Question

In a reaction, $2A+B\to 3C$, the concentration of $A$ decreases from $0.5mol{L}^{-1}$ to $0.3mol{L}^{-1}$ in $10minutes$. The rate of production of ${}^{\prime }{C}^{\prime }$ during this period is:

• A. $0.01mol{L}^{-1}mi{n}^{-1}$
• B. $0.04mol{L}^{-1}mi{n}^{-1}$
• C. $0.05mol{L}^{-1}mi{n}^{-1}$
• D. $0.03mol{L}^{-1}mi{n}^{-1}$
• E. $0.02mol{L}^{-1}mi{n}^{-1}$

Answer 1

The correct option is D $0.03mol{L}^{-1}mi{n}^{-1}$

Rate of decrease in concentration of $A$,
$-\left(\frac{dA}{dt}\right)=\frac{0.5-0.3}{10}=\frac{0.2}{10}=0.02$

From the equation,
$2A+B\to 3C$

$-\frac{d\left[A\right]}{2\times dt}=-\frac{d\left[B\right]}{dt}=+\frac{d\left[C\right]}{3dt}$

Thus, the rate of production of $C$,

$\frac{d\left[C\right]}{dt}=\frac{3}{2}\times -\frac{d\left[A\right]}{dt}$

$=\frac{3}{2}\times$ rate of decrease in concentration of $A$

$=\frac{3}{2}\times 0.02=0.03mol{L}^{-1}mi{n}^{-1}$

Hence, option D is correct.