Question

Observe the following reaction : $2A+B\to C.$ The rate of formation of $C$ is $2.2\times {10}^{-3}$ $mol{l}^{-1}$ $mi{n}^{-1}$. The is the value of $\frac{-d\left[A\right]}{dt}$ (in $mol{l}^{-1}mi{n}^{-1}$):

• A. $2.2\times {10}^{-3}$
• B. $1.1\times {10}^{-3}$
• C. $4.4\times {10}^{-3}$
• D. $5.5\times {10}^{-3}$

Answer 1

The correct option is C $4.4\times {10}^{-3}$

For the given equation ; $2A+B\to C$

Rate equation can be written as; $r=-\frac{1}{2}\frac{d\left[A\right]}{dt}=-\frac{d\left[B\right]}{dt}=\frac{d\left[C\right]}{dt}$

Given that,

$\frac{d\left[C\right]}{dt}$ $=2.2\times {10}^{-3}$ & $-\frac{1}{2}\frac{d\left[A\right]}{dt}=\frac{d\left[C\right]}{dt}$

Therefore, $-\frac{1}{2}\frac{d\left[A\right]}{dt}$ $=2.2\times {10}^{-3}$

$-\frac{d\left[A\right]}{dt}$ $=2\times 2.2\times {10}^{-3}$ $=4.4\times {10}^{-3}$$mol{L}^{-1}mi{n}^{-1}$