Rate of reaction $=\frac{d\left[A\right]}{dt}=+\frac{d\left[B\right]}{dt}=+\frac{1}{2}\frac{d\left[C\right]}{dt}$
Also, rate of reaction
$=-\frac{d\left[A\right]}{dt}=K[A{]}^1={10}^{-2}\times \frac{5}{2}$

$=2.5\times {10}^{-2}mol{L}^{-1}{t}^{-1}$

$\because P=CRT$
$\therefore \frac{-d\left[A\right]}{dt}=-\frac{1}{RT}\cdot \frac{d\left[P_A\right]}{dt}$

$\therefore \frac{-d\left[P_A\right]}{dt}=RT[-\frac{d\left[A\right]}{dt}]=0.0821\times 300\times 2.5\times {10}^{-2}$

$=6.15\times {10}^{-1}atm{t}^{-1}$

$-\frac{d\left[A\right]}{dt}=-\frac{d\left[n\right]}{V\cdot dt}$

$\therefore -\frac{d\left[n_A\right]}{dt}=V\cdot [-\frac{d\left[A\right]}{dt}]=2\times 2.5\times {10}^{-2}$

$=2.0\times {10}^{-2}mol{t}^{-1}$